算法总览

第一次整体的总结一下我的模板。

数学

高斯消元

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;
const double MIN=0.0000001;
int n;
bool zero,all;
double a[114][114],x[114];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n+1;j++)scanf("%lf",&a[i][j]);
	}
	int cnt=1;
	for(int i=1;i<=n;i++){
		if(cnt>n)break;
		for(int j=i;j<=n;j++){
			if(abs(a[j][cnt])>abs(a[i][cnt])){
				for(int k=1;k<=n+1;k++)swap(a[i][k],a[j][k]);
			}
		}
		if(abs(a[i][cnt])<MIN){
			all=true;cnt++;i--;continue;
		}
		for(int j=i+1;j<=n;j++){
			double t=a[j][cnt]/a[i][cnt];
			for(int k=cnt;k<=n+1;k++){
				a[j][k]-=a[i][k]*t;
			}
		}
		cnt++;
	}cnt=n;
	for(int i=n;i>=1;i--){
		if(cnt<1)break;
		if(abs(a[i][cnt])<MIN)continue;
		for(int j=1;j<=i-1;j++){
			double t=a[j][cnt]/a[i][cnt];
			for(int k=cnt;k<=n+1;k++){
				a[j][k]-=a[i][k]*t;
			}
		}
		cnt--;
	}
	for(int i=n;i>=1;i--){
		zero=false;
		for(int j=1;j<=n;j++){
			if(abs(a[i][j])>MIN)zero=true;
		}
		if(!zero&&abs(a[i][n+1])>MIN){
			printf("-1");return 0;
		}
		if(abs(a[i][i])>MIN){
			if(!all)x[i]=a[i][n+1]/a[i][i];
		}
	}
	if(all){
		printf("0");return 0;
	}
	for(int i=1;i<=n;i++){
		if(abs(x[i])<MIN)printf("x%d=0\n",i);
		else printf("x%d=%.2lf\n",i,x[i]);
	}
	return 0;
}

拉格朗日插值

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, mod = 1e9 + 7;

int n, k, tab[N], p[N], pcnt, f[N], pre[N], suf[N], fac[N], inv[N], ans;

int qpow(int x, int y) {
  int ans = 1;
  for (; y; y >>= 1, x = 1LL * x * x % mod)
    if (y & 1) ans = 1LL * ans * x % mod;
  return ans;
}

void sieve(int lim) {
  f[1] = 1;
  for (int i = 2; i <= lim; i++) {
    if (!tab[i]) {
      p[++pcnt] = i;
      f[i] = qpow(i, k);
    }
    for (int j = 1; j <= pcnt && 1LL * i * p[j] <= lim; j++) {
      tab[i * p[j]] = 1;
      f[i * p[j]] = 1LL * f[i] * f[p[j]] % mod;
      if (!(i % p[j])) break;
    }
  }
  for (int i = 2; i <= lim; i++) f[i] = (f[i - 1] + f[i]) % mod;
}

int main() {
  scanf("%d%d", &n, &k);
  sieve(k + 2);
  if (n <= k + 2) return printf("%d\n", f[n]) & 0;
  pre[0] = suf[k + 3] = 1;
  for (int i = 1; i <= k + 2; i++) pre[i] = 1LL * pre[i - 1] * (n - i) % mod;
  for (int i = k + 2; i >= 1; i--) suf[i] = 1LL * suf[i + 1] * (n - i) % mod;
  fac[0] = inv[0] = fac[1] = inv[1] = 1;
  for (int i = 2; i <= k + 2; i++) {
    fac[i] = 1LL * fac[i - 1] * i % mod;
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
  }
  for (int i = 2; i <= k + 2; i++) inv[i] = 1LL * inv[i - 1] * inv[i] % mod;
  for (int i = 1; i <= k + 2; i++) {
    int P = 1LL * pre[i - 1] * suf[i + 1] % mod;
    int Q = 1LL * inv[i - 1] * inv[k + 2 - i] % mod;
    int mul = ((k + 2 - i) & 1) ? -1 : 1;
    ans = (ans + 1LL * (Q * mul + mod) % mod * P % mod * f[i] % mod) % mod;
  }
  printf("%d\n", ans);
  return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int p=998244353;
typedef long long ll;
ll pow(ll a,ll b){
	ll res=1;
	while(b){
		if(b&1)res=res*a%p;
		a=a*a%p;
		b>>=1;
	}
	return res;
}
int x[2010],y[2010];
int main(){
	int n,k;
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)scanf("%d%d",&x[i],&y[i]);
	ll res=0;
	for(int i=1;i<=n;i++){
		ll up=1,down=1;
		for(int j=1;j<=n;j++){
			if(i^j){
				up=up*(k-x[j])%p;
				down=down*(x[i]-x[j])%p;
			}
		}
		res=(res+y[i]*up%p*pow(down,p-2)%p)%p;
	}
	printf("%lld\n",(res+p)%p);
	return 0;
}

原根

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+10;
#define rint register int
#define rll register long long 
typedef long long ll;
ll fpow(ll a,ll b,ll p){
	rll res=1;
	for(;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
int pri[N];
bool npri[N];
int p[N],tot,g[N],cnt;
int get(int x){
	rint res=x;
	for(rint i=1;pri[i]*pri[i]<=x;i++){
		if(x%pri[i]==0){
			res=res/pri[i]*(pri[i]-1);
			while(x%pri[i]==0)x/=pri[i];
		}
	}
	if(x>1)res=res/x*(x-1);
	return res;
}
void init(){
	rint n=1e6;
	for(rint i=2;i<=n;i++){
		if(!npri[i])pri[++pri[0]]=i;
		for(rint j=1;j<=pri[0]&&i*pri[j]<=n;j++){
			npri[i*pri[j]]=1;
			if(i%pri[j]==0)break;
		}
	}
}
int gcd(int x,int y){
	return y?gcd(y,x%y):x;
}
bool check(int x,int n){
	for(rint i=1;i<=cnt;i++)
		if(fpow(x,p[i],n)==1)return 0;
	return 1;
}
int main(){
	init();
	int T;
	scanf("%d",&T);
	while(T--){
		int n,d;
		scanf("%d%d",&n,&d);
		cnt=tot=0;
		int phi=get(n);
		for(rint i=2;i*i<=phi;i++)
			if(phi%i==0)p[++cnt]=i,p[++cnt]=phi/i;
		if(phi!=1)p[++cnt]=1;
		rint res=0;
		for(rint i=1;i<=n;i++){
			if(gcd(i,n)==1&&check(i,n)){
				res=i;
				break;
			}
		}
		if(res){
			for(rint i=1;i<=phi;i++){
				if(gcd(i,phi)==1)g[++tot]=fpow(res,i,n);
			}
		}
		sort(g+1,g+tot+1);
		printf("%d\n",tot);
		for(rint i=d;i<=tot;i+=d)
			printf("%d ",g[i]);
		puts("");
	}
	return 0;
}

bsgs(ex)

#include<map>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
map<ll,ll> mp;
ll fpow(ll a,ll b,ll p){
	ll res=1;
	while(b){
		if(b&1)res=res*a%p;
		a=a*a%p;
		b>>=1;
	}
	return res;
}
int main(){
	ll p,x,y;
	scanf("%lld%lld%lld",&p,&x,&y);
	ll sqr=sqrt(p)+1,now=y%p,tmp=1;
	for(int i=0;i<sqr;i++){
		mp[now]=i;
		now=now*x%p;
		tmp=tmp*x%p;
	}
	now=tmp;
	for(int i=1;i<=sqr;i++){
		if(mp.find(now)!=mp.end())return printf("%lld\n",i*sqr-mp[now]),0;
		now=now*tmp%p;
	}
	puts("no solution");
	return 0;
}


#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long LL;

const int INF = 0x3f3f3f3f;

int a, b, p;
unordered_map<int, int> hs;

int exgcd(int a, int b, int &x, int &y) {
    if (!b) {
        x = 1, y = 0;
        return a;
    }
    int d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}

int BSGS(int a, int b, int p) {
    if (1 % p == b % p) return 0;
    int k = sqrt(p) + 1;
    hs.clear();
    for (int y = 0, r = b % p; y < k; y++) {
        hs[r] = y;
        r = (LL)r * a % p;
    }
    int ak = 1;
    for (int i = 1; i <= k; i++) ak = (LL)ak * a % p;
    for (int x = 1, l = ak; x <= k; x++) {
        if (hs.count(l)) return k * x - hs[l];
        l = (LL)l * ak % p;
    }
    return -INF;
}

int exBSGS(int a, int b, int p) {
    b = (b % p + p) % p;
    if (1 % p == b % p) return 0;
    int x, y;
    int d = exgcd(a, p, x, y);
    if (d > 1) {
        if (b % d) return -INF;
        exgcd(a / d, p / d, x, y);
        return exBSGS(a, (LL)b / d * x % (p / d), p / d) + 1;
    }
    return BSGS(a, b, p);
}

int main() {
    while (~scanf("%d%d%d", &a, &p, &b), a || b || p) {
        int res = exBSGS(a, b, p);
        if (res < 0) puts("No Solution");
        else printf("%d\n", res);
    }
    return 0;
}

lucas(ex)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2e5+10;
ll p,fac[N];
ll fpow(ll a,ll b){
	ll res=1;
	while(b){
		if(b&1)res=res*a%p;
		a=a*a%p;
		b>>=1;
	}
	return res;
}
ll C(int n,int m){
	if(n<m)return 0;
	return fac[n]*fpow(fac[m],p-2)%p*fpow(fac[n-m],p-2)%p;
}
ll lucas(int n,int m){
	if(!m)return 1;
	return lucas(n/p,m/p)*C(n%p,m%p)%p;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		int n,m;
		scanf("%d%d%lld",&n,&m,&p);
		fac[0]=fac[1]=1;
		for(int i=2;i<p;i++)
			fac[i]=fac[i-1]*i%p;
		printf("%lld\n",lucas(n+m,m));
	}
	return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+2;
typedef long long ll;
#define rint register int
#define rll register long long
ll A[N],B[N],n,n1,n2,m,pri[N],prik[N],val[20][N];
int rk[N];
ll fpow(rll a,rll b,rll p){
	rll res=1;
	for(;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
void exgcd(rll a,rll b,rll &x,rll &y){
	if(!b){
		x=1;y=0;return ;
	}
	exgcd(b,a%b,y,x);
	y-=a/b*x;
}
ll Inv(rll a,rll p){
	rll x,y;
	exgcd(a,p,x,y);
	return (x%p+p)%p;
}
ll calc(rll x,rll p,rll pk){
	if(!x)return 1;
	rll res=fpow(val[rk[p]][pk],x/pk,pk)*val[rk[p]][x%pk]%pk;
	return calc(x/p,p,pk)*res%pk;
}
ll calc2(rll x,rll p){
	if(!x)return 0;
	return calc2(x/p,p)+x/p;
}
ll exlc(rll n,rll m,rll p,rll pk){
	if(n<m)return 0;
	rll v1=calc(n,p,pk),v2=calc(m,p,pk)*calc(n-m,p,pk)%pk;
	return v1*Inv(v2,pk)%pk*fpow(p,calc2(n,p)-calc2(m,p)-calc2(n-m,p),pk)%pk;
}
ll solve(rll p,rll pk){
	rint Max=1<<n1,cnt=0;
	rll res=0,s=0;
	for(rint sta=0;sta<Max;sta++){
		s=cnt=0;
		for(rint i=1;i<=n1;i++)
			if(sta&(1<<i-1))s+=A[i],cnt++;
		if(cnt&1)res-=exlc(m-s-1,n-1,p,pk);
		else res+=exlc(m-s-1,n-1,p,pk);
	}
	return (res%pk+pk)%pk;
}
int main(){
	rint p;rll n,m;
	scanf("%lld%lld%d",&n,&m,&p);
	rint x=p;
	for(rint i=2;i*i<=x;i++){
		if(x%i==0){
			pri[++pri[0]]=i;prik[pri[0]]=1;
			while(x%i==0)x/=i,prik[pri[0]]*=i;
		}
	}
	if(x>1)pri[++pri[0]]=x,prik[pri[0]]=x;
	for(rint i=1;i<=pri[0];i++){
		val[i][0]=1;rk[pri[i]]=i;
		for(rint j=1;j<=prik[i];j++){
			if(j%pri[i])val[i][j]=val[i][j-1]*j%prik[i];
			else val[i][j]=val[i][j-1];
		}
	}
	rll res=0;
	for(rint i=1;i<=pri[0];i++){
		ll x=exlc(n,m,pri[i],prik[i]);
		ll x_=Inv(p/prik[i],prik[i]);
		res=(res+x*x_%p*(p/prik[i])%p)%p;
	}
	res=(res%p+p)%p;
	printf("%lld\n",res);
	return 0;
}

crt(excrt)

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=20;
typedef long long ll;
ll a[N],b[N];
void exgcd(ll a,ll b,ll &x,ll &y){
	if(!b){
		x=1;y=0;
		return ;
	}
	exgcd(b,a%b,y,x);
	y-=a/b*x;
}
int main(){
	int n;
	scanf("%d",&n);
	ll M=1;
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		M*=a[i];
	}
	ll res=0;
	for(int i=1;i<=n;i++){
		ll x,y;
		exgcd(M/a[i],a[i],x,y);
		if(x<0)x+=a[i];
		res+=b[i]*M/a[i]*x;
	}
	printf("%lld\n",res%M);
}


#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+10;
#define int long long
int exgcd(int a,int b,int &x,int &y){
	if(!b){
		x=1;y=0;
		return a;
	}
	int g=exgcd(b,a%b,y,x);
	y-=a/b*x;
	return g;
}
int Mul(int a,int b,int p){
	int res=0,f=1;
	if(b<0)f=-1,b=-b;
	while(b){
		if(b&1)res=(res+a)%p;
		a=(a+a)%p;
		b>>=1;
	}
	return res*f;
}
signed main(){
	int n;
	while(~scanf("%lld",&n)){
		int ans,M,a,b,x,y,tag=0;
		scanf("%lld%lld",&M,&ans);
		for(int i=2;i<=n;i++){
			scanf("%lld%lld",&b,&a);
			if(tag)continue;
			int g=exgcd(M,b,x,y);
			int c=(a-ans%b+b)%b;
			if(c%g)tag=1;
			else{
				int bg=b/g;
				x=Mul(x,c/g,bg);
				ans+=x*M;
				M=M*bg;
				ans=(ans%M+M)%M;
			}
		}
		if(tag)puts("-1");
		else printf("%lld\n",ans);
	}
	return 0;
}

二次剩余

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
#define rint register int
#define rll register long long
ll I,n,p;
inline ll Add(rll x,rll y){return (x+y)%p;}
inline ll Del(rll x,rll y){return ((x-y)%p+p)%p;}
inline ll Mul(rll x,rll y){x*=y;return x>=p?x%p:x;}
struct cp{
	ll x,y;
	cp operator * (const cp&A)const{return (cp){Add(Mul(x,A.x),Mul(Mul(I,y),A.y)),Add(Mul(x,A.y),Mul(y,A.x))};}
};
ll fpow(rll a,rll b){
	rll res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
cp fpow(cp a,rll b){
	cp res=(cp){1,0};
	for(;b;b>>=1,a=a*a)
		if(b&1)res=res*a;
	return res;
}
inline bool check(rll x){
	if(fpow(x,p-1>>1)==1)return 1;
	return 0;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%lld%lld",&n,&p);
		if(n==0)printf("0\n");
		else if(!check(n))printf("Hola!\n");
		else {
			rll a;I=0;
			while(I==0||check(I))a=rand()%p,I=Del(Mul(a,a),n);
			rll res1=fpow((cp){a,1},p+1>>1).x;
			rll res2=p-res1;
			if(res1==res2)printf("%lld ",res1);
			else printf("%lld %lld\n",min(res1,res2),max(res1,res2));
		}
	}
	return 0;
}

杜教筛，min25筛，洲阁筛

约数个数

if(i%prime[j]==0){
	f[i*prime[j]]=f[i]*2ll-f[i/prime[j]];
	break;
}
f[i*prime[j]]=f[i]*2;

常用卷积

\varphi=id*\mu

\varphi*I=id

\mu*I=\varepsilon

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<unordered_map>
using namespace std;
const int N=1e7+10;
typedef long long ll;
bool npri[N];
int pri[N];
ll mu[N],phi[N];
void get(int n){
	mu[1]=phi[1]=1;
	for(int i=2;i<=n;i++){
		if(!npri[i])pri[++pri[0]]=i,mu[i]=-1,phi[i]=i-1;
		for(int j=1;j<=pri[0]&&i*pri[j]<=n;j++){
			npri[i*pri[j]]=1;
			if(i%pri[j]==0){
				phi[i*pri[j]]=phi[i]*pri[j];
				break;
			}
			mu[i*pri[j]]=-mu[i];
			phi[i*pri[j]]=phi[i]*phi[pri[j]];
		}
	}
	for(int i=1;i<=n;i++)phi[i]+=phi[i-1],mu[i]+=mu[i-1];
}
unordered_map<int,ll> ans1,ans2;
ll getphi(int n){
	if(n<=1e7)return phi[n];
	if(ans1.find(n)!=ans1.end())return ans1[n];
	ll res=0;
	for(ll l=2,r;l<=n;l=r+1){
		r=n/(n/l);
		res-=getphi(n/l)*(r-l+1);
	}
	return ans1[n]=(unsigned long long)n*(n+1ll)/2+res;
}
ll getmu(int n){
	if(n<=1e7)return mu[n];
	if(ans2.find(n)!=ans2.end())return ans2[n];
	ll res=1;
	for(ll l=2,r;l<=n;l=r+1){
		r=n/(n/l);
		res-=getmu(n/l)*(r-l+1);
	}
	return ans2[n]=res;
}
int main(){
	get(1e7);
	int T;
	scanf("%d",&T);
	while(T--){
		int n;
		scanf("%d",&n);
		printf("%lld %lld\n",getphi(n),getmu(n));
	}
	return 0;
}

可以用来求 $i^k\varphi_i$

以 $i\varphi_i$ 为例，

写出卷积式 $\sum_{d|n}d\varphi_{d}g(n/d)$

容易发现令 $g(n)=n$ 原式即为 $\sum_{d|n}\varphi_{d}=n$

其余次方的也类似

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+10;
const int p=1e9+7;
typedef long long ll;
ll fpow(ll a,ll b){
	ll res;
	for(res=1;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
ll Add(ll x,ll y){return x+y>=p?x+y-p:x+y;}
ll Del(ll x,ll y){return x-y>=0?x-y:x-y+p;}
ll n,sqr;
bool npri[N];
ll pri[N],s1[N],s2[N];
void get(int n){
	for(int i=2;i<=n;i++){
		if(!npri[i]){
			pri[++pri[0]]=i;
			s1[pri[0]]=Add(s1[pri[0]-1],i);
			s2[pri[0]]=Add(s2[pri[0]-1],(ll)i*i%p);
		}
		for(int j=1;j<=pri[0]&&i*pri[j]<=n;j++){
			npri[i*pri[j]]=1;
			if(i%pri[j]==0)break;
		}
	}
}
ll g1[N],g2[N],w[N],v1[N],v2[N],tot;
ll calc(ll x,int y){
	if(y&&x<=pri[y])return 0;
	ll now=x<=sqr?v1[x]:v2[n/x];
	ll res=Add(Del(g2[now],g1[now]),Del(s1[y],s2[y]));
	for(int i=y+1;pri[i]*pri[i]<=x;i++){
		ll np=pri[i];
		for(int j=1;np<=x;j++,np=np*pri[i]){
			ll tmp=np%p;
			res=Add(res,(tmp-1)*tmp%p*Add(calc(x/np,i),j!=1)%p);
		}
	}
	return res;
}
int main(){
	scanf("%lld",&n);
	sqr=sqrt(n);
	get(1e6);
	ll inv2=fpow(2,p-2),inv6=fpow(6,p-2);
	for(ll l=1,r;l<=n;l=r+1){
		r=n/(n/l);
		ll now=n/l;
		w[++tot]=now;
		if(now<=sqr)v1[now]=tot;
		else v2[n/now]=tot;
		now%=p;
		g1[tot]=Del((now+1)*now%p*inv2%p,1);
		g2[tot]=Del((now+1)*now%p*(now*2+1)%p*inv6%p,1);
	}
	for(int i=1;pri[i]*pri[i]<=n;i++){
		for(int j=1;j<=tot&&pri[i]*pri[i]<=w[j];j++){
			ll now=w[j]/pri[i];
			now=now<=sqr?v1[now]:v2[n/now];
			g1[j]=Del(g1[j],pri[i]*Del(g1[now],s1[i-1])%p);
			g2[j]=Del(g2[j],pri[i]*pri[i]%p*Del(g2[now],s2[i-1])%p);
		}
	}
	printf("%lld\n",Add(1,calc(n,0)));
	return 0;
}

Miller-Rabin,Pollard-rho

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int pri[]={2,3,5,7,11,13,17,19,23};
inline ll Mul(rll x,rll y,rll p){return ((unsigned long long)x*y-(unsigned long long)((long double)x/p*y)*p+2*p)%p;}
inline ll fpow(rll a,rll b,rll p){
	rll res=1;
	for(;b;b>>=1,a=Mul(a,a,p))
		if(b&1)res=Mul(res,a,p);
	return res;
}
bool check(rll x){
	rll t1=x-1,t2=0;
	while(~t1&1)t1>>=1,t2++;
	for(rint i=0;i<9;i++){
		if(x==pri[i])return 1;
		if(x<pri[i])return 0;
		rll t=fpow(pri[i],t1,x),lst;
		for(rint j=0;j<t2;j++){
			lst=t;t=Mul(t,t,x);
			if(t==1&&lst!=1&&lst!=x-1)return 0;
		}
		if(t!=1)return 0;
	}
	return 1;
}
set<ll> res;
inline ll getnxt(rll x){
	if(x==4)return 2;
	rll v1=rand()%x+1,v2=rand()%x+1,v3=0,v4=1,v5=1;
	for(rint i=0,j=2;v4==1;i++){
		v1=(Mul(v1,v1,x)+v2)%x;
		v5=Mul(v5,abs(v1-v3),x);
		if(i%127==0&&v5!=1){
			v4=__gcd(x,v5);
			v5=1;
		}
		if(i==j){
			j<<=1;v4=__gcd(x,v5);
			v5=1;v3=v1;
		}
	}
	return v4;
}
void dfs(rll x){
	if(x==1)return ;
	if(check(x)){
		res.insert(x);
		return ;
	}
	rll nxt=x;
	while(nxt==x)nxt=getnxt(nxt);
	while(x%nxt==0)x/=nxt;
	dfs(x);dfs(nxt);
}
int main(){
	rint T;
	scanf("%d",&T);
	while(T--){
		rll x;
		scanf("%lld",&x);
		if(check(x))puts("Prime");
		else {
			res.clear();
			dfs(x);
			printf("%lld\n",*res.rbegin());
		}
	}
	return 0;
}

类欧几里得

给定 $n,\,a,\,b,\,c$ ，分别求 $\sum\limits_{i=0}^{n}\lfloor \frac{ai+b}{c} \rfloor\,,\ \sum\limits_{i=0}^{n}{\lfloor \frac{ai+b}{c} \rfloor}^2\,,\ \sum\limits_{i=0}^{n}i\lfloor \frac{ai+b}{c} \rfloor$ ，答案对 $998244353$ 取模。多组数据。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int MAXN=22,P=998244353;
ll t,p,q,r,l;
struct Node {
	Node () {cntu=cntr=sumi=sums=sqrs=prod=0;}
	ll cntu,cntr,sumi,sums,sqrs,prod;
	Node operator + (Node b) {
		Node c;
		c.cntu=(cntu+b.cntu)%P,c.cntr=(cntr+b.cntr)%P;
		c.sumi=(sumi+b.sumi+cntr*b.cntr)%P;
		c.sums=(sums+b.sums+cntu*b.cntr)%P;
		c.sqrs=(sqrs+b.sqrs+((cntu*cntu)%P)*b.cntr+(2*cntu*b.sums)%P)%P;
		c.prod=((prod+b.prod+((cntu*cntr)%P)*b.cntr)%P+cntu*b.sumi+cntr*b.sums)%P;
		return c;
	}
}nu,nr,ans;
Node qpow (Node a,ll k) {
	Node res;
	while (k) {
		if (k&1) {res=res+a;}
		a=a+a,k>>=1;
	}
	return res;
}
ll div (ll a,ll b,ll c,ll d) {return ((long double)1.0*a*b+c)/d;}
Node solve (ll p,ll q,ll r,ll l,Node a,Node b) {
	if (!l) {return Node();}
	if (p>=q) {return solve(p%q,q,r,l,a,qpow(a,p/q)+b);}
	ll m=div(l,p,r,q);
	if (!m) {return qpow(b,l);}
	ll cnt=l-div(q,m,-r-1,p);
	return qpow(b,(q-r-1)/p)+a+solve(q,p,(q-r-1)%p,m-1,b,a)+qpow(b,cnt);
}
int main () {
	scanf("%lld",&t);
	for (int ii=1;ii<=t;ii++) {
		scanf("%lld%lld%lld%lld",&l,&p,&r,&q);
		nu.cntu=1,nu.cntr=0,nu.sumi=0,nu.sums=0,nu.sqrs=0,nu.prod=0;
		nr.cntu=0,nr.cntr=1,nr.sumi=1,nr.sums=0,nr.sqrs=0,nr.prod=0;
		ans=qpow(nu,r/q)+solve(p,q,r%q,l,nu,nr);
		printf("%lld %lld %lld\n",(ans.sums+r/q)%P,(ans.sqrs+((r/q)%P)*((r/q)%P))%P,ans.prod);
	}
	return 0;
}

线性基

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
struct xor_base{
	ll b[55];
	bool insert(rll x){
		for(rint i=50;~i;i--){
			if(~x>>i&1)continue;
			if(!b[i]){
				b[i]=x;
				return 1;
			}
			x^=b[i];
		}
		return 1;
	}
	ll getmax(){
		rll res=0;
		for(rint i=50;~i;i--)
			if((res^b[i])>res)res^=b[i];
		return res;
	}
}b;
int main(){
	rint n;
	scanf("%d",&n);
	while(n--){
		rll x;
		scanf("%lld",&x);
		b.insert(x);
	}
	printf("%lld\n",b.getmax());
	return 0;
}

矩阵求逆

Mat getinv(Mat a){
	Mat res;res.init();
	for(int i=0;i<3;i++){
		for(int j=i;j<3;j++){
			if(a.mt[j][i]){
				swap(a.mt[j],a.mt[i]);
				swap(res.mt[j],res.mt[i]);
				break;
			}
		}
		assert(a.mt[i][i]!=0);
		for(int j=0;j<3;j++){
			if(!a.mt[j][i])continue;
			if(i==j)continue;
			ll w=a.mt[j][i]*fpow(a.mt[i][i],p-2)%p;
			for(int k=0;k<3;k++){
				a.mt[j][k]=(a.mt[j][k]-a.mt[i][k]*w%p+p)%p;
				res.mt[j][k]=(res.mt[j][k]-res.mt[i][k]*w%p+p)%p;
			}
		}
	}
	for(int i=0;i<3;i++)
		for(int j=0;j<3;j++)
			res.mt[i][j]=res.mt[i][j]*fpow(a.mt[i][i],p-2)%p;
	return res;
}

行列式,矩阵树定理

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=310;
const int p=1e9+7;
typedef long long ll;
ll fpow(ll a,ll b){
	ll res;
	for(res=1;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
ll a[N][N];
ll solve(int n){
	ll res=1;
	for(int i=2;i<=n;i++){
		if(a[i][i]==0){
			for(int j=i+1;j<=n;j++)
				if(a[j][i]){
					swap(a[i],a[j]);
					res=-res;goto L;
				}
			return 0;
		}L:;
		for(int j=i+1;j<=n;j++){
			ll r=a[j][i]*fpow(a[i][i],p-2)%p;
			for(int k=i;k<=n;k++)
				a[j][k]=(a[j][k]+p-a[i][k]*r%p)%p;
		}
	}
	for(int i=2;i<=n;i++)res=res*a[i][i]%p;
	return (res+p)%p;
}
int main(){
	int n,m,t;
	scanf("%d%d%d",&n,&m,&t);
	if(t){
		while(m--){
			int x,y,w;
			scanf("%d%d%d",&x,&y,&w);
			a[y][y]=(a[y][y]+w)%p;a[x][y]=(a[x][y]-w+p)%p;
		}
	}else{
		while(m--){
			int x,y,w;
			scanf("%d%d%d",&x,&y,&w);
			a[x][y]=(a[x][y]+p-w)%p;
			a[y][x]=(a[y][x]+p-w)%p;
			a[x][x]=(a[x][x]+w)%p;
			a[y][y]=(a[y][y]+w)%p;
		}
	}
	printf("%lld\n",solve(n));
	return 0;
}

组合计数

莫比乌斯反演

n=\sum_{d|n}\phi(d)

f(x)=\sum_{d|x}g(d)

那么有

g(x)=\sum_{d|x}\mu(d)f(\frac{x}d)

二项式反演

F_i=\sum_{j=i}^n\binom{j}{i}G_{j}

那么一定有

G_i=\sum_{j=i}^n\binom{j}{i}F_j(-1)^{j-i}

F_i=\sum_{j=0}^i\binom{i}{j}G_{j}

G_i=\sum_{j=0}^i\binom{i}{j}F_j(-1)^{i-j}

Min-Max容斥

很多时候能够求出的最小值是一个期望，我们要求最大值的期望，期望只有在加减的时候才有线性性，所以并不能直接取 $max$ ，我们需要一个能够直接加减得到最大值的式子。

具体的，有

\max(S) = \sum_{T \in S \&\&T\neq\emptyset} (-1)^{|T|-1}\min(T)

子集反演

G(S) = \sum _{T \in S}(-1)^{|S-T|}F(T)

伯努利数

有一系列数 $B_i$ ，首先有 $B_0=1$ ，对于别的，有

\sum_{i=0}^k\binom{k+1}{i}B_i=0

这就是伯努利数，~~看起来没什么用~~。

由上式可以看出可以 $O(n^2)$ 预处理出伯努利数，然后呢？

~~再次扔出一个式子~~

\sum_{i=0}^{n-1} i^k=\frac{1}{k+1}\sum_{i=0}^k\binom{k+1}{i}B_in^{k+1-i}

typedef long long ll;
const int maxn = 10000;
const int mod = 1e9 + 7;
ll B[maxn];        // 伯努利数
ll C[maxn][maxn];  // 组合数
ll inv[maxn];      // 逆元（计算伯努利数）

void init() {
  // 预处理组合数
  for (int i = 0; i < maxn; i++) {
    C[i][0] = C[i][i] = 1;
    for (int k = 1; k < i; k++) {
      C[i][k] = (C[i - 1][k] % mod + C[i - 1][k - 1] % mod) % mod;
    }
  }
  // 预处理逆元
  inv[1] = 1;
  for (int i = 2; i < maxn; i++) {
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  }
  // 预处理伯努利数
  B[0] = 1;
  for (int i = 1; i < maxn; i++) {
    ll ans = 0;
    if (i == maxn - 1) break;
    for (int k = 0; k < i; k++) {
      ans += C[i + 1][k] * B[k];
      ans %= mod;
    }
    ans = (ans * (-inv[i + 1]) % mod + mod) % mod;
    B[i] = ans;
  }
}

斯特林数

{n \brack m}={n-1 \brack m-1}+(n-1){n-1 \brack m}

$n!=\sum_{i=0}^n{n \brack i}$

$x^{\overline n}=\sum_{i=0}^n{n \brack i}x^i$

{n \brace m}={n-1 \brace m-1}+m{n-1 \brace m}

$m^n=\sum_{i=1}^m\binom{m}{i}{n \brace i}i!$

\begin{aligned} \sum_{k=0}^n\sum_{i=0}^ma_ik^ix^k\binom{n}{k}&=\sum_{k=0}^n \sum_{i=0}^m a_ix^k\binom{n}{k} \sum_{j=0}^i \binom{k}{j}j! {i \brace j}\\ &=\sum_{i=0}^m a_i \sum_{j=0}^i {i \brace j}j! \sum_{k=0}^n x^k \binom{n}{k}\binom{k}{j} \end{aligned}

\begin{aligned} \sum_{i=0}^m a_i \sum_{j=0}^i {i \brace j}j! \sum_{k=0}^n x^k \binom{n}{k}\binom{k}{j}&= \sum_{i=0}^m a_i \sum_{j=0}^i {i \brace j}j! \sum_{k=0}^n x^k \binom{n}{j}\binom{n-j}{k-j} \\ &=\sum_{i=0}^m a_i \sum_{j=0}^i {i \brace j}j!\binom{n}{j} \sum_{k=j}^n x^k \binom{n-j}{k-j}\\ &=\sum_{i=0}^m a_i \sum_{j=0}^i {i \brace j}\frac{n!}{(n-j)!} x^j\sum_{k=0}^{n-j} x^k \binom{n-j}{k} \\&= \sum_{i=0}^m a_i \sum_{j=0}^i {i \brace j}\frac{n!}{(n-j)!} x^j(x+1)^{n-j} \end{aligned}

第一类（行）

#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<19;
const int p=167772161;
#define rint register int
#define ll long long
#define rll register long long
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline ll Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
int r[N],w[N],pd,t;
ll fpow(rll a,rll b){
	rll res=1;
	for(;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd!=n){
		pd=n;
		for(rint i=0;i<n;i++)
			r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
	}
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint i=0,R=mid<<1;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=Mul(a[i+j+mid],w[mid+j]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(~typ)return ;
	rint finv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)
		a[i]=Mul(a[i],finv);
}
ll fac[N],inv[N];
vector<int> vec[N<<2];
void dfs(rint x,rint l,rint r){
	if(l==r){
		vec[x].resize(2);
		vec[x][0]=l;
		vec[x][1]=1;
		return ;
	}
	rint mid=l+r>>1;mid-=r%2==0;
	dfs(x<<1,l,mid);
	rint Max=1;while(Max<r-l+2)Max<<=1;
	vector<int> f;f.resize(Max);vec[x<<1|1].resize(Max);
	for(rint i=0,pw=1;i<=mid+1;i++){
		f[i]=Mul(fac[i],vec[x<<1][i]);
		vec[x<<1|1][i]=Mul(pw,inv[i]);
		pw=Mul(pw,mid+1);
	}
	ntt(Max,f,-1);ntt(Max,vec[x<<1|1],1);
	for(rint i=0;i<Max;i++)
		vec[x<<1|1][i]=Mul(vec[x<<1|1][i],f[i]);
	ntt(Max,vec[x<<1|1],1);
	for(rint i=0;i<Max;i++)
		vec[x<<1|1][i]=Mul(vec[x<<1|1][i],inv[i]);
	vec[x<<1|1].resize(mid+2);
	ntt(Max,vec[x<<1],1);ntt(Max,vec[x<<1|1],1);
	vec[x].resize(Max);
	for(rint i=0;i<Max;i++)
		vec[x][i]=Mul(vec[x<<1][i],vec[x<<1|1][i]);
	ntt(Max,vec[x],-1);
	if(r%2==0){
		for(rint i=Max;i;i--){
			vec[x][i]=Add(Mul(vec[x][i],r),vec[x][i-1]);
		}
		vec[x][0]=Mul(vec[x][0],r);
	}
}
int main(){
	w[N/2]=1;w[N/2+1]=t=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],t);
	for(rint i=N/2-1;i;i--)w[i]=w[i<<1];
	rint n;
	scanf("%d",&n);
	fac[0]=fac[1]=inv[0]=inv[1]=1;
	for(rint i=2;i<=n;i++){
		fac[i]=fac[i-1]*i%p;
		inv[i]=p-p/i*inv[p%i]%p;
	}
	for(rint i=2;i<=n;i++)
		inv[i]=inv[i-1]*inv[i]%p;
	if(n==1)vec[1].resize(1),vec[1][0]=1;
	else dfs(1,0,n-1);
	for(rint i=0;i<=n;i++)
		printf("%d ",vec[1][i]);
	return 0;
}

第二类（行）

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+10;
const int p=167772161;
typedef long long ll;
ll fpow(ll a,ll b){
	ll res;
	for(res=1;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
ll A[N],B[N],fac[N];
#define Add(x,y) (x+y>=p?x+y-p:x+y)
#define Del(x,y) (x-y>=0?x-y:x-y+p)
int r[N];
void ntt(int n,ll *a,int typ){
	for(int i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(int mid=1;mid<n;mid<<=1){
		ll Wn=fpow(3,(p-1)/(mid*2));
		if(typ==-1)Wn=fpow(Wn,p-2);
		for(int i=0,R=mid<<1;i<n;i+=R){
			ll w=1;
			for(int j=0;j<mid;j++,w=w*Wn%p){
				const ll x=a[i+j],y=w*a[i+j+mid]%p;
				a[i+j]=Add(x,y);
				a[i+j+mid]=Del(x,y);
			}
		}
	}
}
int main(){
	int n;
	scanf("%d",&n);
	fac[0]=1;
	for(int i=1;i<=n;i++)fac[i]=fac[i-1]*i%p;
	fac[n]=fpow(fac[n],p-2);
	for(int i=n;i;i--)fac[i-1]=fac[i]*i%p;
	for(int i=0;i<=n;i++){
		A[i]=fpow(i,n)*fac[i]%p;
		B[i]=(i&1)?p-fac[i]:fac[i];
	}
	int Max=1,l=0;
	while(Max<=n+n)Max<<=1,l++;
	for(int i=0;i<Max;i++)
		r[i]=(r[i>>1]>>1|((i&1)<<l-1));
	ntt(Max,A,1);ntt(Max,B,1);
	for(int i=0;i<Max;i++)
		A[i]=A[i]*B[i]%p;
	ntt(Max,A,-1);
	ll inv=fpow(Max,p-2);
	for(int i=0;i<Max;i++)
		A[i]=A[i]*inv%p;
	for(int i=0;i<=n;i++)
		printf("%lld ",A[i]);
	return 0;
}

第二类（列）

#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<18;
const int p=167772161;
#define rint register int
#define ll long long
#define rll register long long
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline ll Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
int r[N],w[N],pd,t;
ll fpow(rll a,rll b){
	rll res=1;
	for(;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd!=n){
		pd=n;
		for(rint i=0;i<n;i++)
			r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
	}
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint i=0;i<n;i+=mid<<1)
	for(rint j=0;j<mid;j++)
		t=Mul(w[mid+j],a[i+j+mid]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(~typ)return ;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)
		a[i]=Mul(a[i],inv);
}
vector<int> vec[N<<2];
void dfs(rint x,rint l,rint r){
	if(l==r){
		vec[x].resize(2);
		vec[x][0]=1;
		vec[x][1]=p-l;
		return ;
	}
	rint mid=l+r>>1;
	dfs(x<<1,l,mid);
	dfs(x<<1|1,mid+1,r);
	rint Max=1;
	while(Max<r-l+2)Max<<=1;
	ntt(Max,vec[x<<1],1);ntt(Max,vec[x<<1|1],1);
	vec[x].resize(Max);
	for(rint i=0;i<Max;i++)
		vec[x][i]=Mul(vec[x<<1][i],vec[x<<1|1][i]);
	ntt(Max,vec[x],-1);
}
void dfs_inv(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){
		f.resize(1);
		f[0]=fpow(g[0],p-2);
		return ;
	}
	dfs_inv(f,g,n+1>>1);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);f.resize(n);
}
int ans[N],tot;
int main(){
	w[N/2]=1;w[N/2+1]=t=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],t);
	for(rint i=N/2-1;i;i--)w[i]=w[i<<1];
	rint n,k;
	scanf("%d%d",&n,&k);
	dfs(1,1,k);dfs_inv(vec[1],vec[1],n+1);
	for(rint i=1;i<=k;i++)
		ans[tot++]=0;
	for(rint i=0;i<=n;i++)
		ans[tot++]=vec[1][i];
	for(rint i=0;i<=n;i++)
		printf("%d ",ans[i]);
	return 0;
}

欧拉数

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=1<<19;
const int p=998244353;
int w[N],r[N],pd;
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline ll Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
inline ll fpow(rll a,rll b){
	rll res=1;
	for(;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
unsigned long long A[N],t;
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd!=n){
		for(rint i=0;i<n;i++)
			r[i]=(r[i>>1]>>1|((i&1)?n>>1:0));
		pd=n;
	}
	for(rint i=0;i<n;i++)
		A[i]=a[r[i]];
	for(rint mid=1;mid<n;mid<<=1)
	for(rint i=0;i<n;i+=mid<<1)
	for(rint j=0;j<mid;j++)
		t=w[mid+j]*A[i+j+mid]%p,A[i+j+mid]=A[i+j]-t+p,A[i+j]+=t;
	for(rint i=0;i<n;i++)
		a[i]=A[i]%p;
	if(~typ)return ;
	reverse(a.begin()+1,a.end());
	rint inv=fpow(n,p-2);
	for(rint i=0;i<n;i++)
		a[i]=Mul(a[i],inv);
}
ll fac[N],inv[N];
ll C(rint n,rint m){
	return fac[n]*inv[m]%p*inv[n-m]%p;
}
int main(){
	w[N/2]=1;w[N/2+1]=t=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],t);
	for(rint i=N/2-1;i;i--)w[i]=w[i<<1];
	rint n;
	cin >> n;
	fac[0]=1;
	for(rint i=1;i<=n+1;i++)
		fac[i]=fac[i-1]*i%p;
	inv[n+1]=fpow(fac[n+1],p-2);
	for(rint i=n+1;i;i--)
		inv[i-1]=inv[i]*i%p;
	vector<int> f,g;
	f.resize(n+1);g.resize(n+1);
	for(rint i=0;i<=n;i++)
		f[i]=fpow(i+1,n);
	for(rint i=0;i<=n;i++)
		g[i]=(i&1)?p-C(n+1,i):C(n+1,i);
	rint Max=1;
	while(Max<=n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);
	for(rint i=0;i<=n;i++)
		printf("%d ",f[i]);
	return 0;
}

如何优雅的求和

计算 $\sum_{k=0}^{n}f(k)\binom{n}{k}x^k(1-x)^{n-k}$

#include <bits/stdc++.h>
#define rep(i, l, r) for (int i = (l); i <= (r); i++)
#define per(i, r, l) for (int i = (r); i >= (l); i--)
#define mem(a, b) memset(a, b, sizeof a)
#define For(i, l, r) for (int i = (l), i##e = (r); i < i##e; i++)
#define pb push_back

using namespace std;
using ll = long long;

const int N = 2e6 + 5, P = 998244353;

int n, m, x, f[N], a[N], fac[N], ifac[N];

ll Pow(ll a, int n, ll r = 1)
{
	for (; n; n >>= 1, a = a * a % P)
		if (n & 1)
			r = r * a % P;
	return r;
}
int main()
{
	cin.tie(0)->sync_with_stdio(0);
	int T;
	cin >> T;
	while (T--)
	{
		memset(a, 0, sizeof(a));
		int A, B;
		cin >> n >> m >> A >> B;
		f[0] = 0;
		x = A * Pow(B, P - 2) % P;
		for (int i = 1; i <= m + 2; i++)
		{
			f[i] = (f[i - 1] + Pow(i, m)) % P;
		}
		++m;
		m = min(n, m);
		rep(i, 0, m) fac[i] = i ? (ll)fac[i - 1] * i % P : 1;
		per(i, m + 1, 1) ifac[i - 1] = i > m ? Pow(fac[m], P - 2) : (ll)ifac[i] * i % P;
		rep(i, 1, m) ifac[i] = (ll)ifac[i] * fac[i - 1] % P;
		int p = 1, ans = 0;
		if (n == m)
			per(i, m, 0) a[i] = p, p = p * ll(P + 1 - x) % P;
		else
			per(i, m, 0)
			{
				a[i] = (a[i + 1] * ll(P + 1 - x) + p) % P;
				p = p * ll(n - i) % P * ifac[m - i + 1] % P * (P - x) % P;
			}
		p = 1;
		rep(i, 0, m)
		{
			ans = (ans + (ll)f[i] * p % P * a[i]) % P;
			p = p * ll(n - i) % P * ifac[i + 1] % P * x % P;
		}
		cout << ans << '\n';
	}
	return 0;
}

数据结构

lct维护子树信息

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=2e5+10;
struct Lct{
	int ch[N][2],fa[N],siz[N],siz2[N],tag[N];
	void up(int x){siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1+siz2[x];}
	bool nrt(int x){
		return ch[fa[x]][0]==x||ch[fa[x]][1]==x;
	}
	void rotate(int x){
		int y=fa[x];int z=fa[y];
		int k=ch[y][1]==x;
		if(nrt(y))ch[z][ch[z][1]==y]=x;
		fa[x]=z;ch[y][k]=ch[x][!k];
		fa[ch[x][!k]]=y;ch[x][!k]=y;
		fa[y]=x;up(y);up(x);
	}
	void down(int x){
		if(tag[x]){
			swap(ch[x][0],ch[x][1]);
			tag[ch[x][0]]^=1;
			tag[ch[x][1]]^=1;
			tag[x]=0;
		}
	}
	void update(int x){
		if(nrt(x))update(fa[x]);
		down(x);
	}
	void splay(int x){
		update(x);
		for(int y=fa[x],z=fa[y];nrt(x);rotate(x),y=fa[x],z=fa[y])
			if(nrt(y))(ch[y][1]==x)==(ch[z][1]==y)?rotate(y):rotate(x);
	}
	void access(int x){
		for(int t=0;x;t=x,x=fa[x]){
			splay(x);
			siz2[x]+=siz[ch[x][1]]-siz[t];
			ch[x][1]=t;
			up(x);
		}
	}
	void mkrt(int x){
		access(x);
		splay(x);
		tag[x]^=1;
	}
	void link(int x,int y){
		mkrt(x);mkrt(y);
		fa[x]=y;
		siz2[y]+=siz[x];
		up(y);
	}
	void split(int x,int y){
		mkrt(x);
		access(y);
		splay(y);
	}
}T;
int main(){
	int n,q;
	scanf("%d%d",&n,&q);
	for(int i=1;i<=n;i++)T.siz[i]=1;
	while(q--){
		char op;
		int x,y;
		scanf(" %c%d%d",&op,&x,&y);
		if(op=='A')T.link(x,y);
		else {
			T.split(x,y);
			printf("%lld\n",1ll*(T.siz2[x]+1)*(T.siz2[y]+1));
		}
	}
	return 0;
}

莫队

带奇偶性优化

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=5e4+10;
int col[N],a[N],cnt[N];
ll up[N],down[N];
struct Node{
    int l,r,id;
    bool operator < (const Node&A)const{
        return col[l]==col[A.l]?col[l]&1?r<A.r:r>A.r:col[l]<col[A.l];
    }
}q[N];
int main(){
    rint n,m;
    scanf("%d%d",&n,&m);
    rint sqr=sqrt(n);
    for(rint i=1;i<=n;i++){
        scanf("%d",&a[i]);
        col[i]=(i-1)/sqr+1;
    }
    for(rint i=1;i<=m;i++){
        scanf("%d%d",&q[i].l,&q[i].r);
        q[i].id=i;
    }
    sort(q+1,q+m+1);
    rll res=0;
    for(rint i=1,cl=1,cr=0;i<=m;i++){
        //先扩再缩，不然会出现问题
        while(cr<q[i].r)res+=2*cnt[a[++cr]]++;
        while(cl>q[i].l)res+=2*cnt[a[--cl]]++;
        while(cr>q[i].r)res-=2*--cnt[a[cr--]];
        while(cl<q[i].l)res-=2*--cnt[a[cl++]];
        if(res){
            up[q[i].id]=res;
            down[q[i].id]=1ll*(cr-cl+1)*(cr-cl);
        }else{
            up[q[i].id]=0;
            down[q[i].id]=1;
        }
    }
    for(rint i=1;i<=m;i++){
        rll g=__gcd(up[i],down[i]);
        printf("%lld/%lld\n",up[i]/g,down[i]/g);
    }
    return 0;
}

带修莫队

本质上是三维莫队

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=1e6+10;
int sqr;
struct Node{
    int l,r,t,id;
    bool operator < (const Node&A)const{
        if(l/sqr!=A.l/sqr)return l/sqr<A.l/sqr;
        if(r/sqr!=A.r/sqr)return r/sqr<A.r/sqr;
        return t/sqr<A.t/sqr;
    }
}q[N];
struct xg{
    int idx,x;
}r[N];
int cnt1,cnt2,col[N],cl,cr,ct,cnt[N],res,ans[N];
void Add(rint x){
    if(cl<=r[x].idx&&r[x].idx<=cr){
        res-=!--cnt[col[r[x].idx]];
        res+=!cnt[r[x].x]++;
    }
    swap(r[x].x,col[r[x].idx]);
}
int main(){
    rint n,m;
    scanf("%d%d",&n,&m);
    sqr=pow(n,2.0/3);
    for(rint i=1;i<=n;i++)
        scanf("%d",&col[i]);
    for(rint i=1;i<=m;i++){
        char s[10];
        scanf("%s",s);
        if(s[0]=='Q'){
            ++cnt1;
            scanf("%d%d",&q[cnt1].l,&q[cnt1].r);
            q[cnt1].t=cnt2;
            q[cnt1].id=cnt1;
        }else {
            ++cnt2;
            scanf("%d%d",&r[cnt2].idx,&r[cnt2].x);
        }
    }
    sort(q+1,q+cnt1+1);cr=ct=0;cl=1;
    for(rint i=1;i<=cnt1;i++){
        while(cr<q[i].r)res+=!cnt[col[++cr]]++;
        while(cl>q[i].l)res+=!cnt[col[--cl]]++;
        while(cr>q[i].r)res-=!--cnt[col[cr--]];
        while(cl<q[i].l)res-=!--cnt[col[cl++]];
        while(ct<q[i].t)Add(++ct);
        while(ct>q[i].t)Add(ct--);
        ans[q[i].id]=res;
    }
    for(rint i=1;i<=cnt1;i++)
        printf("%d\n",ans[i]);
    return 0;
}

回滚莫队

#include<bits/stdc++.h>
using namespace std;
#define rint register int
const int N=2e5+10;
unordered_map<int,int> mp;
int sqr,cnt,L[N],R[N],col[N];
struct Node{
	int l,r,id;
	bool operator < (const Node&A)const{
		return col[l]==col[A.l]?r<A.r:col[l]<col[A.l];
	}
}q[N];
int a[N],pos1[N],pos2[N],ans[N];
int main(){
	rint n;
	scanf("%d",&n);
	sqr=sqrt(n);
	for(rint i=1;i<=n;i++){
		scanf("%d",&a[i]);
		if(!mp[a[i]])mp[a[i]]=++cnt;
		a[i]=mp[a[i]];
		col[i]=(i-1)/sqr+1;
		if(!L[col[i]])L[col[i]]=i;
		R[col[i]]=i;
	}
	rint m;
	scanf("%d",&m);
	for(rint i=1;i<=m;i++){
		scanf("%d%d",&q[i].l,&q[i].r);
		q[i].id=i;
	}
	sort(q+1,q+m+1);
	for(rint c=1,k=1;c<=col[n];c++){
		rint l=R[c]+1,r=R[c],nowr=R[c],now=0;
		for(;col[q[k].l]==c;k++){
			if(col[q[k].r]==c){
				for(rint i=q[k].l;i<=q[k].r;i++){
					if(!pos1[a[i]])pos1[a[i]]=i;
					now=max(now,i-pos1[a[i]]);
				}
				for(rint i=q[k].l;i<=q[k].r;i++)
					pos1[a[i]]=0;
				ans[q[k].id]=now;
				now=0;
			}else {
				while(r<q[k].r){
					++r;
					if(!pos1[a[r]])pos1[a[r]]=r;
					pos2[a[r]]=r;
					now=max(now,r-pos1[a[r]]);
				}
				rint tmp=now;
				while(l>q[k].l){
					--l;
					if(!pos2[a[l]])pos2[a[l]]=l;
					else now=max(now,pos2[a[l]]-l);
				}
				ans[q[k].id]=now;
				now=tmp;
				while(l<=nowr){
					if(pos2[a[l]]==l)pos2[a[l]]=0;
					l++;
				}
			}
		}
		memset(pos1,0,sizeof(pos1));
		memset(pos2,0,sizeof(pos2));
	}
	for(rint i=1;i<=m;i++)
		printf("%d\n",ans[i]);
	return 0;
}

树上莫队

带修改了

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=3e5+10;
struct Edge{
	int to,nxt;
}e[N<<1];
int h[N],idx;
void Ins(int a,int b){
	e[++idx].to=b;e[idx].nxt=h[a];h[a]=idx;
}
int fir[N],sec[N],rk[N],fa[N],tp[N],d[N],son[N],siz[N],v[N],w[N],c[N],Time;
void dfs1(rint u){
	fir[u]=++Time;rk[Time]=u;
	siz[u]=1;
	for(rint i=h[u];i;i=e[i].nxt){
		rint v=e[i].to;
		if(v==fa[u])continue;
		fa[v]=u;
		d[v]=d[u]+1;
		dfs1(v);
		siz[u]+=siz[v];
		if(siz[v]>siz[son[u]])son[u]=v;
	}
	sec[u]=++Time;rk[Time]=u;
}
void dfs2(rint u,rint t){
	tp[u]=t;
	if(son[u])dfs2(son[u],t);
	for(rint i=h[u];i;i=e[i].nxt){
		rint v=e[i].to;
		if(v==fa[u]||v==son[u])continue;
		dfs2(v,v);
	}
}
int lca(rint x,rint y){
	while(tp[x]^tp[y]){
		if(d[tp[x]]<d[tp[y]])y=fa[tp[y]];
		else x=fa[tp[x]];
	}
	return d[x]>d[y]?y:x;
}
int sqr,bel[N];
struct Ask{
	int l,r,h,id,lc;
	bool operator < (const Ask&A)const{
		return bel[l]==bel[A.l]?bel[r]==bel[A.r]?h<A.h:r<A.r:l<A.l;
	}
}q[N];
int c1,c2,v1[N],v2[N],cnt[N];
long long ans[N],res;
bool vis[N];
void upd(rint x){
	if(vis[x]){
		res-=1ll*w[cnt[c[x]]]*v[c[x]];
		cnt[c[x]]--;
	}else {
		cnt[c[x]]++;
		res+=1ll*w[cnt[c[x]]]*v[c[x]];
	}
	vis[x]^=1;
}
void xg(rint x){
	if(vis[v1[x]]){
		upd(v1[x]);
		swap(c[v1[x]],v2[x]);
		upd(v1[x]);
	}else {
		swap(c[v1[x]],v2[x]);
	}
}
int main(){
	rint n,m,ask;
	scanf("%d%d%d",&n,&m,&ask);
	for(rint i=1;i<=m;i++)scanf("%d",&v[i]);
	for(rint i=1;i<=n;i++)scanf("%d",&w[i]);
	for(rint i=2;i<=n;i++){
		rint a,b;
		scanf("%d%d",&a,&b);
		Ins(a,b);Ins(b,a);
	}
	for(rint i=1;i<=n;i++)scanf("%d",&c[i]);
	dfs1(1);
	dfs2(1,1);
	sqr=pow(Time,2.0/3.0);
	for(rint i=1;i<=Time;i++)bel[i]=(i-1)/sqr+1;
	for(rint i=1;i<=ask;i++){
		rint opt;
		scanf("%d",&opt);
		if(opt==0){
			c1++;
			scanf("%d%d",&v1[c1],&v2[c1]);
		}else {
			c2++;
			rint x,y;
			scanf("%d%d",&x,&y);
			rint lc=lca(x,y);
			if(lc==x||lc==y){
				if(lc==x){
					q[c2].l=fir[x];
					q[c2].r=fir[y];
					q[c2].h=c1;
					q[c2].id=c2;
					q[c2].lc=-1;
				}else {
					q[c2].l=fir[y];
					q[c2].r=fir[x];
					q[c2].h=c1;
					q[c2].id=c2;
					q[c2].lc=-1;
				}
			}else {
				if(fir[x]>sec[y]){
					q[c2].l=sec[y];
					q[c2].r=fir[x];
					q[c2].h=c1;
					q[c2].id=c2;
					q[c2].lc=lc;
				}else {
					q[c2].l=sec[x];
					q[c2].r=fir[y];
					q[c2].h=c1;
					q[c2].id=c2;
					q[c2].lc=lc;
				}
			}
		}
	}
	sort(q+1,q+c2+1);
	for(rint i=1,x=1,y=0,z=0;i<=c2;i++){
		while(y<q[i].r)upd(rk[++y]);
		while(x>q[i].l)upd(rk[--x]);
		while(y>q[i].r)upd(rk[y--]);
		while(x<q[i].l)upd(rk[x++]);
		while(z<q[i].h)xg(++z);
		while(z>q[i].h)xg(z--);
		if(q[i].lc!=-1)upd(q[i].lc);
		ans[q[i].id]=res;
		if(q[i].lc!=-1)upd(q[i].lc);
	}
	for(rint i=1;i<=c2;i++)
		printf("%lld\n",ans[i]);
	return 0;
}

二次离线莫队

二次离线是指，莫队是询问的一次离线，由于移动莫队端点时更新这个点的贡献可能比较困难，所以把移动的贡献重新算。

所有的贡献可以表示为 $r+1$ 在 $[l,r]$ 产生的贡献的类型，把它转化为 $r+1$ 在 $[1,r]$ 和 $[1,l-1]$ 的类型，之后离线这种询问，统一处理。

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
#define rint register int
#define ll long long
#define rll register long long
int cnt[N],a[N],col[N];
struct Ask{
	int l,r,id;
	bool operator < (const Ask&A)const{
		return col[l]==col[A.l]?col[l]&1?r<A.r:r>A.r:col[l]<col[A.l];
	}
}q[N];
vector<int> vec;
struct Node{
	int l,r,typ,id;
};
vector<Node> vec2[N];
ll ans[N],tmpans[N],tmpans2[N];
int main(){
	rint n,m,k;
	scanf("%d%d%d",&n,&m,&k);
	for(rint i=0;i<16384;i++){
		rint c=0;
		for(rint x=i;x;x&=x-1)c++;
		if(c==k)vec.push_back(i);
	}
	rint sqr=sqrt(n);
	for(rint i=1;i<=n;i++){
		col[i]=(i-1)/sqr+1;
		scanf("%d",&a[i]);
	}
	for(rint i=1;i<=m;i++){
		scanf("%d%d",&q[i].l,&q[i].r);
		q[i].id=i;
	}
	sort(q+1,q+m+1);
	for(rint i=1;i<=n;i++){
		for(rint v:vec)
			cnt[v^a[i]]++;
		tmpans[i+1]=cnt[a[i+1]];
		tmpans2[i]=cnt[a[i]];
	}
	for(rint i=1;i<=n;i++)
		tmpans[i]+=tmpans[i-1],tmpans2[i]+=tmpans2[i-1];
	for(rint i=1,cl=1,cr=0;i<=m;i++){
		if(cr<q[i].r){
			vec2[cl-1].push_back((Node){cr+1,q[i].r,0,i});
			ans[i]+=tmpans[q[i].r]-tmpans[cr];
			cr=q[i].r;
		}
		if(cl>q[i].l){
			vec2[cr].push_back((Node){q[i].l,cl-1,1,i});
			ans[i]-=tmpans2[cl-1]-tmpans2[q[i].l-1];
			cl=q[i].l;
		}
		if(cr>q[i].r){
			vec2[cl-1].push_back((Node){q[i].r+1,cr,1,i});
			ans[i]-=tmpans[cr]-tmpans[q[i].r];
			cr=q[i].r;
		}
		if(cl<q[i].l){
			vec2[cr].push_back((Node){cl,q[i].l-1,0,i});
			ans[i]+=tmpans2[q[i].l-1]-tmpans2[cl-1];
			cl=q[i].l;
		}
	}
	memset(cnt,0,sizeof(cnt));
	for(rint i=1;i<=n;i++){
		for(rint v:vec)
			cnt[v^a[i]]++;
		for(Node v:vec2[i]){
			if(v.typ){
				for(rint j=v.l;j<=v.r;j++)
					ans[v.id]+=cnt[a[j]];
			}else {
				for(rint j=v.l;j<=v.r;j++)
					ans[v.id]-=cnt[a[j]];
			}
		}
	}
	for(rint i=1;i<=m;i++){
		ans[i]+=ans[i-1];
		tmpans[q[i].id]=i;
	}
	for(rint i=1;i<=m;i++)
		printf("%lld\n",ans[tmpans[i]]);
	return 0;
}

Splay

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=1e5+10;
int fa[N],ch[N][2],rt,tot,siz[N],val[N],cnt[N];
void up(rint x){
    siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+cnt[x];
}
void rotate(rint x){
    rint y=fa[x],z=fa[y];
    rint k=ch[y][1]==x;
    ch[z][ch[z][1]==y]=x;
    fa[x]=z;
    ch[y][k]=ch[x][k^1];
    fa[ch[x][k^1]]=y;
    ch[x][k^1]=y;
    fa[y]=x;
    up(y);up(x);
}
void splay(rint x,rint t){
    for(rint y=fa[x],z=fa[y];fa[x]!=t;rotate(x),y=fa[x],z=fa[y])
        if(z!=t)((ch[z][1]==y)==(ch[y][1]==x))?rotate(y):rotate(x);
    if(!t)rt=x;
}
void find(rint x){
    rint u=rt;
    if(!u)return ;
    while(ch[u][x>val[u]]&&x!=val[u])
        u=ch[u][x>val[u]];
    splay(u,0);
}
void insert(rint x){
    rint u=rt,f=0;
    while(u&&val[u]!=x){
        f=u;
        u=ch[u][x>val[u]];
    }
    if(u)cnt[u]++;
    else {
        u=++tot;
        if(f)ch[f][x>val[f]]=u;
        fa[tot]=f;
        val[tot]=x;
        cnt[tot]=siz[tot]=1;
        ch[tot][0]=ch[tot][1]=0;
    }
    splay(u,0);
}
int pre(rint x){
    find(x);
    if(val[rt]<x)return rt;
    int u=ch[rt][0];
    while(ch[u][1])u=ch[u][1];
    return u;
}
int nxt(rint x){
    find(x);
    if(val[rt]>x)return rt;
    int u=ch[rt][1];
    while(ch[u][0])u=ch[u][0];
    return u;
}
void del(rint x){
    int l_x=pre(x),r_x=nxt(x);
    splay(l_x,0);
    splay(r_x,l_x);
    rint u=ch[r_x][0];
    if(cnt[u]>1){
        cnt[u]--;splay(u,0);
    }else {
        ch[r_x][0]=0;
    }
}
int qrk(rint x){
    rint u=rt;
    if(siz[u]<x)return -1;
    while(1){
        int v=ch[u][0];
        if(x>siz[v]+cnt[u]){
            x-=siz[v]+cnt[u];
            u=ch[u][1];
        }else if(siz[v]>=x)u=v;
        else return val[u];
    }
    return -1;
}
const int INF=0x3f3f3f3f;
int main(){
    int n;
    scanf("%d",&n);
    insert(INF);insert(-INF);
    while(n--){
        rint opt,x;
        scanf("%d%d",&opt,&x);
        if(opt==1)insert(x);
        else if(opt==2)del(x);
        else if(opt==3){find(x);printf("%d\n",siz[ch[rt][0]]);}
        else if(opt==4){printf("%d\n",qrk(x+1));}
        else if(opt==5){printf("%d\n",val[pre(x)]);}
        else printf("%d\n",val[nxt(x)]);
    }
    return 0;
}

无旋Treap

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=1e5+10;
int cnt,rt,ch[N][2],siz[N],val[N],rnd[N];
mt19937 trnd(time(0));
uniform_int_distribution<int> dis(1,1e9);
int New(rint x){
    ++cnt;val[cnt]=x;rnd[cnt]=dis(trnd);siz[cnt]=1;
    return cnt;
}
void up(rint x){siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;}
void split(rint rt,rint k,rint &x,rint &y){
    if(!rt){x=0;y=0;return;}
    if(val[rt]<=k)x=rt,split(ch[x][1],k,ch[x][1],y);
    else y=rt,split(ch[y][0],k,x,ch[y][0]);
    up(rt);
}
int merge(rint x,rint y){
    if(!x||!y)return x+y;
    if(rnd[x]<rnd[y])return ch[x][1]=merge(ch[x][1],y),up(x),x;
    return ch[y][0]=merge(x,ch[y][0]),up(y),y;
}
int kth(rint x,rint k){
    while(1){
        if(k<=siz[ch[x][0]])x=ch[x][0];
        else if(k==siz[ch[x][0]]+1)return x;
        else k-=siz[ch[x][0]]+1,x=ch[x][1];
    }
    return -1;
}
int main(){
    rint T,x,y,z;
    scanf("%d",&T);
    while(T--){
        rint opt,w;
        scanf("%d%d",&opt,&w);
        if(opt==1){
            split(rt,w,x,y);
            rt=merge(x,merge(New(w),y));
        }else if(opt==2){
            split(rt,w,x,y);
            split(x,w-1,x,z);
            z=merge(ch[z][0],ch[z][1]);
            rt=merge(merge(x,z),y);
        }else if(opt==3){
            split(rt,w-1,x,y);
            printf("%d\n",siz[x]+1);
            rt=merge(x,y);
        }else if(opt==4){
            printf("%d\n",val[kth(rt,w)]);
        }else if(opt==5){
            split(rt,w-1,x,y);
            printf("%d\n",val[kth(x,siz[x])]);
            rt=merge(x,y);
        }else {
            split(rt,w,x,y);
            printf("%d\n",val[kth(y,1)]);
            rt=merge(x,y);
        }
    }
    return 0;
}

可持久化平衡树

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+10;
const int maxn=1e7+10;
#define rint register int
struct fhq{
	int ch[2],fa,siz,val,rnd;
}t[maxn<<2];
int rt[N],cnt;
mt19937 rnd(time(0));
uniform_int_distribution<int> dis(1,1e9);
int New(rint v){
	t[++cnt].val=v;t[cnt].rnd=dis(rnd);t[cnt].siz=1;
	return cnt;
}
void up(rint x){
	t[x].siz=t[t[x].ch[0]].siz+t[t[x].ch[1]].siz+1;
}
void split(rint rt,rint k,rint &x,rint &y){
	if(!rt){x=y=0;return;}
	if(t[rt].val<=k){
		x=++cnt;t[x]=t[rt];
		split(t[x].ch[1],k,t[x].ch[1],y);
		up(x);
	}else {
		y=++cnt;t[y]=t[rt];
		split(t[y].ch[0],k,x,t[y].ch[0]);
		up(y);
	}
}
int merge(rint x,rint y){
	if(!x||!y)return x+y;
	if(t[x].rnd>t[y].rnd){
		rint p=++cnt;t[p]=t[x];
		t[p].ch[1]=merge(t[p].ch[1],y);
		up(p);return p;
	}else {
		rint p=++cnt;t[p]=t[y];
		t[p].ch[0]=merge(x,t[p].ch[0]);
		up(p);return p;
	}
}
int x,y,z;
void insert(int &rt,int v){
	split(rt,v,x,y);
	rt=merge(merge(x,New(v)),y);
}
void del(int &rt,int v){
	split(rt,v,x,y);
	split(x,v-1,x,z);
	if(z)z=merge(t[z].ch[0],t[z].ch[1]);
	rt=merge(merge(x,z),y);
}
int kth(int rt,int k){
	while(1){
		if(k<=t[t[rt].ch[0]].siz)rt=t[rt].ch[0];
		else if(k==t[t[rt].ch[0]].siz+1)return rt;
		else k-=t[t[rt].ch[0]].siz+1,rt=t[rt].ch[1];
	}
}
int per(int &rt,int k){
	split(rt,k-1,x,y);
	rint res=kth(x,t[x].siz);
	rt=merge(x,y);
	return res;
}
int nxt(int &rt,int k){
	split(rt,k,x,y);
	rint res=kth(y,1);
	rt=merge(x,y);
	return res;
}
void getk(int &rt,int k){
	split(rt,k-1,x,y);
	printf("%d\n",t[x].siz);
	rt=merge(x,y);
}
int main(){
	rint n,tim=0;
	scanf("%d",&n);
	rt[0]=merge(New(-0x7fffffff),New(0x7fffffff));
	while(n--){
		rint v,opt,a;
		scanf("%d%d%d",&v,&opt,&a);
		rt[++tim]=rt[v];
		if(opt==1)insert(rt[tim],a);
		else if(opt==2)del(rt[tim],a);
		else if(opt==3)getk(rt[tim],a);
		else if(opt==4)printf("%d\n",t[kth(rt[tim],a+1)].val);
		else if(opt==5)printf("%d\n",t[per(rt[tim],a)].val);
		else printf("%d\n",t[nxt(rt[tim],a)].val);
	}
	return 0;
}

线段树合并

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
struct Edge{
	int to,nxt;
}e[N<<1];
int h[N],idx;
void Ins(int a,int b){
	e[++idx].to=b;e[idx].nxt=h[a];h[a]=idx;
}
int dep[N],fa[N],siz[N],son[N];
void dfs1(int u){
	siz[u]=1;
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa[u])continue;
		fa[v]=u;
		dep[v]=dep[u]+1;
		dfs1(v);
		siz[u]+=siz[v];
		if(siz[v]>siz[son[u]])son[u]=v;
	}
}
int top[N];
void dfs2(int u,int tt){
	top[u]=tt;
	if(son[u])dfs2(son[u],tt);
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa[u]||v==son[u])continue;
		dfs2(v,v);
	}
}
int lca(int x,int y){
	while(top[x]!=top[y]){
		if(dep[top[x]]<dep[top[y]])swap(x,y);
		x=fa[top[x]];
	}
	return dep[x]>dep[y]?y:x;
}
int cnt,lc[N*80],rc[N*80],val[N*80],ide[N*80];
void pushup(int rt){
	int ls=lc[rt],rs=rc[rt];
	if(val[ls]>=val[rs]){
		val[rt]=val[ls];
		ide[rt]=ide[ls];
	}else{
		val[rt]=val[rs];
		ide[rt]=ide[rs];
	}
}
void modify(int &rt,int l,int r,int pos,int w){
	if(!rt)rt=++cnt;
	if(l==r){
		val[rt]+=w;
		ide[rt]=l;
		return;
	}
	int mid=l+r>>1;
	if(pos<=mid)modify(lc[rt],l,mid,pos,w);
	else modify(rc[rt],mid+1,r,pos,w);
	pushup(rt);
}
int merge(int ls,int rs,int l,int r){
	if(!ls||!rs)return ls+rs;
	if(l==r){
		val[ls]+=val[rs];
		return ls;
	}
	int mid=l+r>>1;
	lc[ls]=merge(lc[ls],lc[rs],l,mid);
	rc[ls]=merge(rc[ls],rc[rs],mid+1,r);
	val[ls]+=val[rs];
	pushup(ls);
	return ls;
}
int ans[N];
int rt[N];
void dfs(int u){
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa[u])continue;
		dfs(v);
		rt[u]=merge(rt[u],rt[v],1,N-10);
	}
	ans[u]=ide[rt[u]];
	if(val[rt[u]]==0)ans[u]=0;
}
inline int read(){
	int x=0,f=1;
	char ch=getchar();
	while(ch<'0'||ch>'9'){
		if(ch=='-')f=-1;
		ch=getchar();
	}
	while(ch<='9'&&ch>='0'){
		x=x*10+ch-'0';
		ch=getchar();
	}
	return x*f;
}
int main(){
	int n=read(),m=read();
	for(int i=1;i<n;i++){
		int a=read(),b=read();
		Ins(a,b);Ins(b,a);
	}
	dfs1(1);
	dfs2(1,1);
	for(int i=1;i<=m;i++){
		int x=read(),y=read(),z=read();
		modify(rt[x],1,N-10,z,1);
		modify(rt[y],1,N-10,z,1);
		int t=lca(x,y);
		modify(rt[t],1,N-10,z,-1);
		modify(rt[fa[t]],1,N-10,z,-1);
	}
	dfs(1);
	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
    return 0;
}

线段树分裂

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=1e6+10;
const int maxn=N<<2;
int rt[N],lc[maxn],rc[maxn],cnt;
ll val[maxn];
void up(rint x){
	val[x]=val[lc[x]]+val[rc[x]];
}
void Build(rint &rt,rint l,rint r){
	if(!rt)rt=++cnt;
	if(l==r){
		scanf("%lld",&val[rt]);
		return ;
	}
	rint mid=l+r>>1;
	Build(lc[rt],l,mid);
	Build(rc[rt],mid+1,r);
	up(rt);
}
void split(rint &r1,rint &r2,rint l,rint r,rint cl,rint cr){
	if(!r1){
		r2=0;
		return;
	}
	if(cl<=l&&r<=cr){
		r2=r1;r1=0;
		return ;
	}
	r2=++cnt;
	rint mid=l+r>>1;
	if(cl<=mid)split(lc[r1],lc[r2],l,mid,cl,cr);
	if(cr>mid)split(rc[r1],rc[r2],mid+1,r,cl,cr);
	up(r1);up(r2);
}
int merge(rint x,rint y){
	if(!x||!y)return x+y;
	lc[x]=merge(lc[x],lc[y]);
	rc[x]=merge(rc[x],rc[y]);
	val[x]+=val[y];
	return x;
}
void update(rint &rt,rint l,rint r,rint p,rint v){
	if(!rt)rt=++cnt;
	val[rt]+=v;
	if(l==r)return;
	rint mid=l+r>>1;
	if(p<=mid)update(lc[rt],l,mid,p,v);
	else update(rc[rt],mid+1,r,p,v);
}
ll query(rint rt,rint l,rint r,rint cl,rint cr){
	if(!rt)return 0;
	if(cl<=l&&r<=cr)return val[rt];
	rint mid=l+r>>1;rll res=0;
	if(cl<=mid)res+=query(lc[rt],l,mid,cl,cr);
	if(cr>mid)res+=query(rc[rt],mid+1,r,cl,cr);
	return res;
}
int kth(rint rt,rint l,rint r,rint k){
	if(k>val[rt])return -1;
	if(l==r)return l;
	int mid=l+r>>1;
	if(k<=val[lc[rt]])return kth(lc[rt],l,mid,k);
	else return kth(rc[rt],mid+1,r,k-val[lc[rt]]);
}
int main(){
	rint n,m,cnt=1;
	scanf("%d%d",&n,&m);
	Build(rt[1],1,n);
	while(m--){
		rint opt;
		scanf("%d",&opt);
		if(opt==0){
			rint p,x,y;
			scanf("%d%d%d",&p,&x,&y);++cnt;
			split(rt[p],rt[cnt],1,n,x,y);
		}else if(opt==1){
			rint x,y;
			scanf("%d%d",&x,&y);
			rt[x]=merge(rt[x],rt[y]);
		}else if(opt==2){
			rint p,x,y;
			scanf("%d%d%d",&p,&x,&y);
			update(rt[p],1,n,y,x);
		}else if(opt==3){
			rint p,x,y;
			scanf("%d%d%d",&p,&x,&y);
			printf("%lld\n",query(rt[p],1,n,x,y));
		}else if(opt==4){
			rint p,k;
			scanf("%d%d",&p,&k);
			printf("%d\n",kth(rt[p],1,n,k));
		}
	}
	return 0;
}

势能分析线段树

区间与/或一个数

#include<bits/stdc++.h>
using namespace std;
namespace lin4xu{

#define re register
#define rint re int
#define ll long long
#define rll re ll
#define db double
#define rdb re db
#define rch re char

template <typename T> inline T read()
{
	re T ans=0;rint f=0;rch ch=getchar();
	for(;!isdigit(ch);ch=getchar()) f=ch=='-';
	for(;isdigit(ch);ch=getchar()) ans=ans*10+(ch&15);
	return f?-ans:ans;
}

const int maxn=5e5+10;

int sum1[maxn<<2];
int sum2[maxn<<2];
int tag[maxn<<2];
int mn[maxn<<2];

inline void push_up(rint x)
{
	rint lx=x<<1,rx=lx|1;
	mn[x]=min(mn[lx],mn[rx]);
	sum1[x]=sum1[lx]&sum1[rx];
	sum2[x]=sum2[lx]|sum2[rx];
}

inline void push_down(rint x)
{
	if(!tag[x]) return;
	rint lx=x<<1,rx=lx|1;
	sum1[lx]+=tag[x],sum2[lx]+=tag[x],tag[lx]+=tag[x],mn[lx]+=tag[x];
	sum1[rx]+=tag[x],sum2[rx]+=tag[x],tag[rx]+=tag[x],mn[rx]+=tag[x];
	tag[x]=0;
}

int left,right,val;

void build(rint x,rint l,rint r)
{
	if(l==r) return sum1[x]=sum2[x]=mn[x]=read<int>(),void();
	rint mid=l+r>>1;
	build(x<<1,l,mid);
	build(x<<1|1,mid+1,r);
	push_up(x);
}

void modify1(rint x,rint l,rint r)
{
	if(l>=left&&r<=right&&sum1[x]-(sum1[x]&val)==sum2[x]-(sum2[x]&val)) return tag[x]+=(sum1[x]&val)-sum1[x],mn[x]+=(sum1[x]&val)-sum1[x],sum1[x]&=val,sum2[x]&=val,void();
	rint mid=l+r>>1;push_down(x);
	if(left<=mid) modify1(x<<1,l,mid);
	if(right>mid) modify1(x<<1|1,mid+1,r);
	push_up(x);
}

void modify2(rint x,rint l,rint r)
{
	if(l>=left&&r<=right&&sum1[x]-(sum1[x]|val)==sum2[x]-(sum2[x]|val)) return tag[x]+=(sum1[x]|val)-sum1[x],mn[x]+=(sum1[x]|val)-sum1[x],sum1[x]|=val,sum2[x]|=val,void();
	rint mid=l+r>>1;push_down(x);
	if(left<=mid) modify2(x<<1,l,mid);
	if(right>mid) modify2(x<<1|1,mid+1,r);
	push_up(x);
}

int query(rint x,rint l,rint r)
{
	if(l>=left&&r<=right) return mn[x];
	rint mid=l+r>>1,ans=0x7fffffff;push_down(x);
	if(left<=mid) ans=min(ans,query(x<<1,l,mid));
	if(right>mid) ans=min(ans,query(x<<1|1,mid+1,r));
	return ans;
}

int main()
{
	rint n=read<int>(),m=read<int>();
	build(1,1,n);
	for(rint tt=1;tt<=m;tt++)
	{
		rint opt=read<int>();left=read<int>(),right=read<int>();
		if(opt==1) val=read<int>(),modify1(1,1,n);
		else if(opt==2) val=read<int>(),modify2(1,1,n);
		else printf("%d\n",query(1,1,n));
	}
	return 0;
}};

int main(){return lin4xu::main();}

区间最值

以区间和一个数 $v$ 取 $\min$ 为例，我们记录一个最大值 $Max$ ，记录一个次大值 $sMax$ ，如果 $v>Max$ 那么可以直接返回，如果 $v>sMax$ 那么只需要改最大值，否则直接递归，这样做是对的，原因在于每次往下递归一定会合并至少两个不同的数，合并次数最多为 $n$ ，单次合并是 $\log$ 。

点分治/树

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=4e4+10;
struct Edge{
	int to,val,nxt;
}e[N<<1];
int h[N],idx;
void Ins(int a,int b,int c){
	e[++idx].to=b;e[idx].val=c;
	e[idx].nxt=h[a];h[a]=idx;
}
int Mx[N],siz[N],rt,Tsiz;
bool vis[N];
void get(int u,int fa){
	siz[u]=1;Mx[u]=0;
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(vis[v]||v==fa)continue;
		get(v,u);
		siz[u]+=siz[v];
		Mx[u]=max(Mx[u],siz[v]);
	}
	Mx[u]=max(Mx[u],Tsiz-siz[u]);
	if(Mx[u]<Mx[rt])rt=u;
}
int d[N],tt,k;
ll ans;
void dfs(int u,int fa,int dis){
	d[++tt]=dis;
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa||vis[v])continue;
		dfs(v,u,dis+e[i].val);
	}
}
ll getans(int rt,int st){
	tt=0;
	dfs(rt,0,st);
	sort(d+1,d+tt+1);
	int i=1,j=tt;
	ll tmp=0;
	while(i<=j){
		while(i<j&&d[i]+d[j]>k)j--;
		tmp+=j-i;
		i++;
	}
	return tmp;
}
void div(int rt){
	ans+=getans(rt,0);
	vis[rt]=1;
	for(int i=h[rt];i;i=e[i].nxt){
		int v=e[i].to;
		if(vis[v])continue;
		ans-=getans(v,e[i].val);
		Tsiz=siz[v];rt=0;
		get(v,0);
		div(rt);
	}
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<n;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		Ins(a,b,c);Ins(b,a,c);
	}
	scanf("%d",&k);
	Tsiz=n;Mx[0]=0x7fffffff;
	get(1,0);
	div(rt);
	printf("%lld\n",ans);
	return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
typedef long long ll;
unordered_map<int,int> dis[N];
struct Edge{
	int to,nxt,val;
}e[N<<1];
int h[N],idx;
void Ins(int a,int b,int c){
	e[++idx].to=b;e[idx].nxt=h[a];
	h[a]=idx;e[idx].val=c;
}
struct Node{
	int x,y,w;
	Node(){}
	Node(int a,int b,int c){x=a;y=b;w=c;}
	bool operator < (const Node&A)const{
		return y<A.y;
	}
};
vector<Node> vec[N],rc[N];
vector<ll> s[N],s2[N];
int x[N],rt,Tsiz,Max[N],siz[N];
bool vis[N];
void getrt(int u,int fa){
	siz[u]=1;Max[u]=0;
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa||vis[v])continue;
		getrt(v,u);
		siz[u]+=siz[v];
		Max[u]=max(Max[u],siz[v]);
	}
	Max[u]=max(Max[u],Tsiz-siz[u]);
	if(Max[rt]>Max[u])rt=u;
}
int fa[N];
void cal(int u,int ff,int now){
	vec[now].push_back(Node(u,x[u],dis[now][u]));
	s[now].push_back(0);
	if(fa[now]){
		rc[now].push_back(Node(u,x[u],dis[fa[now]][u]));
		s2[now].push_back(0);
	}
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==ff||vis[v])continue;
		dis[now][v]=dis[now][u]+e[i].val;
		cal(v,u,now);
	}
}
void dfs(int u){
	vis[u]=1;
	dis[u][u]=0;
	vec[u].push_back(Node(0,-1,0));
	vec[u].push_back(Node(0,2e9,0));
	s[u].push_back(0);
	s[u].push_back(0);
	rc[u].push_back(Node(0,-1,0));
	rc[u].push_back(Node(0,2e9,0));
	s2[u].push_back(0);
	s2[u].push_back(0);
	cal(u,0,u);
	sort(vec[u].begin(),vec[u].end());
	for(int i=1;i<vec[u].size();i++)s[u][i]=s[u][i-1]+vec[u][i].w;
	if(fa[u]){
		sort(rc[u].begin(),rc[u].end());
		for(int i=1;i<vec[u].size();i++)s2[u][i]=s2[u][i-1]+rc[u][i].w;
	}
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(vis[v])continue;
		Tsiz=siz[v];rt=0;
		getrt(v,u);
		fa[rt]=u;
		dfs(rt);
	}
}
int main(){
	int n,m,A;
	scanf("%d%d%d",&n,&m,&A);
	for(int i=1;i<=n;i++)
		scanf("%d",&x[i]);
	for(int i=2;i<=n;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		Ins(a,b,c);Ins(b,a,c);
	}
	Max[0]=1e9;Tsiz=n;
	getrt(1,0);
	dfs(rt);
	ll res=0;
	while(m--){
		int x,l,r;
		scanf("%d%d%d",&x,&l,&r);
		l=(l+res)%A,r=(r+res)%A;
		if(l>r)swap(l,r);
		ll tmp1,posl=lower_bound(vec[x].begin(),vec[x].end(),Node(0,l,0))-vec[x].begin()-1;
		ll tmp2,posr=upper_bound(vec[x].begin(),vec[x].end(),Node(0,r,0))-vec[x].begin()-1;
		res=s[x][posr]-s[x][posl];
		for(int i=x;fa[i];i=fa[i]){
			posl=lower_bound(vec[fa[i]].begin(),vec[fa[i]].end(),Node(0,l,0))-vec[fa[i]].begin()-1;
			posr=upper_bound(vec[fa[i]].begin(),vec[fa[i]].end(),Node(0,r,0))-vec[fa[i]].begin()-1;
			tmp1=s[fa[i]][posr]-s[fa[i]][posl];
			tmp2=posr-posl;
			res+=tmp1+tmp2*dis[fa[i]][x];
			posl=lower_bound(rc[i].begin(),rc[i].end(),Node(0,l,0))-rc[i].begin()-1;
			posr=upper_bound(rc[i].begin(),rc[i].end(),Node(0,r,0))-rc[i].begin()-1;
			tmp1=s2[i][posr]-s2[i][posl];
			tmp2=posr-posl;
			res-=tmp1+tmp2*dis[fa[i]][x];
		}
		printf("%lld\n",res);
	}
	return 0;
}

边分治/树

#include<bits/stdc++.h>
using namespace std;
#define rint register int 
#define rll register long long
typedef long long ll;
const int N=3e5+10;
struct Edge{
	int to,nxt;bool vis;
	ll val;
};
struct Node{
	int x,y;
	ll dis;
};
struct T3{
	Edge e[N<<1];
	int h[N],idx;
	void Ins(rint a,rint b,rll c){
		e[++idx].to=b;e[idx].nxt=h[a];
		h[a]=idx;e[idx].val=c;
	}
	int tp[N],son[N],fa[N],siz[N],d[N];
	ll dis[N];
	void dfs1(rint u){
		siz[u]=1;
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(v==fa[u])continue;
			dis[v]=dis[u]+e[i].val;
			d[v]=d[u]+1;fa[v]=u;
			dfs1(v);
			siz[u]+=siz[v];
			if(siz[son[u]]<siz[v])son[u]=v;
		}
	}
	void dfs2(rint u,rint t){
		tp[u]=t;
		if(son[u])dfs2(son[u],t);
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(v==fa[u]||v==son[u])continue;
			dfs2(v,v);
		}
	}
	int lca(rint x,rint y){
		while(tp[x]^tp[y]){
			if(d[tp[x]]<d[tp[y]])y=fa[tp[y]];
			else x=fa[tp[x]];
		}
		return d[x]<d[y]?x:y;
	}
	ll dist(rint x,rint y){
		return dis[x]+dis[y]-2*dis[lca(x,y)];
	}
	void solve(){
		dfs1(1);
		dfs2(1,1);
	}
}tr3;
ll val[N],ext;
int col[N],dfn[N];
bool cmp(rint a,rint b){
	return dfn[a]<dfn[b];
}
Node merge(Node u,Node v){
	if(u.x==0)return v;
	if(v.x==0)return u;
	Node res=u;
	ll tmp=tr3.dist(u.x,v.x)+val[u.x]+val[v.x];
	if(tmp>res.dis)res=(Node){u.x,v.x,tmp};
	tmp=tr3.dist(u.x,v.y)+val[u.x]+val[v.y];
	if(tmp>res.dis)res=(Node){u.x,v.y,tmp};
	tmp=tr3.dist(u.x,u.y)+val[u.x]+val[u.y];
	if(tmp>res.dis)res=(Node){u.x,u.y,tmp};
	tmp=tr3.dist(v.x,v.y)+val[v.x]+val[v.y];
	if(tmp>res.dis)res=(Node){v.x,v.y,tmp};
	tmp=tr3.dist(v.y,u.y)+val[v.y]+val[u.y];
	if(tmp>res.dis)res=(Node){v.y,u.y,tmp};
	tmp=tr3.dist(v.x,u.y)+val[v.x]+val[u.y];
	if(tmp>res.dis)res=(Node){v.x,u.y,tmp};
	return res;
}
ll merge2(Node u,Node v){
	if(u.x==0)return -1e16;
	if(v.x==0)return -1e16;
	Node res=(Node){0,0,0};
	ll tmp=tr3.dist(u.x,v.x)+val[u.x]+val[v.x];
	if(tmp>res.dis)res=(Node){u.x,v.x,tmp};
	tmp=tr3.dist(u.x,v.y)+val[u.x]+val[v.y];
	if(tmp>res.dis)res=(Node){u.x,v.y,tmp};
	tmp=tr3.dist(v.y,u.y)+val[v.y]+val[u.y];
	if(tmp>res.dis)res=(Node){v.y,u.y,tmp};
	tmp=tr3.dist(v.x,u.y)+val[v.x]+val[u.y];
	if(tmp>res.dis)res=(Node){v.x,u.y,tmp};
	return res.dis;
}
ll res;
struct T2{
	Edge e[N<<1];
	int h[N],idx;
	void Ins(rint a,rint b,rll c){
		e[++idx].to=b;e[idx].nxt=h[a];
		h[a]=idx;e[idx].val=c;
	}
	int Time,top[N],son[N],fa[N],siz[N],d[N];
	ll dis[N];
	void dfs1(rint u){
		siz[u]=1;dfn[u]=++Time;
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(v==fa[u])continue;
			dis[v]=dis[u]+e[i].val;
			d[v]=d[u]+1;fa[v]=u;
			dfs1(v);
			siz[u]+=siz[v];
			if(siz[son[u]]<siz[v])son[u]=v;
		}
	}
	void dfs2(rint u,rint t){
		top[u]=t;
		if(son[u])dfs2(son[u],t);
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(v==fa[u]||v==son[u])continue;
			dfs2(v,v);
		}
		h[u]=0;
	}
	int lca(rint x,rint y){
		while(top[x]^top[y]){
			if(d[top[x]]<d[top[y]])y=fa[top[y]];
			else x=fa[top[x]];
		}
		return d[x]<d[y]?x:y;
	}
	ll dist(rint x,rint y){
		return dis[x]+dis[y]-2*dis[lca(x,y)];
	}
	int stk[N],tp;
	void insert(rint x){
		if(!tp){
			stk[++tp]=x;
			return;
		}
		int lc=lca(x,stk[tp]);
		if(lc==stk[tp])stk[++tp]=x;
		else {
			while(tp>1&&d[stk[tp-1]]>=d[lc]){
				Ins(stk[tp-1],stk[tp],0);
				tp--;
			}
			if(stk[tp]!=lc){
				Ins(lc,stk[tp],0);
				stk[tp]=lc;
			}
			stk[++tp]=x;
		}
	}
	Node f[3][N];
	void dfs(rint u){
		stk[++tp]=u;
		if(col[u])f[col[u]][u]=(Node){u,u,val[u]};
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			dfs(v);
			res=max(res,merge2(f[1][u],f[2][v])+ext-2*dis[u]);
			res=max(res,merge2(f[2][u],f[1][v])+ext-2*dis[u]);
			f[1][u]=merge(f[1][u],f[1][v]);
			f[2][u]=merge(f[2][u],f[2][v]);
		}
	}
	void calc(rint *a,rint n){
		sort(a+1,a+n+1,cmp);
		if(a[1]!=1)stk[++tp]=1;
		for(rint i=1;i<=n;i++)insert(a[i]),val[a[i]]+=dis[a[i]];
		while(tp>1)Ins(stk[tp-1],stk[tp],0),tp--;
		tp=0;
		dfs(1);
		for(rint i=1;i<=tp;i++)
			h[stk[i]]=f[2][stk[i]].x=f[2][stk[i]].y=f[2][stk[i]].dis=f[1][stk[i]].x=f[1][stk[i]].y=f[1][stk[i]].dis=0;
		tp=idx=0;
	}
	void solve(){
		dfs1(1);
		dfs2(1,1);
		idx=0;
	}
}tr2;
const int INF=0x7f7f7f7f;
struct T1{
	Edge e[N<<1];
	int h[N],idx;
	void Ins(rint a,rint b,rll c){
		e[++idx].to=b;e[idx].nxt=h[a];
		h[a]=idx;e[idx].val=c;e[idx].vis=0;
	}
	Edge e2[N<<1];
	int h2[N],idx2,cnt,n;
	void Ins2(rint a,rint b,rll c){
		e2[++idx2].to=b;e2[idx2].nxt=h2[a];
		h2[a]=idx2;e2[idx2].val=c;
	}
	void rebuild(rint u,rint fa){
		rint lst=0;
		for(rint i=h2[u];i;i=e2[i].nxt){
			rint v=e2[i].to;
			if(v==fa)continue;
			rebuild(v,u);
			if(!lst){
				lst=u;
			}else {
				++cnt;
				Ins(lst,cnt,0);Ins(cnt,lst,0);
				lst=cnt;
			}
			Ins(lst,v,e2[i].val);Ins(v,lst,e2[i].val);
		}
	}
	int siz[N],Tsiz,Max,rt;
	void getsiz(rint u,rint fa){
		siz[u]=1;
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(v==fa||e[i].vis)continue;
			getsiz(v,u);
			siz[u]+=siz[v];
		}
	}
	void getrt(rint u,rint fa){
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(v==fa||e[i].vis)continue;
			getrt(v,u);
			rint now=max(siz[v],Tsiz-siz[v]);
			if(Max>now)Max=now,rt=i;
		}
	}
	int s[N],tt;
	void dfs(rint u,rint fa,rint *f,rint typ,rint &tt,rll *g,rll dis){
		if(u<=n){
			f[++tt]=u;col[u]=typ;g[u]=dis;
		}
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(v==fa||e[i].vis)continue;
			dfs(v,u,f,typ,tt,g,dis+e[i].val);
		}
	}
	void dfs(){
		if(!rt)return ;
		e[rt].vis=e[rt^1].vis=1;
		rint v1=e[rt].to,v2=e[rt^1].to;
		tt=0;ext=e[rt].val;
		dfs(v1,0,s,1,tt,val,0);
		dfs(v2,0,s,2,tt,val,0);
		tr2.calc(s,tt);
		for(rint i=1;i<=tt;i++)col[s[i]]=0,val[s[i]]=0;
		getsiz(v1,0);Tsiz=siz[v1];Max=INF;rt=0;getrt(v1,0);dfs();
		getsiz(v2,0);Tsiz=siz[v2];Max=INF;rt=0;getrt(v2,0);dfs();
	}
	void solve(rint x){
		cnt=n=x;idx=1;
		rebuild(1,0);
		getsiz(1,0);
		rt=0;Tsiz=n;Max=INF;
		getrt(1,0);
		dfs();
	}
}tr1;
int main(){
	rint n;
	scanf("%d",&n);
	for(rint i=2;i<=n;i++){
		rint a,b;rll w;
		scanf("%d%d%lld",&a,&b,&w);
		tr1.Ins2(a,b,w);tr1.Ins2(b,a,w);
	}
	for(rint i=2;i<=n;i++){
		rint a,b;rll w;
		scanf("%d%d%lld",&a,&b,&w);
		tr2.Ins(a,b,w);tr2.Ins(b,a,w);
	}
	for(rint i=2;i<=n;i++){
		rint a,b;rll w;
		scanf("%d%d%lld",&a,&b,&w);
		tr3.Ins(a,b,w);tr3.Ins(b,a,w);
	}
	tr2.solve();
	tr3.solve();
	tr1.solve(n);
	printf("%lld\n",res);
	return 0;
}

#include<bits/stdc++.h>
#define int long long
using namespace std;
int const N=366671;
template<unsigned M>struct graph
{
    int target[2*M],pre[2*M],last[M],tot,w[2*M];
    void add(int x,int y,int z)
    {
        target[++tot]=y;
        pre[tot]=last[x];
        last[x]=tot;
        w[tot]=z;
    }
};
graph<N>g1,g3;graph<N<<1>g2;
int n,now,siz[N<<1],nows,ed,li,cnt,mx[N<<5|1][2],ch[N<<5|1][2],last[N],dis[N],
tmp,ans=-1e18,rt[N];
bool del[N<<2];
void dfs(int x,int fa)
{
    int las=x;
    for(int i=g1.last[x];i;i=g1.pre[i])
    {
        int tar=g1.target[i];
        if(tar==fa)continue;
        g2.add(las,++now,0),g2.add(now,las,0),g2.add(now,tar,g1.w[i]),g2.add(tar,now,g1.w[i]);
        dis[tar]=dis[x]+g1.w[i];las=now;dfs(tar,x);
    }
}
void dfs2(int x,int fa,int nowd,int op)
{
    if(!op)tmp++;
    if(x<=n)
    {
        cnt++;
        if(!last[x])rt[x]=last[x]=cnt,cnt++;
        ch[last[x]][op]=cnt;
        mx[last[x]][op]=dis[x]+nowd;
        last[x]=cnt;
    }
    for(int i=g2.last[x];i;i=g2.pre[i])
    {
        int tar=g2.target[i];
        if(tar==fa||del[i])continue;
        dfs2(tar,x,nowd+g2.w[i],op);
    }
}
void get(int x,int fa)
{
    siz[x]=1;
    for(int i=g2.last[x];i;i=g2.pre[i])
    {
        int tar=g2.target[i];
        if(tar==fa||del[i])continue;
        get(tar,x);siz[x]+=siz[tar];
        if(max(siz[tar],nows-siz[tar])<li)li=max(siz[tar],nows-siz[tar]),ed=(i+1)>>1;
    }
}
void solve(int x,int s)
{
    if(s==1)return;
    ed=li=1e9;nows=s;
    get(x,0);
    int r1=g2.target[(ed<<1)-1],r2=g2.target[ed<<1];
    del[(ed<<1)-1]=del[ed<<1]=1;tmp=0;
    dfs2(r1,r2,0,0),dfs2(r2,r1,g2.w[ed<<1],1);
    int tt=siz[x]-tmp;
    solve(r1,tmp);solve(r2,tt);
}
int merge(int x,int y,int t)
{
    if((!x)||(!y))return x+y;
    ans=max(ans,max(mx[x][0]+mx[y][1],mx[y][0]+mx[x][1])+2*t);
    mx[x][0]=max(mx[x][0],mx[y][0]),mx[x][1]=max(mx[x][1],mx[y][1]);
    ch[x][0]=merge(ch[x][0],ch[y][0],t),ch[x][1]=merge(ch[x][1],ch[y][1],t);
    return x;
}
void dfs3(int x,int fa,int nowd)
{
    ans=max(ans,2*(dis[x]-nowd));
    for(int i=g3.last[x];i;i=g3.pre[i])
    {
        int tar=g3.target[i];
        if(tar==fa)continue;
        dfs3(tar,x,nowd+g3.w[i]);
        rt[x]=merge(rt[x],rt[tar],-nowd);
    }
}
signed main()
{
    memset(mx,0xc0,sizeof(mx));
    int x,y,z;
    scanf("%lld",&n);now=n;
    for(int i=1;i<n;i++)scanf("%lld%lld%lld",&x,&y,&z),g1.add(x,y,z),g1.add(y,x,z);
    dfs(1,0);
    solve(1,now);
    for(int i=1;i<n;i++)scanf("%lld%lld%lld",&x,&y,&z),g3.add(x,y,z),g3.add(y,x,z);
    dfs3(1,0,0);
    printf("%lld",ans>>1);
    return 0;
}

cdq分治

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=5e5+10;
int tr[N],maxn;
void Add(int x,int w){
    for(;x<=maxn;x+=x&-x)
        tr[x]+=w;
}
int ans[N],tmp[N],tt;
int query(int x){
	int cnt=0;
    for(;x;x&=x-1)cnt+=tr[x];
	return cnt;
}
struct Node{
	int s,c,m,num,val;
	bool operator < (const Node&A)const{
		if(s==A.s){
			if(c==A.c)return m<A.m;
			return c<A.c;
		}
		return s<A.s;
	}
}p1[N],p2[N],t[N];
void cdq(int l,int r){
	if(l==r)return;
	int mid=l+r>>1;
	cdq(l,mid);cdq(mid+1,r);
	int cl=l,cr=mid+1,nl=l;
	while(cl<=mid&&cr<=r){
		if(p2[cl].c<=p2[cr].c)Add(p2[cl].m,p2[cl].num),t[nl++]=p2[cl++];
		else p2[cr].val+=query(p2[cr].m),t[nl++]=p2[cr++];
	}
	while(cl<=mid)Add(p2[cl].m,p2[cl].num),t[nl++]=p2[cl++];
	while(cr<=r)p2[cr].val+=query(p2[cr].m),t[nl++]=p2[cr++];
	for(int i=l;i<=mid;i++)Add(p2[i].m,-p2[i].num);
	for(int i=l;i<=r;i++)p2[i]=t[i];
}
int main(){
	int n;
	scanf("%d%d",&n,&maxn);
	for(int i=1;i<=n;i++)
		scanf("%d%d%d",&p1[i].s,&p1[i].c,&p1[i].m);
	int x=0;
	sort(p1+1,p1+n+1);
	for(int i=1,s=0;i<=n;i++){
		s++;
		if(p1[i].s!=p1[i+1].s||p1[i].c!=p1[i+1].c||p1[i].m!=p1[i+1].m){
			p2[++x]=p1[i];
			p2[x].num=s;
			s=0;
		}
	}
	cdq(1,x);
	for(int i=1;i<=x;i++)
		ans[p2[i].val+p2[i].num-1]+=p2[i].num;
	for(int i=0;i<n;i++)
		printf("%d\n",ans[i]);
	return 0;
}

树状数组套主席树

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
const int maxn=9e6+10;
#define rint register int 
int rt[N],val[maxn],lc[maxn],rc[maxn],a[N],stk[N],tp,cnt;
int New(){
	if(tp)return stk[tp--];
	return ++cnt;
}
void update(int &rt,int l,int r,int p,int v){
	if(!rt)rt=New();
	val[rt]+=v;
	if(l==r)return;
	rint mid=l+r>>1;
	if(p<=mid){
		update(lc[rt],l,mid,p,v);
		if(!val[lc[rt]])stk[++tp]=lc[rt],lc[rt]=0;
	}else {
		update(rc[rt],mid+1,r,p,v);
		if(!val[rc[rt]])stk[++tp]=rc[rt],rc[rt]=0;
	}
}
int q1[N],tt1,q2[N],tt2;
int kth(int l,int r,int k){
	if(l==r)return l;
	rint now=0;
	for(rint i=1;i<=tt1;i++)now+=val[lc[q1[i]]];
	for(rint i=1;i<=tt2;i++)now-=val[lc[q2[i]]];
	rint mid=l+r>>1;
	if(k<=now){
		for(rint i=1;i<=tt1;i++)q1[i]=lc[q1[i]];
		for(rint i=1;i<=tt2;i++)q2[i]=lc[q2[i]];
		return kth(l,mid,k);
	}else {
		for(rint i=1;i<=tt1;i++)q1[i]=rc[q1[i]];
		for(rint i=1;i<=tt2;i++)q2[i]=rc[q2[i]];
		return kth(mid+1,r,k-now);
	}
}
int query(rint l,rint r,rint k,rint res){
	if(l==r)return res;
	rint mid=l+r>>1;
	if(k<=mid){
		for(rint i=1;i<=tt1;i++)q1[i]=lc[q1[i]];
		for(rint i=1;i<=tt2;i++)q2[i]=lc[q2[i]];
		return query(l,mid,k,res);
	}else {
		for(rint i=1;i<=tt1;i++)res+=val[lc[q1[i]]],q1[i]=rc[q1[i]];
		for(rint i=1;i<=tt2;i++)res-=val[lc[q2[i]]],q2[i]=rc[q2[i]];
		return query(mid+1,r,k,res);
	}
}
int n,m;
int getk(rint l,rint r,rint k){
	tt1=tt2=0;
	for(rint i=r;i;i-=i&-i)q1[++tt1]=rt[i];
	for(rint i=l-1;i;i-=i&-i)q2[++tt2]=rt[i];
	return query(0,1e8,k,1);
}
int main(){
	scanf("%d%d",&n,&m);
	for(rint i=1;i<=n;i++){
		scanf("%d",&a[i]);
		for(rint j=i;j<=n;j+=j&-j)update(rt[j],0,1e8,a[i],1);
	}
	while(m--){
		rint opt;
		scanf("%d",&opt);
		if(opt==1){
			rint l,r,k;
			scanf("%d%d%d",&l,&r,&k);
			printf("%d\n",getk(l,r,k));
		}else if(opt==2){
			rint l,r,k;
			scanf("%d%d%d",&l,&r,&k);
			tt1=tt2=0;
			for(rint j=r;j;j-=j&-j)q1[++tt1]=rt[j];
			for(rint j=l-1;j;j-=j&-j)q2[++tt2]=rt[j];
			printf("%d\n",kth(0,1e8,k));
		}else if(opt==3){
			rint pos,k;
			scanf("%d%d",&pos,&k);
			for(rint j=pos;j<=n;j+=j&-j){
				update(rt[j],0,1e8,a[pos],-1);
				update(rt[j],0,1e8,k,1);
			}
			a[pos]=k;
		}else if(opt==4){
			rint l,r,k;
			scanf("%d%d%d",&l,&r,&k);
			k=getk(l,r,k);
			tt1=tt2=0;
			for(rint j=r;j;j-=j&-j)q1[++tt1]=rt[j];
			for(rint j=l-1;j;j-=j&-j)q2[++tt2]=rt[j];
			if(k==1)puts("-2147483647");
			else printf("%d\n",kth(0,1e8,k-1));
		}else {
			rint l,r,k;
			scanf("%d%d%d",&l,&r,&k);
			k=getk(l,r,k);
			for(rint i=1;i<=tt1;i++)k+=val[q1[i]];
			for(rint i=1;i<=tt2;i++)k-=val[q2[i]];
			tt1=tt2=0;
			for(rint j=r;j;j-=j&-j)q1[++tt1]=rt[j];
			for(rint j=l-1;j;j-=j&-j)q2[++tt2]=rt[j];
			if(k>r-l+1)puts("2147483647");
			else printf("%d\n",kth(0,1e8,k));
		}
	}
	return 0;
}

线段树套平衡树

#include<bits/stdc++.h>
using namespace std;

#define re register
#define rint re int
#define ll long long
#define rll re ll
#define db double
#define rdb re db
#define rch re char
#define un unsigned
#define ull un ll
#define rull re ull
#define lx t[x].l
#define rx t[x].r
#define ly t[y].l
#define ry t[y].r
#define lz t[z].l
#define rz t[z].r

template <typename T> T read()
{
	re T ans=0;rint f=1;rch ch=getchar();
	for(;!isdigit(ch);ch=getchar())
		ch=='-'?f=-1:0;
	for(;isdigit(ch);ch=getchar())
		ans=ans*10+(ch&15);
	return ans*f;
}

const int maxn=5e4+10;
const int maxnlg=1e6+10;

struct Node
{
	int l,r,val,size,key;
}t[maxnlg<<2];
int rt[maxn<<2];
int a[maxn];
int t_cnt;

inline int get_point(rint val)
{
	t[++t_cnt]=(Node){0,0,val,1,rand()};
	return t_cnt;
}

inline void push_up(rint x)
{
	t[x].size=t[lx].size+t[rx].size+1;
}

void spilt(rint z,rint val,rint &x,rint &y)
{
	if(!z) return x=y=0,void();
	if(t[z].val<=val) x=z,spilt(rz,val,rx,y),push_up(x);
	else y=z,spilt(lz,val,x,ly),push_up(y);
}

int merge(rint x,rint y)
{
	if(!x||!y) return x|y;
	if(t[x].key>t[y].key) return rx=merge(rx,y),push_up(x),x;
	else return ly=merge(x,ly),push_up(y),y;
}

void insert(rint &rt,rint val)
{
	rint x,y;
	spilt(rt,val,x,y);
	rt=merge(x,merge(get_point(val),y));
}

void del(rint &rt,rint val)
{
	rint x,y,z;
	spilt(rt,val-1,x,y);
	spilt(y,val,y,z);
	rt=merge(x,merge(merge(ly,ry),z));
}

void build(rint x,rint l,rint r)
{
	for(rint i=l;i<=r;i++) insert(rt[x],a[i]);
	if(l==r) return;
	rint mid=l+r>>1;
	build(x<<1,l,mid),build(x<<1|1,mid+1,r);
}

int query(rint &rt,rint val)
{
	rint x,y;
	spilt(rt,val-1,x,y);
	rint ans=t[x].size;
	rt=merge(x,y);
	return ans;
}

int query(rint x,rint l,rint r,rint left,rint right,rint val)
{
	if(l>=left&&r<=right) return query(rt[x],val);
	rint mid=l+r>>1,ans=0;
	if(left<=mid) ans+=query(x<<1,l,mid,left,right,val);
	if(right>mid) ans+=query(x<<1|1,mid+1,r,left,right,val);
	return ans;
}

void modify(rint x,rint l,rint r,rint pos,rint val)
{
	del(rt[x],a[pos]),insert(rt[x],val);
	if(l==r) return a[l]=val,void();
	rint mid=l+r>>1;
	if(pos<=mid) modify(x<<1,l,mid,pos,val);
	else modify(x<<1|1,mid+1,r,pos,val);
}

int find_kth(rint x,rint kth)
{
	if(!kth) return -2147483647;
	if(kth>t[x].size) return 2147483647;
	while(1)
	{
		if(kth<=t[lx].size) x=lx;
		else if(kth==t[lx].size+1) return t[x].val;
		else kth-=t[lx].size+1,x=rx;
	}
}

int query_pre(rint &rt,rint val)
{
	rint x,y;
	spilt(rt,val-1,x,y);
	rint ans=find_kth(x,t[x].size);
	rt=merge(x,y);
	return ans;
}

int query_nxt(rint &rt,rint val)
{
	rint x,y;
	spilt(rt,val,x,y);
	rint ans=find_kth(y,1);
	rt=merge(x,y);
	return ans;
}

int query_pre(rint x,rint l,rint r,rint left,rint right,rint val)
{
	if(l>=left&&r<=right) return query_pre(rt[x],val);
	rint mid=l+r>>1,ans=-2147483647;
	if(left<=mid) ans=max(ans,query_pre(x<<1,l,mid,left,right,val));
	if(right>mid) ans=max(ans,query_pre(x<<1|1,mid+1,r,left,right,val));
	return ans;
}

int query_nxt(rint x,rint l,rint r,rint left,rint right,rint val)
{
	if(l>=left&&r<=right) return query_nxt(rt[x],val);
	rint mid=l+r>>1,ans=2147483647;
	if(left<=mid) ans=min(ans,query_nxt(x<<1,l,mid,left,right,val));
	if(right>mid) ans=min(ans,query_nxt(x<<1|1,mid+1,r,left,right,val));
	return ans;
}

int main()
{
	rint n=read<int>(),m=read<int>();
	for(rint i=1;i<=n;i++) a[i]=read<int>();
	build(1,1,n);
	for(rint tt=1;tt<=m;tt++)
	{
		rint opt=read<int>();
		if(opt==1)
		{
			rint l=read<int>(),r=read<int>();
			printf("%d\n",query(1,1,n,l,r,read<int>())+1);
		}
		if(opt==2)
		{
			rint l=read<int>(),r=read<int>(),K=read<int>();
			rint left=0,right=1e8;
			while(left<=right)
			{
				rint mid=left+right>>1;
				if(query(1,1,n,l,r,mid)+1<=K) left=mid+1;
				else right=mid-1;
			}
			printf("%d\n",right);
		}
		if(opt==3)
		{
			rint pos=read<int>();
			modify(1,1,n,pos,read<int>());
		}
		if(opt==4)
		{
			rint l=read<int>(),r=read<int>();
			printf("%d\n",query_pre(1,1,n,l,r,read<int>()));
		}
		if(opt==5)
		{
			rint l=read<int>(),r=read<int>();
			printf("%d\n",query_nxt(1,1,n,l,r,read<int>()));
		}
	}
	return 0;
}

虚树

#include<bits/stdc++.h>
using namespace std;
const int N=5e5+10;
typedef long long ll;
struct Edge{
	int to,nxt,val;
}e[N<<1];
int h[N],idx;
void Ins(int a,int b,int c){
	e[++idx].to=b;e[idx].nxt=h[a];
	e[idx].val=c;h[a]=idx;
}
int top[N],dep[N],fa[N],dfn[N],Time,son[N],siz[N];
ll Min[N];
void dfs1(int u){
	dfn[u]=++Time;
	siz[u]=1;
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa[u])continue;
		fa[v]=u;
		dep[v]=dep[u]+1;
		Min[v]=min(Min[u],1ll*e[i].val);
		dfs1(v);
		siz[u]+=siz[v];
		if(siz[v]>siz[son[u]])son[u]=v;
	}
}
void dfs2(int u,int t){
	top[u]=t;
	if(son[u])dfs2(son[u],t);
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa[u]||v==son[u])continue;
		dfs2(v,v);
	}
}
int lca(int x,int y){
	while(top[x]^top[y]){
		if(dep[top[x]]>dep[top[y]])x=fa[top[x]];
		else y=fa[top[y]];
	}
	return dep[x]<dep[y]?x:y;
}
int arr[N],stk[N],tp;
ll f[N];
bool cmp(int a,int b){
	return dfn[a]<dfn[b];
}
void insert(int x){
	int lc=lca(x,stk[tp]);
	if(lc==stk[tp]){
		stk[++tp]=x;
		return;
	}
	while(dep[lc]<dep[stk[tp-1]]){
		Ins(stk[tp-1],stk[tp],0);
		tp--;
	}
	if(lc==stk[tp-1])Ins(stk[tp-1],stk[tp],0),tp--;
	else Ins(lc,stk[tp],0),stk[tp]=lc;
	stk[++tp]=x;
}
bool vis[N];
void dfs3(int u){
	ll s=0;
	f[u]=Min[u];
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		dfs3(v);
		s+=f[v];
	}
	if(!vis[u])f[u]=min(s,f[u]);
	h[u]=vis[u]=0;
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=2;i<=n;i++){
		int u,v,w;
		scanf("%d%d%d",&u,&v,&w);
		Ins(u,v,w);Ins(v,u,w);
	}
	Min[1]=1e18;
	dfs1(1);
	dfs2(1,1);
	int m;
	scanf("%d",&m);
	memset(h,0,sizeof(h));
	idx=0;
	while(m--){
		int k;
		scanf("%d",&k);
		for(int i=1;i<=k;i++)
			scanf("%d",&arr[i]),vis[arr[i]]=1;
		sort(arr+1,arr+k+1,cmp);
		stk[++tp]=1;
		for(int i=(arr[1]==1)?2:1;i<=k;i++)insert(arr[i]);
		while(tp>1)Ins(stk[tp-1],stk[tp],0),tp--;
		dfs3(1);
		printf("%lld\n",f[1]);
		tp=idx=0;
	}
	return 0;
}

圆方树

#include<bits/stdc++.h>
using namespace std;
const int N=2e6+10;
#define rint register int
struct Edge{
	int to,nxt;
}e[N<<1],e2[N<<1];
int h[N],idx;
void Ins(rint a,rint b){
	e[++idx].to=b;e[idx].nxt=h[a];h[a]=idx;
}
int h2[N],idx2;
void Ins2(rint a,rint b){
	e2[++idx2].to=b;e2[idx2].nxt=h2[a];h2[a]=idx2;
}
int stk[N],tp,low[N],dfn[N],Time,fang;
void tarjan(rint u,rint fa){
	stk[++tp]=u;
	low[u]=dfn[u]=++Time;
	for(rint i=h[u];i;i=e[i].nxt){
		rint v=e[i].to;
		if(v==fa)continue;
		if(!dfn[v]){
			tarjan(v,u);
			low[u]=min(low[u],low[v]);
			if(low[v]>=dfn[u]){
				++fang;
				while(1){
					rint x=stk[tp--];
					Ins2(x,fang);Ins2(fang,x);
					if(x==v)break;
				}
				Ins2(fang,u);Ins2(u,fang);
			}
		}else low[u]=min(low[u],dfn[v]);
	}
}
int top[N],fa[N],son[N],siz[N],d[N],dis[N];
void dfs1(rint u){
	siz[u]=1;
	for(rint i=h2[u];i;i=e2[i].nxt){
		rint v=e2[i].to;
		if(v==fa[u])continue;
		fa[v]=u;
		d[v]=d[u]+1;
		dis[v]+=dis[u];
		dfs1(v);
		siz[u]+=siz[v];
		if(siz[v]>siz[son[u]])son[u]=v;
	}
}
void dfs2(rint u,rint t){
	top[u]=t;
	if(son[u])dfs2(son[u],t);
	for(rint i=h2[u];i;i=e2[i].nxt){
		rint v=e2[i].to;
		if(v==fa[u]||v==son[u])continue;
		dfs2(v,v);
	}
}
int lca(rint x,rint y){
	while(top[x]^top[y]){
		if(d[top[x]]>d[top[y]])x=fa[top[x]];
		else y=fa[top[y]];
	}
	return d[x]<d[y]?x:y;
}
int main(){
	rint n,m;
	scanf("%d%d",&n,&m);fang=n;
	for(rint i=1;i<=m;i++){
		rint a,b;
		scanf("%d%d",&a,&b);
		Ins(a,b);Ins(b,a);
	}
	tarjan(1,0);
	for(rint i=1;i<=n;i++)dis[i]=1;
	dfs1(1);
	dfs2(1,1);
	rint q;
	scanf("%d",&q);
	while(q--){
		rint a,b;
		scanf("%d%d",&a,&b);
		rint lc=lca(a,b);
		printf("%d\n",dis[a]+dis[b]-dis[lc]-dis[fa[lc]]);
	}
	return 0;
}

长链剖分

#include<bits/stdc++.h>
using namespace std;
const int N=1e6+10;
#define rint register int
#define rll register long long
struct Edge{
	int to,nxt;
}e[N<<1];
int h[N],idx;
void Ins(rint a,rint b){
	e[++idx].to=b;e[idx].nxt=h[a];h[a]=idx;
}
int buf[N<<2],*p,*f[N];
int len[N],son[N];
void dfs1(rint u,rint fa){
	len[u]=1;
	for(rint i=h[u];i;i=e[i].nxt){
		rint v=e[i].to;
		if(v==fa)continue;
		dfs1(v,u);
		len[u]=max(len[u],len[v]+1);
		if(len[v]>len[son[u]])son[u]=v;
	}
}
int ans[N];
void dfs2(rint u,rint fa){
	if(son[u]){
		f[son[u]]=f[u]+1;
		dfs2(son[u],u);
	}else {
		f[u][0]=1;
		return;
	}
	f[u][0]=1;
	if(f[son[u]][ans[son[u]]]==1)ans[u]=0;
	else ans[u]=ans[son[u]]+1;
	for(rint i=h[u];i;i=e[i].nxt){
		rint v=e[i].to;
		if(v==fa||v==son[u])continue;
		f[v]=p;p+=len[v];
		dfs2(v,u);
		for(rint j=0;j<len[v];j++){
			f[u][j+1]+=f[v][j];
			if(f[u][j+1]>f[u][ans[u]]||(f[u][j+1]==f[u][ans[u]]&&j+1<ans[u]))ans[u]=j+1;
		}
	}
}
int main(){
	rint n;
	scanf("%d",&n);
	for(rint i=2;i<=n;i++){
		rint a,b;
		scanf("%d%d",&a,&b);
		Ins(a,b);Ins(b,a);
	}
	dfs1(1,0);p=buf;
	f[1]=p;p+=len[1];
	dfs2(1,0);
	for(rint i=1;i<=n;i++)
		printf("%d\n",ans[i]);
	return 0;
}

KD-Tree

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=(5e5+10)*32;
const double alh=0.75;
int typ;
struct Node{
	int x[2];
	bool operator < (const Node&A)const{
		return x[typ]<A.x[typ];
	}
}a[N];
struct KdT{
	int ls,rs,siz,Max[2],Min[2];
	Node v;
}T[N];
#define ls T[rt].ls
#define rs T[rt].rs
int cnt,krt[N],lc[N],rc[N];
void up(int rt){
	for(int i=0;i<2;i++){
		T[rt].Max[i]=T[rt].Min[i]=T[rt].v.x[i];
		if(ls){
			T[rt].Max[i]=max(T[rt].Max[i],T[ls].Max[i]);
			T[rt].Min[i]=min(T[rt].Min[i],T[ls].Min[i]);
		}
		if(rs){
			T[rt].Max[i]=max(T[rt].Max[i],T[rs].Max[i]);
			T[rt].Min[i]=min(T[rt].Min[i],T[rs].Min[i]);
		}
	}
	T[rt].siz=T[ls].siz+T[rs].siz+1;
}
int stk[N],tt;
void get(int rt){
	stk[++tt]=rt;
	a[tt]=T[rt].v;
	if(ls)get(ls);
	if(rs)get(rs);
}
void Build(int &rt,int l,int r,int d){
	rt=0;
	if(l>r)return;
	rt=stk[tt--];
	int mid=l+r>>1;typ=d;
	nth_element(a+l,a+mid,a+r+1);
	T[rt].v=a[mid];
	Build(ls,l,mid-1,d^1);
	Build(rs,mid+1,r,d^1);
	up(rt);
}
void insert(int &rt,Node v,int d){
	if(!rt){
		rt=++cnt;
		T[rt].v=v;
		up(rt);
		return;
	}
	if(max(T[ls].siz,T[rs].siz)>=T[rt].siz*alh){
		get(rt);
		Build(rt,1,tt,d);
	}
	if(v.x[d]<=T[rt].v.x[d])insert(ls,v,d^1);
	else insert(rs,v,d^1);
	up(rt);
}
int dfs(int rt,Node v1,Node v2){
	if(!rt)return 0;
	if(T[rt].Max[0]<v1.x[0])return 0;
	if(T[rt].Max[1]<v1.x[1])return 0;
	if(T[rt].Min[0]>v2.x[0])return 0;
	if(T[rt].Min[1]>v2.x[1])return 0;
	if(v1.x[0]<=T[rt].Min[0]&&T[rt].Max[0]<=v2.x[0]
	&& v1.x[1]<=T[rt].Min[1]&&T[rt].Max[1]<=v2.x[1])
		return T[rt].siz;
	int res=0;
	if(v1.x[0]<=T[rt].v.x[0]&&T[rt].v.x[0]<=v2.x[0]
	&& v1.x[1]<=T[rt].v.x[1]&&T[rt].v.x[1]<=v2.x[1])
		res++;
	if(ls)res+=dfs(ls,v1,v2);
	if(rs)res+=dfs(rs,v1,v2);
	return res;
}
int tot;
void update(int &rt,int l,int r,int pos,Node v){
	if(!rt)rt=++tot;
	insert(krt[rt],v,1);
	if(l==r)return;
	int mid=l+r>>1;
	if(pos<=mid)update(lc[rt],l,mid,pos,v);
	else update(rc[rt],mid+1,r,pos,v);
}
int query(int rt,int l,int r,int k,Node v1,Node v2){
	if(l==r)return l;
	int rval=dfs(krt[rc[rt]],v1,v2);
	int mid=l+r>>1;
	if(k<=rval)return query(rc[rt],mid+1,r,k,v1,v2);
	else return query(lc[rt],l,mid,k-rval,v1,v2);
}
int main(){
	int n,q,rt=0,res=0;
	scanf("%d%d",&n,&q);
	while(q--){
		int opt;
		scanf("%d",&opt);
		if(opt==1){
			Node v;int val;
			scanf("%d%d%d",&v.x[0],&v.x[1],&val);
			v.x[0]^=res;v.x[1]^=res;val^=res;
			update(rt,1,1e9,val,v);
		}else{
			Node v1,v2;int k;
			scanf("%d%d%d%d%d",&v1.x[0],&v1.x[1],&v2.x[0],&v2.x[1],&k);
			v1.x[0]^=res;v1.x[1]^=res;v2.x[0]^=res;v2.x[1]^=res;k^=res;
			res=0;
			if(dfs(krt[rt],v1,v2)<k)puts("NAIVE!ORZzyz.");
			else printf("%d\n",res=query(rt,1,1e9,k,v1,v2));
		}
	}
	return 0;
}

LCT

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+10;
const int p=51061;
typedef long long ll;
int ch[N][2],fa[N],tag[N],siz[N];
ll sum[N],taga[N],tagm[N],val[N];
void up(int x){
	sum[x]=(sum[ch[x][0]]+sum[ch[x][1]]+val[x])%p;
	siz[x]=siz[ch[x][0]]+siz[ch[x][1]]+1;
}
void upd1(int x,ll v){
	if(!x)return;
	val[x]=val[x]*v%p;
	sum[x]=sum[x]*v%p;
	taga[x]=taga[x]*v%p;
	tagm[x]=tagm[x]*v%p;
}
void upd2(int x,ll v){
	if(!x)return;
	taga[x]=(taga[x]+v)%p;
	val[x]=(val[x]+v)%p;
	sum[x]=(sum[x]+v*siz[x]%p)%p;
}
void down(int x){
	if(tag[x]){
		tag[ch[x][0]]^=1;tag[ch[x][1]]^=1;
		swap(ch[x][0],ch[x][1]);
		tag[x]=0;
	}
	if(tagm[x]!=1){
		upd1(ch[x][0],tagm[x]);
		upd1(ch[x][1],tagm[x]);
		tagm[x]=1;
	}
	if(taga[x]){
		upd2(ch[x][0],taga[x]);
		upd2(ch[x][1],taga[x]);
		taga[x]=0;
	}
}
bool nrt(int x){return ch[fa[x]][1]==x||ch[fa[x]][0]==x;}
void rotate(int x){
	int y=fa[x];int z=fa[y];
	int k=ch[y][1]==x;
	if(nrt(y))ch[z][ch[z][1]==y]=x;
	fa[x]=z;ch[y][k]=ch[x][!k];
	fa[ch[x][!k]]=y;ch[x][!k]=y;
	fa[y]=x;up(y);up(x);
}
void update(int x){
	if(nrt(x))update(fa[x]);
	down(x);
}
void splay(int x){
	update(x);
	for(int y=fa[x],z=fa[y];nrt(x);rotate(x),y=fa[x],z=fa[y])
		if(nrt(y))(ch[y][1]==x)==(ch[z][1]==y)?rotate(y):rotate(x);
}
void access(int x){
	for(int t=0;x;t=x,x=fa[x])
		splay(x),ch[x][1]=t,up(x);
}
void mkrt(int x){
	access(x);
	splay(x);
	tag[x]^=1;
}
void split(int x,int y){
	mkrt(x);
	access(y);
	splay(y);
}
void link(int x,int y){
	mkrt(x);
	fa[x]=y;
}
void cut(int x,int y){
	split(x,y);
	if(ch[y][0]==x&&ch[x][1]==0)
		ch[y][0]=0,fa[x]=0,up(y);
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		siz[i]=val[i]=tagm[i]=1;
	for(int i=1;i<n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		link(x,y);
	}
	while(m--){
		char str[10];int x,y;
		scanf("%s%d%d",str,&x,&y);
		if(str[0]=='+'){
			int v;
			scanf("%d",&v);
			split(x,y);
			upd2(y,v);
		}else if(str[0]=='-'){
			cut(x,y);
			scanf("%d%d",&x,&y);
			link(x,y);
		}else if(str[0]=='*'){
			int v;
			scanf("%d",&v);
			split(x,y);
			upd1(y,v);
		}else {
			split(x,y);
			printf("%lld\n",sum[y]);
		}
	}
	return 0;
}

左偏树（可并堆）

#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int dis[N],val[N],lc[N],rc[N],f[N];
int find(int x){
	return x==f[x]?x:(f[x]=find(f[x]));
}
int merge(int lt,int rt){
	if(!lt||!rt)return lt+rt;
	if(val[lt]>val[rt]||(val[lt]==val[rt]&&lt>rt))swap(lt,rt);
	rc[lt]=merge(rc[lt],rt);
	if(dis[lc[lt]]<dis[rc[lt]])swap(lc[lt],rc[lt]);
	dis[lt]=dis[rc[lt]]+1;
	f[lc[lt]]=f[rc[lt]]=f[lt]=lt;
	return lt;
}
void pop(int x){
	x=find(x);
	printf("%d\n",val[x]);
	f[lc[x]]=lc[x];
	f[rc[x]]=rc[x];
	f[x]=merge(lc[x],rc[x]);
	val[x]=-1;
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
		scanf("%d",&val[i]),f[i]=i;
	for(int i=1;i<=m;i++){
		int op,x;
		scanf("%d%d",&op,&x);
		if(op==1){
			int y;
			scanf("%d",&y);
			if(find(x)==find(y)||val[x]==-1||val[y]==-1)continue;
			f[find(x)]=f[find(y)]=merge(find(x),find(y));
		}else {
			if(val[x]==-1)printf("-1\n");
			else pop(x);
		}
	}
}

猫树

const int MAXN=(1<<19)+7;

static int n,m,a[MAXN],len,Log2[MAXN],pos[MAXN];

namespace Meow_Tree
{
	static int p[21][MAXN],s[21][MAXN];//区间最大子段和，区间最大前缀和
	void build_tree(int h,int l,int r,int dps)
	{
		if(l==r){pos[l]=h;return;}
		int mid=(l+r)>>1,prep,sm;
		p[dps][mid]=s[dps][mid]=sm=a[mid];
		prep=a[mid];
		if(sm<0)sm=0;
		Repe(i,mid-1,l)
		{
			prep+=a[i];sm+=a[i];
			s[dps][i]=max(s[dps][i+1],prep);
			p[dps][i]=max(p[dps][i+1],sm);
			if(sm<0)sm=0;
		}

		p[dps][mid+1]=s[dps][mid+1]=sm=a[mid+1];
		prep=a[mid+1];
		if(sm<0)sm=0;
		Rep(i,mid+2,r)
		{
			prep+=a[i];sm+=a[i];
			s[dps][i]=max(s[dps][i-1],prep);
			p[dps][i]=max(p[dps][i-1],sm);
			if(sm<0)sm=0;
		}
		build_tree(h<<1,l,mid,dps+1);
		build_tree(h<<1|1,mid+1,r,dps+1);
	}

	inline int query(int l,int r)
	{
		if(l==r)return a[l];
		static int dps;dps=Log2[pos[l]]-Log2[pos[l]^pos[r]];
		return max(max(p[dps][l],p[dps][r]),s[dps][l]+s[dps][r]);
	}
}

using namespace Meow_Tree;

inline void init()
{
	read(n);
	Rep(i,1,n)read(a[i]);
	len=2;while(len<n)len<<=1;
	Rep(i,2,len<<1)Log2[i]=Log2[i>>1]+1;
	build_tree(1,1,len,1);
}

支配树

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
namespace dominator{

const int N=2e5+10,M=3e5+10;
struct Edge{
    int to,nxt;
}e[3][M];
int h[3][N],idx[3];
// 0 原图
// 1 反图
void Ins(rint t,rint x,rint y){
    e[t][++idx[t]].to=y;e[t][idx[t]].nxt=h[t][x];h[t][x]=idx[t];
}
int dfn[N],rk[N],Time,fa[N],f[N],mn[N],sdom[N],idom[N],ans[N];
// sdom 半支配点
// idom 支配点
void dfs(int u){
	f[u]=mn[u]=sdom[u]=u;
	dfn[u]=++Time,rk[Time]=u;
	for(rint i=h[0][u];i;i=e[0][i].nxt){
        rint v=e[0][i].to;
        if(!dfn[v])
            fa[v]=u,dfs(v);
    }
}
int find(rint u){
	if(u==f[u])return u;
	int ret=find(f[u]);
	if(dfn[sdom[mn[f[u]]]]<dfn[sdom[mn[u]]])
        mn[u]=mn[f[u]];
	return f[u]=ret;
}
void work(rint S){
	dfs(S);
	for(rint i=Time;i>1;i--){
		rint u=rk[i];
		for(rint j=h[1][u];j;j=e[1][j].nxt){
            rint v=e[1][j].to;
			if(!dfn[v])continue;
            // 有可能S走不到v
			find(v);
			if(dfn[sdom[mn[v]]]<dfn[sdom[u]])
                sdom[u]=sdom[mn[v]];
		}
		f[u]=fa[u];
		Ins(2,sdom[u],u);
		u=fa[u];
		for(rint j=h[2][u];j;j=e[2][j].nxt){
            rint v=e[2][j].to;
			find(v);
			idom[v]=(u==sdom[mn[v]]?u:mn[v]);
		}
		h[2][u]=0;
	}
	for(int i=2;i<=Time;i++){
		int u=rk[i];
		if(idom[u]!=sdom[u])idom[u]=idom[idom[u]];
	}
	for(int i=Time;i;i--){
		int u=rk[i];ans[u]++;
		ans[idom[u]]+=ans[u];
	}
}

}
int main(){
    rint n,m;
    scanf("%d%d",&n,&m);
	for(rint i=1;i<=m;i++){
		rint x,y;
        scanf("%d%d",&x,&y);
		dominator::Ins(0,x,y);
        dominator::Ins(1,y,x);
	}
	dominator::work(1);
	for(int i=1;i<=n;i++)
        printf("%d ",dominator::ans[i]);
	return 0;
}

李超树

#include<map>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+10;
typedef double db;
#define rint register int
int tr[N<<2],cnt;
db k[N],b[N];
void New(rint x,rint y,rint xx,rint yy){
	++cnt;
	if(x==xx)k[cnt]=0,b[cnt]=max(y,yy);
	else k[cnt]=1.0*(yy-y)/(xx-x),b[cnt]=yy-k[cnt]*xx;
}
#define ls rt<<1
#define rs rt<<1|1
void update(rint rt,rint l,rint r,rint cl,rint cr,rint v){
	rint mid=l+r>>1;
	if(cl<=l&&r<=cr){
		rint id=tr[rt];
		if(!id){
			tr[rt]=v;
			return;
		}
		db v1=mid*k[id]+b[id],v2=mid*k[v]+b[v];
		if(l==r){
			if(v2>v1)tr[rt]=v;
		}else if(k[id]==k[v]){
			if(v2>v1)tr[rt]=v;
		}else if(k[id]<k[v]){
			if(v2>v1){
				tr[rt]=v;
				update(ls,l,mid,cl,cr,id);
			}else update(rs,mid+1,r,cl,cr,v);
		}else {
			if(v2>v1){
				tr[rt]=v;
				update(rs,mid+1,r,cl,cr,id);
			}else update(ls,l,mid,cl,cr,v);
		}
		return;
	}
	if(cl<=mid)update(ls,l,mid,cl,cr,v);
	if(cr>mid)update(rs,mid+1,r,cl,cr,v);
}
int res1;
db res2;
void query(rint rt,rint l,rint r,rint p){
	rint v=tr[rt];
	if(v){
		db now=k[v]*p+b[v];
		if(!res1)res1=v,res2=now;
		else if(now>res2)res1=v,res2=now;
		else if(now==res2&&res1>v)res1=v;
	}
	if(l==r)return;
	rint mid=l+r>>1;
	if(p<=mid)query(ls,l,mid,p);
	else query(rs,mid+1,r,p);
}
const int Max=39989;
int main(){
	rint n;
	scanf("%d",&n);
	rint lst=0;
	while(n--){
		rint opt;
		scanf("%d",&opt);
		if(opt==0){
			rint x;
			scanf("%d",&x);
			x=(x+lst-1)%Max+1;res1=0;
			query(1,1,Max,x);
			printf("%d\n",lst=res1);
		}else {
			rint x,y,xx,yy;
			scanf("%d%d%d%d",&x,&y,&xx,&yy);
			x=(x+lst-1)%Max+1,xx=(xx+lst-1)%Max+1;
			y=(y+lst-1)%1000000000+1;
			yy=(yy+lst-1)%1000000000+1;
			New(x,y,xx,yy);
			update(1,1,Max,min(x,xx),max(x,xx),cnt);
		}
	}
	return 0;
}

划分树

#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
typedef long long lld;

int getint()  // 本题没有负数输入，因此快读不需判断负号。
{
  char ch;
  while ((ch = getchar()) < '!')
    ;
  int x = ch ^ '0';
  while ((ch = getchar()) > '!') x = (x * 10) + (ch ^ '0');
  return x;
}

void putll(lld x) {  // 输出long long
  if (x / 10) putll(x / 10);
  putchar((x % 10) ^ '0');
}

int ti[2][17][131073];

int lowbit(int x) { return x & (-x); }

void fixup(int k, int x) {
  for (; lowbit(x) < 1 << k; x += lowbit(x)) {
    ++ti[0][k - 1][x];
  }
  ++ti[0][k - 1][x];
}

void fixdown(int k, int x) {
  for (; ((x ^ lowbit(x)) - 1) >> k == (x - 1) >> k; x ^= lowbit(x)) {
    ++ti[1][k - 1][x];
  }
  ++ti[1][k - 1][x];
}

int queryup(int k, int x) {
  int res = 0;
  for (; lowbit(x) < 1 << k; x += lowbit(x)) {
    res += ti[1][k - 1][x];
  }
  return res + ti[1][k - 1][x];
}

int querydown(int k, int x) {
  int res = 0;
  for (; ((x ^ lowbit(x)) - 1) >> k == (x - 1) >> k; x ^= lowbit(x)) {
    res += ti[0][k - 1][x];
  }
  return res + ti[0][k - 1][x];
}

int ai[100005];
int tx[100005];

lld mgx(int* npi, int* nai, int* rxi, int* lxi, int al, int ar, int bl,
        int br) {
  int rx = 1;
  int lx = ar - al;
  lld res = 0;
  for (; al != ar || bl != br; ++npi, ++nai, ++rxi, ++lxi) {
    if (al != ar && (bl == br || tx[al] < tx[bl])) {
      --lx;
      *rxi = rx;
      *nai = tx[al];
      *npi = al;
      ++al;
    } else {
      ++rx;
      *lxi = lx;
      res += ar - al;
      *nai = tx[bl];
      *npi = bl;
      ++bl;
    }
  }
  return res;
}

int npi[1700005];
int nri[1700005];
int nli[1700005];

int main() {
  const int n = getint();
  const int m = getint();
  for (int i = 1; i <= n; ++i) {
    ai[i] = getint();
  }

  if (n == 1) {
    for (int i = 1; i <= m; ++i) {
      putchar('0');
      putchar('\n');
    }
    return 0;
  }

  const int logn = 31 - __builtin_clz(n - 1);

  lld ans = 0;
  for (int i = logn; i >= 0; --i) {
    memcpy(tx, ai, (n + 1) * sizeof(int));
    for (int j = 1; j <= n; j += 1 << (logn - i + 1)) {
      ans += mgx(npi + n * i + j, ai + j, nri + n * i + j, nli + n * i + j, j,
                 min(n + 1, j + (1 << (logn - i))),
                 min(n + 1, j + (1 << (logn - i))),
                 min(n + 1, j + (1 << (logn - i + 1))));
    }
  }

  putll(ans);
  putchar('\n');

  for (int asdf = 1; asdf < m; ++asdf) {
    int x = getint();
    for (int i = 0, p = 0; i <= logn; ++i, p += n) {
      if (nri[p + x]) {
        ans -= nri[p + x] - querydown(logn - i + 1, x) - 1;
        fixdown(logn - i + 1, x);
      } else {
        ans -= nli[p + x] - queryup(logn - i + 1, x);
        fixup(logn - i + 1, x);
      }
      x = npi[p + x];
    }
    putll(ans);
    putchar('\n');
  }
}

笛卡尔树

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define rint register int
const int N=1e5+10;
inline int read(){
	rint x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
int stk[N],tp,a[N],ch[N][2];
int main(){
	rint n=read(),m=read();
	for(rint i=1;i<=n;i++){
		a[i]=read();
		while(tp&&a[stk[tp]]<=a[i])ch[i][0]=stk[tp--];
		ch[stk[tp]][1]=i;
		stk[++tp]=i;
	}
	rint rt=stk[1];
	while(m--){
		rint l=read(),r=read();
		for(rint x=rt;;x=x>r?ch[x][0]:ch[x][1])
			if(l<=x&&x<=r){
				printf("%d\n",a[x]);
				break;
			}
	}
}

分块/树分块

#include<bits/stdc++.h>
using namespace std;
namespace lin4xu{

#define re register
#define rint re int
#define ll long long
#define rll re ll
#define db double
#define rdb re db
#define rch re char
#define ldb long db
#define rldb re ldb
#define un unsigned
#define ull un ll
#define rull re ull

template <typename T> inline T read()
{
	re T ans=0;rint f=0;rch ch=getchar();
	for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=1;
	for(;isdigit(ch);ch=getchar()) ans=ans*10+(ch&15);
	return f?-ans:ans;
}

const int maxn=4e4+10;
const int S=1000;
const int nS=maxn/S+10;

struct edge
{
	int to,next;
}e[maxn<<1];
bitset<maxn> mapp[nS][nS];
bitset<maxn> b;
int head[maxn];
int sta[nS],rk[nS];
int fa[maxn];
int top[maxn];
int deep[maxn];
int id[maxn];
int a[maxn];
int lsh[maxn];
int up[maxn];
int size[maxn];
int son[maxn];
int cnt,s_cnt,sta_top;

inline void add(rint x,rint y)
{
	e[++cnt].to=y;
	e[cnt].next=head[x];
	head[x]=cnt;
}

inline int dfs(rint x)
{
	rint tmp=0;size[x]=1;
	for(rint i=head[x];i;i=e[i].next)
	{
		rint v=e[i].to;
		if(v==fa[x]) continue;
		deep[v]=deep[x]+1,fa[v]=x,tmp=max(tmp,dfs(v)),size[x]+=size[v];
		if(size[son[x]]<size[v]) son[x]=v;
	}
	if(++tmp==S) rk[id[x]=++s_cnt]=x,tmp=0;
	return tmp;
}

void dfs(rint x,rint t)
{
	top[x]=t;
	if(id[x])
	{
		up[x]=sta[sta_top];
		sta[++sta_top]=id[x];
		mapp[id[x]][id[x]][a[x]]=1;
		for(rint i=up[x],pre=id[x];i;pre=i,i=up[rk[i]])
		{
			mapp[id[x]][i]=mapp[id[x]][pre],mapp[id[x]][i][a[rk[i]]]=1;
			for(rint j=fa[rk[pre]];j!=rk[i];j=fa[j]) mapp[id[x]][i][a[j]]=1;
		}
	}
	if(son[x]) dfs(son[x],t);
	for(rint i=head[x];i;i=e[i].next)
	{
		rint v=e[i].to;
		if(v==fa[x]||v==son[x]) continue;
		dfs(v,v);
	}
	if(id[x]) sta_top--;
	else up[x]=sta[sta_top];
}

inline int query(rint x,rint y)
{
	while(top[x]!=top[y]) deep[top[x]]<deep[top[y]]?y=fa[top[y]]:x=fa[top[x]];
	return deep[x]<deep[y]?x:y;
}

int main()
{
	rint n=read<int>(),m=read<int>();
	for(rint i=1;i<=n;i++) a[i]=lsh[i]=read<int>();
	sort(lsh+1,lsh+n+1);rint N=unique(lsh+1,lsh+n+1)-lsh-1;
	for(rint i=1;i<=n;i++) a[i]=lower_bound(lsh+1,lsh+N+1,a[i])-lsh;
	for(rint i=1;i<n;i++)
	{
		rint x=read<int>(),y=read<int>();
		add(x,y),add(y,x);
	}
	deep[1]=1,dfs(1),dfs(1,1);
	rint lastans=0;
	for(rint tt=1;tt<=m;tt++)
	{
		rint x=read<int>()^lastans,y=read<int>(),lca=query(x,y);
		if(deep[rk[up[x]]]>=deep[lca])
		{
			rint fax=rk[up[x]];
			for(rint i=x;i!=fax;i=fa[i]) b[a[i]]=1;
			for(;deep[rk[up[fax]]]>=deep[lca];fax=rk[up[fax]]);
			b|=mapp[up[x]][id[fax]];
			if(fax!=lca)
			{
				for(rint i=fa[fax];i!=lca;i=fa[i]) b[a[i]]=1;
				b[a[lca]]=1;
			}
		}
		else
		{
			for(rint i=x;i!=lca;i=fa[i]) b[a[i]]=1;
			b[a[lca]]=1;
		}
		if(deep[rk[up[y]]]>deep[lca])
		{
			rint fay=rk[up[y]];
			for(rint i=y;i!=fay;i=fa[i]) b[a[i]]=1;
			for(;deep[rk[up[fay]]]>deep[lca];fay=rk[up[fay]]);
			b|=mapp[up[y]][id[fay]];
			for(rint i=fa[fay];i!=lca;i=fa[i]) b[a[i]]=1;
		}
		else for(rint i=y;i!=lca;i=fa[i]) b[a[i]]=1;
		printf("%d\n",lastans=b.count()),b.reset();
	}
	return 0;
}};

int main(){return lin4xu::main();}

克鲁斯卡尔重构树

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=1e6+10;
struct Edge{
	int to,nxt,val,val2;
	bool operator < (const Edge&A)const{
		return val2>A.val2;
	}
}e[N<<1],e2[N];
int h[N],idx;
void Ins(int a,int b,int c){
	e[++idx].to=b;e[idx].nxt=h[a];
	h[a]=idx;e[idx].val=c;
}
int cnt,lst[N],f[N];
int find(int x){
	return x==f[x]?x:(f[x]=find(f[x]));
}
struct Node{
	int id;ll val;
	bool operator < (const Node&A)const{
		return val>A.val;
	}
};
priority_queue<Node> q;
bool vis[N];
ll dis[N];
void dij(){
	memset(vis,0,sizeof(vis));
	memset(dis,0x3f,sizeof(dis));
	q.push((Node){1,0});dis[1]=0;
	while(!q.empty()){
		Node u=q.top();q.pop();
		if(vis[u.id])continue;
		vis[u.id]=1;
		for(int i=h[u.id];i;i=e[i].nxt){
			int v=e[i].to;
			if(dis[v]>dis[u.id]+e[i].val){
				dis[v]=dis[u.id]+e[i].val;
				q.push((Node){v,dis[v]});
			}
		}
	}
}
int dfn[N],Time,w[N],siz[N],p[N][25];
ll rk[N];
void dfs(int u){
	dfn[u]=++Time;
	rk[Time]=dis[u];
	siz[u]=1;
	for(int i=0;p[u][i];i++)
		p[u][i+1]=p[p[u][i]][i];
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		p[v][0]=u;
		dfs(v);
		siz[u]+=siz[v];
	}
}
struct Tree{
	int l,r;
	ll Min;
}tr[N<<2];
#define ls rt<<1
#define rs rt<<1|1
void Build(int rt,int l,int r){
	tr[rt].l=l;tr[rt].r=r;
	if(l==r){
		tr[rt].Min=rk[l];
		return;
	}
	int mid=l+r>>1;
	Build(ls,l,mid);
	Build(rs,mid+1,r);
	tr[rt].Min=min(tr[ls].Min,tr[rs].Min);
}
ll query(int rt,int l,int r){
	if(l<=tr[rt].l&&tr[rt].r<=r)return tr[rt].Min;
	int mid=tr[rt].l+tr[rt].r>>1;
	ll res=dis[0];
	if(l<=mid)res=min(res,query(ls,l,r));
	if(r>mid)res=min(res,query(rs,l,r));
	return res;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		memset(h,0,sizeof(h));
		idx=0;
		int n,m;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=m;i++){
			scanf("%d%d%d%d",&e2[i].nxt,&e2[i].to,&e2[i].val,&e2[i].val2);
			Ins(e2[i].nxt,e2[i].to,e2[i].val);
			Ins(e2[i].to,e2[i].nxt,e2[i].val);
		}
		dij();
		memset(h,0,sizeof(h));cnt=idx=0;
		for(int i=1;i<=n;i++){
			f[i]=i;
			w[i]=0;
			lst[i]=++cnt;
		}
		sort(e2+1,e2+m+1);
		for(int i=1;i<=m;i++){
			int x=find(e2[i].nxt),y=find(e2[i].to);
			if(x!=y){
				++cnt;w[cnt]=e2[i].val2;
				Ins(cnt,lst[x],0);
				Ins(cnt,lst[y],0);
				f[x]=y;dis[cnt]=dis[0];
				lst[y]=cnt;
			}
		}
		memset(p,0,sizeof(p));
		Time=0;
		dfs(cnt);
		Build(1,1,cnt);
		int ask,k,s;
		scanf("%d%d%d",&ask,&k,&s);
		ll res=0;
		while(ask--){
			int v,wv;
			scanf("%d%d",&v,&wv);
			v=(v+k*res-1)%n+1;
			wv=(wv+k*res)%(s+1);
			for(int i=23;~i;i--)
				if(p[v][i]&&w[p[v][i]]>wv)v=p[v][i];
			printf("%lld\n",res=query(1,dfn[v],dfn[v]+siz[v]-1));
		}
	}
	return 0;
}

扫描线

#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=4e5+10;
typedef long long ll;
struct Node{
	int x,y,xx,yy;
}a[N];
struct Add{
	int l,r,v;
	Add(){}
	Add(int a,int b,int c){
		l=a;r=b;v=c;
	}
};
vector<Add> vec[N];
int rk[N],s[N],tmp[N],tot;
int len[N<<2],cnt[N<<2];
void update(int rt,int l,int r,int cl,int cr,int v){
	if(cl<=l&&r<=cr){
		cnt[rt]+=v;
		if(cnt[rt])len[rt]=rk[r+1]-rk[l];
		else if(l!=r)len[rt]=len[rt<<1]+len[rt<<1|1];
		else len[rt]=0;
		return;
	}
	int mid=l+r>>1;
	if(cl<=mid)update(rt<<1,l,mid,cl,cr,v);
	if(cr>mid)update(rt<<1|1,mid+1,r,cl,cr,v);
	if(cnt[rt])len[rt]=rk[r+1]-rk[l];
	else len[rt]=len[rt<<1]+len[rt<<1|1];
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d%d%d%d",&a[i].x,&a[i].y,&a[i].xx,&a[i].yy);
		tmp[++tot]=a[i].y;tmp[++tot]=a[i].yy;
	}
	sort(tmp+1,tmp+tot+1);
	tot=unique(tmp+1,tmp+tot+1)-tmp-1;
	for(int i=1;i<=n;i++){
		int x=lower_bound(tmp+1,tmp+tot+1,a[i].y)-tmp;
		rk[x]=a[i].y;a[i].y=x;
		x=lower_bound(tmp+1,tmp+tot+1,a[i].yy)-tmp;
		rk[x]=a[i].yy;a[i].yy=x;
	}
	int maxy=tot;tot=0;
	for(int i=1;i<=n;i++)tmp[++tot]=a[i].x,tmp[++tot]=a[i].xx;
	sort(tmp+1,tmp+tot+1);
	tot=unique(tmp+1,tmp+tot+1)-tmp-1;
	for(int i=1;i<=n;i++){
		int x=lower_bound(tmp+1,tmp+tot+1,a[i].x)-tmp;
		s[x]=a[i].x;
		vec[x].push_back(Add(a[i].y,a[i].yy-1,1));
		x=lower_bound(tmp+1,tmp+tot+1,a[i].xx)-tmp;
		s[x]=a[i].xx;
		vec[x].push_back(Add(a[i].y,a[i].yy-1,-1));
	}
	ll res=0;
	for(int i=1;i<tot;i++){
		for(int j=0;j<vec[i].size();j++)
			update(1,1,maxy,vec[i][j].l,vec[i][j].r,vec[i][j].v);
		res+=1ll*(s[i+1]-s[i])*len[1];
	}
	printf("%lld\n",res);
	return 0;
}

欧拉游览树 ETT

/*
虽然上文提到过块状链表实现 ETT
在某些情况下可能较简单，但对于此题块状链表复杂度有可能无法通过而且实现较繁琐，所以这份代码采用
FHQ Treap 实现。
*/
#include <bits/stdc++.h>
#define N 1000000
#define int long long
using namespace std;
/*FHQ TREAP*/
int rt, tot, f[N], rnd[N], ls[N], rs[N], siz[N], tag[N], val[N], sum[N], pd[N],
    pds[N];

void pushup(int x) {
  siz[x] = siz[ls[x]] + siz[rs[x]] + 1;
  sum[x] = sum[ls[x]] + sum[rs[x]] + val[x];
  pds[x] = pds[ls[x]] + pds[rs[x]] + pd[x];
}

void link(int x, int c, int y) {
  if (c)
    rs[x] = y;
  else
    ls[x] = y;
  if (y) f[y] = x;
  pushup(x);
}

int newNode(int x, int y) {
  siz[++tot] = 1;
  val[tot] = sum[tot] = x;
  pd[tot] = pds[tot] = y;
  rnd[tot] = rand();
  return tot;
}

void setTag(int x, int v) {
  tag[x] += v;
  sum[x] += v * pds[x];
  val[x] += v * pd[x];
}

void pushdown(int x) {
  if (ls[x]) setTag(ls[x], tag[x]);
  if (rs[x]) setTag(rs[x], tag[x]);
  tag[x] = 0;
}

void split(int now, int k, int &x, int &y) {
  f[now] = 0;
  if (!now) {
    x = y = 0;
    return;
  }
  pushdown(now);
  if (siz[ls[now]] + 1 <= k) {
    x = now;
    split(rs[now], k - siz[ls[now]] - 1, rs[x], y);
    link(x, 1, rs[x]);
  } else {
    y = now;
    split(ls[now], k, x, ls[y]);
    link(y, 0, ls[y]);
  }
}

int merge(int x, int y) {
  if (!x || !y) return x | y;
  if (rnd[x] < rnd[y]) {
    pushdown(x);
    link(x, 1, merge(rs[x], y));
    return x;
  } else {
    pushdown(y);
    link(y, 0, merge(x, ls[y]));
    return y;
  }
}

int rnk(int x) {
  int c = 1, ans = 0;
  while (x) {
    if (c) ans += siz[ls[x]] + 1;
    c = (rs[f[x]] == x);
    x = f[x];
  }
  return ans;
}

/*ETT*/
int s[N], e[N];

void add(int x, int v) {
  int a, b, c;
  split(rt, rnk(s[x]) - 1, a, b);
  split(b, rnk(e[x]) - rnk(s[x]) + 1, b,
        c);  // 这里 b 是我们要进行操作的子树的括号序列。
  setTag(b, v);
  rt = merge(merge(a, b), c);
}

int query(int x) {
  int a, b;
  split(rt, rnk(s[x]), a, b);
  int ans = sum[a];
  rt = merge(a, b);
  return ans;
}

void changeFa(int x, int y) {
  int a, b, c, d;
  split(rt, rnk(s[x]) - 1, a, b);
  split(b, rnk(e[x]) - rnk(s[x]) + 1, b, c);
  a = merge(
      a,
      c);  // 因为我们确定不了要设置为父亲的节点在括号序列中的哪边，所以先把两边合并。
  split(a, rnk(s[y]), a, d);
  rt = merge(merge(a, b), d);  // 把要进行操作的子树放在父亲括号序列的最前面。
}

/*main function*/
int n, m, w[N];
vector<int> v[N];

void dfs(int x) {
  rt = merge(rt, s[x] = newNode(w[x], 1));
  for (auto to : v[x]) dfs(to);
  rt = merge(rt, e[x] = newNode(-w[x], -1));
}

signed main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int f;
    cin >> f;
    v[f].push_back(i);
  }
  for (int i = 1; i <= n; i++) cin >> w[i];
  dfs(1);
  cin >> m;
  for (int i = 1; i <= m; i++) {
    char c;
    cin >> c;
    if (c == 'Q') {
      int d;
      cin >> d;
      cout << query(d) << endl;
    } else if (c == 'C') {
      int x, y;
      cin >> x >> y;
      changeFa(x, y);
    } else {
      int p, q;
      cin >> p >> q;
      add(p, q);
    }
  }
  return 0;
}

Top Tree

Sqrt Tree

SqrtTreeItem op(const SqrtTreeItem &a, const SqrtTreeItem &b);

inline int log2Up(int n) {
  int res = 0;
  while ((1 << res) < n) {
    res++;
  }
  return res;
}

class SqrtTree {
 private:
  int n, lg, indexSz;
  vector<SqrtTreeItem> v;
  vector<int> clz, layers, onLayer;
  vector<vector<SqrtTreeItem> > pref, suf, between;

  inline void buildBlock(int layer, int l, int r) {
    pref[layer][l] = v[l];
    for (int i = l + 1; i < r; i++) {
      pref[layer][i] = op(pref[layer][i - 1], v[i]);
    }
    suf[layer][r - 1] = v[r - 1];
    for (int i = r - 2; i >= l; i--) {
      suf[layer][i] = op(v[i], suf[layer][i + 1]);
    }
  }

  inline void buildBetween(int layer, int lBound, int rBound, int betweenOffs) {
    int bSzLog = (layers[layer] + 1) >> 1;
    int bCntLog = layers[layer] >> 1;
    int bSz = 1 << bSzLog;
    int bCnt = (rBound - lBound + bSz - 1) >> bSzLog;
    for (int i = 0; i < bCnt; i++) {
      SqrtTreeItem ans;
      for (int j = i; j < bCnt; j++) {
        SqrtTreeItem add = suf[layer][lBound + (j << bSzLog)];
        ans = (i == j) ? add : op(ans, add);
        between[layer - 1][betweenOffs + lBound + (i << bCntLog) + j] = ans;
      }
    }
  }

  inline void buildBetweenZero() {
    int bSzLog = (lg + 1) >> 1;
    for (int i = 0; i < indexSz; i++) {
      v[n + i] = suf[0][i << bSzLog];
    }
    build(1, n, n + indexSz, (1 << lg) - n);
  }

  inline void updateBetweenZero(int bid) {
    int bSzLog = (lg + 1) >> 1;
    v[n + bid] = suf[0][bid << bSzLog];
    update(1, n, n + indexSz, (1 << lg) - n, n + bid);
  }

  void build(int layer, int lBound, int rBound, int betweenOffs) {
    if (layer >= (int)layers.size()) {
      return;
    }
    int bSz = 1 << ((layers[layer] + 1) >> 1);
    for (int l = lBound; l < rBound; l += bSz) {
      int r = min(l + bSz, rBound);
      buildBlock(layer, l, r);
      build(layer + 1, l, r, betweenOffs);
    }
    if (layer == 0) {
      buildBetweenZero();
    } else {
      buildBetween(layer, lBound, rBound, betweenOffs);
    }
  }

  void update(int layer, int lBound, int rBound, int betweenOffs, int x) {
    if (layer >= (int)layers.size()) {
      return;
    }
    int bSzLog = (layers[layer] + 1) >> 1;
    int bSz = 1 << bSzLog;
    int blockIdx = (x - lBound) >> bSzLog;
    int l = lBound + (blockIdx << bSzLog);
    int r = min(l + bSz, rBound);
    buildBlock(layer, l, r);
    if (layer == 0) {
      updateBetweenZero(blockIdx);
    } else {
      buildBetween(layer, lBound, rBound, betweenOffs);
    }
    update(layer + 1, l, r, betweenOffs, x);
  }

  inline SqrtTreeItem query(int l, int r, int betweenOffs, int base) {
    if (l == r) {
      return v[l];
    }
    if (l + 1 == r) {
      return op(v[l], v[r]);
    }
    int layer = onLayer[clz[(l - base) ^ (r - base)]];
    int bSzLog = (layers[layer] + 1) >> 1;
    int bCntLog = layers[layer] >> 1;
    int lBound = (((l - base) >> layers[layer]) << layers[layer]) + base;
    int lBlock = ((l - lBound) >> bSzLog) + 1;
    int rBlock = ((r - lBound) >> bSzLog) - 1;
    SqrtTreeItem ans = suf[layer][l];
    if (lBlock <= rBlock) {
      SqrtTreeItem add =
          (layer == 0) ? (query(n + lBlock, n + rBlock, (1 << lg) - n, n))
                       : (between[layer - 1][betweenOffs + lBound +
                                             (lBlock << bCntLog) + rBlock]);
      ans = op(ans, add);
    }
    ans = op(ans, pref[layer][r]);
    return ans;
  }

 public:
  inline SqrtTreeItem query(int l, int r) { return query(l, r, 0, 0); }

  inline void update(int x, const SqrtTreeItem &item) {
    v[x] = item;
    update(0, 0, n, 0, x);
  }

  SqrtTree(const vector<SqrtTreeItem> &a)
      : n((int)a.size()), lg(log2Up(n)), v(a), clz(1 << lg), onLayer(lg + 1) {
    clz[0] = 0;
    for (int i = 1; i < (int)clz.size(); i++) {
      clz[i] = clz[i >> 1] + 1;
    }
    int tlg = lg;
    while (tlg > 1) {
      onLayer[tlg] = (int)layers.size();
      layers.push_back(tlg);
      tlg = (tlg + 1) >> 1;
    }
    for (int i = lg - 1; i >= 0; i--) {
      onLayer[i] = max(onLayer[i], onLayer[i + 1]);
    }
    int betweenLayers = max(0, (int)layers.size() - 1);
    int bSzLog = (lg + 1) >> 1;
    int bSz = 1 << bSzLog;
    indexSz = (n + bSz - 1) >> bSzLog;
    v.resize(n + indexSz);
    pref.assign(layers.size(), vector<SqrtTreeItem>(n + indexSz));
    suf.assign(layers.size(), vector<SqrtTreeItem>(n + indexSz));
    between.assign(betweenLayers, vector<SqrtTreeItem>((1 << lg) + bSz));
    build(0, 0, n, 0);
  }
};

O(n)-O(1) ST表

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int inf=1e9;
int n,m,w[2000000],lg2[1500000];
int pos[1500000],lst[1500000],dep[1500000],id,mm,S,ch[1500000][2],stack[1500000];
struct LIM_RMQ
{
	int w[1500000],bl[1500000],blo,L[150000],R[150000],pos[10000],val[150000],minn[150000],minpos[150000],t[1000],n_st;
	struct ST_node{int f,id;bool operator <(const ST_node &tmp)const{return f<tmp.f;}};
	struct STable
	{
		ST_node a[4][12];//n=1e6的话logn/2只有9...所以放心开
		void make(int w[],int n)
		{
			for(int i=1;i<=n;i++)a[0][i]=(ST_node){w[i],i};
			for(int i=1;(1<<i)<=n;i++)//st表，不过要同时处理出最小值所在的位置
				for(int j=1;j+(1<<i)-1<=n;j++)
					a[i][j]=min(a[i-1][j],a[i-1][j+(1<<(i-1))]);
		}
		ST_node query(int l,int r)
		{
			int len=lg2[r-l+1];
			return min(a[len][l],a[len][r-(1<<len)+1]);
		}
	}st[10000];//这里只有sqrt(n)个，也不用开这么大
	struct STable_block
	{
		ST_node a[20][150000];//这个占空间最大了吧...不过也只是O(n)的
		void make(int w[],int n)
		{
			for(int i=1;i<=n;i++)a[0][i]=(ST_node){w[i],i};
			for(int i=1;(1<<i)<=n;i++)
				for(int j=1;j+(1<<i)-1<=n;j++)
					a[i][j]=min(a[i-1][j],a[i-1][j+(1<<(i-1))]);
		}
		ST_node query(int l,int r)
		{
			int len=lg2[r-l+1];//细节，math库的log2函数不能看做O(1)的，要提前处理
			return min(a[len][l],a[len][r-(1<<len)+1]);
		}
	}st_block;
	void make(int a[],int n)
	{
		for(int i=1;i<=n;i++)w[i]=a[i];
		lg2[1]=0;for(int i=2;i<=n;i++)lg2[i]=lg2[i>>1]+1;//处理log2
		blo=max(lg2[n]>>1,1);//分块
		for(int i=1;i<=n;i++)bl[i]=(i-1)/blo+1;
		for(int i=1;i<=bl[n];i++)L[i]=(i-1)*blo+1,R[i]=min(i*blo,n),minn[i]=inf;
		w[0]=w[1]-1;
		for(int i=1;i<=bl[n];i++)
		{
			int tmp=0,nn=0;
			for(int j=L[i];j<=R[i];j++)
			{
				t[++nn]=w[j],tmp=tmp<<1|(w[j]-w[j-1]==1?1:0);
				if(w[j]<minn[i])minn[i]=w[j],minpos[i]=j;
                //可以用一个状压来表示本质
			}
			if(!pos[tmp])st[pos[tmp]=++n_st].make(t,nn);
			val[i]=pos[tmp];//记下每个块属于哪一个本质
		}
		st_block.make(minn,bl[n]);//块间rmq
	}
	ST_node query_block(int id,int l,int r){ST_node t=st[val[id]].query(l-L[id]+1,r-L[id]+1);return (ST_node){t.f,t.id+L[id]-1};}
    //实际位置=块左端点+块内查询位置-1，如果你把块内查询从0开始写就可以省略-1
	int query(int l,int r)
	{
		int bll=bl[l],blr=bl[r];
		if(bll==blr)return query_block(bll,l,r).id;//一个块
		int ml=query_block(bll,l,R[bll]).id,mr=query_block(blr,L[blr],r).id,mm;
		if(w[ml]<w[mr])mm=ml;else mm=mr;//两端零散块
		if(bll+1<=blr-1)//整块
		{
			int mmid=minpos[st_block.query(bll+1,blr-1).id];
			if(w[mmid]<w[mm])mm=mmid;
		}
		return mm;
	}
}a;
void dfs(int u,int depth)
{
	lst[u]=++id,pos[id]=u;dep[id]=depth;//处理欧拉序列
	for(int i=0;i<=1;i++)
		if(ch[u][i])
		{
			dfs(ch[u][i],depth+1);
			pos[++id]=u,dep[id]=depth;
		}
}
int getin()
{
	int x=0;char ch=getchar();
	while(ch<'0'||ch>'9')ch=getchar();
	while(ch>='0'&&ch<='9')x=x*10+ch-48,ch=getchar();
	return x;
}
int wt[30];
void putout(int x)
{
	if(!x){putchar('0');return;}
	int l=0;
	while(x)wt[++l]=x%10,x/=10;
	while(l)putchar(wt[l--]+48);
	puts("");
}
void build_tree()//笛卡尔树
{
	int top=1;stack[1]=S=1;//开始根为1
	for(int i=2;i<=n;i++)
	{
		int lst=0;
		while(top&&w[i]>w[stack[top]])lst=stack[top--];
		if(lst)ch[i][0]=lst;
		if(top)ch[stack[top]][1]=i;else S=i;
		stack[++top]=i;
	}
	dfs(S,1);
}
int main()
{
	n=getin(),m=getin();
	for(int i=1;i<=n;i++)w[i]=getin();
	build_tree();
	a.make(dep,id);
	for(int i=1;i<=m;i++)
	{
		int l=lst[getin()],r=lst[getin()];
		if(l>r)swap(l,r);//可能第一次出现的位置是反过来的
		putout(w[pos[a.query(l,r)]]);
	}
}

析合树

#include<bits/stdc++.h>
const int N=1e5+10;
char buf[1<<20],*p1,*p2;
char gc(){return p1==p2?p2=buf+fread(p1=buf,1,1<<20,stdin),p1==p2?EOF:*p1++:*p1++;}
inline int read(){
	int x=0,f=1;char ch=gc();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1,ch=gc();}
	while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
	return x*f;
}
int n,a[N],b[N];
int stmx[N][20],stmi[N][20];
void init(){
	for(int i=1;i<=n;i++)stmx[i][0]=stmi[i][0]=a[i];
	for(int j=1;(1<<j)<=n;j++){
		b[1<<j]=j;
		for(int i=1;i+(1<<j)-1<=n;i++){
			stmx[i][j]=std::max(stmx[i][j-1],stmx[i+(1<<j-1)][j-1]);
			stmi[i][j]=std::min(stmi[i][j-1],stmi[i+(1<<j-1)][j-1]);
		}
	}
	for(int i=1;i<=n;i++)if(!b[i])b[i]=b[i-1];
}
int qmx(int l,int r){
	int len=b[r-l+1];
	return std::max(stmx[l][len],stmx[r-(1<<len)+1][len]);
}
int qmi(int l,int r){
	int len=b[r-l+1];
	return std::min(stmi[l][len],stmi[r-(1<<len)+1][len]);
}
struct Tree{
	int l,r,Mi,tar;
}T[N<<2];
#define ls rt<<1
#define rs rt<<1|1
void pushup(int rt){
	T[rt].Mi=std::min(T[ls].Mi,T[rs].Mi);
}
void Build(int rt,int l,int r){
	T[rt].l=l;T[rt].r=r;
	if(l==r)return;
	int mid=l+r>>1;
	Build(ls,l,mid);Build(rs,mid+1,r);
}
void pushdown(int rt){
	T[ls].Mi+=T[rt].tar;
	T[ls].tar+=T[rt].tar;
	T[rs].Mi+=T[rt].tar;
	T[rs].tar+=T[rt].tar;
	T[rt].tar=0;
}
void update(int rt,int l,int r,int w){
	if(l<=T[rt].l&&T[rt].r<=r){
		T[rt].Mi+=w;
		T[rt].tar+=w;
		return ;
	}
	if(T[rt].tar)pushdown(rt);
	int mid=T[rt].l+T[rt].r>>1;
	if(l<=mid)update(ls,l,r,w);
	if(r>mid)update(rs,l,r,w);
	pushup(rt);
}
int query(int rt){
	if(T[rt].l==T[rt].r)return T[rt].l;
	if(T[rt].tar)pushdown(rt);
	if(!T[ls].Mi)return query(ls);
	else return query(rs);
}
int stk[N],stkmx[N],stkmi[N],tp,tp1,tp2;
int lim[N<<1],L[N<<1],R[N<<1],idf[N],cnt;
bool ishe[N<<1];
bool pd(int l,int r){
	return qmx(l,r)-qmi(l,r)==r-l;
}
struct Edge{
	int to,nxt;
}e[N<<1];
int h[N<<1],idx;
void Ins(int a,int b){
	e[++idx].to=b;e[idx].nxt=h[a];h[a]=idx;
}
int p[N<<1][25],dep[N<<1];
void dfs(int u){
	for(int i=0;p[u][i];i++)
		p[u][i+1]=p[p[u][i]][i];
	for(int i=h[u];i;i=e[i].nxt){
		int v=e[i].to;
		if(v==p[u][0])continue;
		p[v][0]=u;
		dep[v]=dep[u]+1;
		dfs(v);
	}
}
int get(int u,int d){
	for(int i=0;d;d>>=1,i++)
		if(d&1)u=p[u][i];
	return u;
}
int lca(int x,int y){
	if(dep[x]<dep[y])std::swap(x,y);
	x=get(x,dep[x]-dep[y]);
	if(x==y)return x;
	for(int i=21;i>=0;i--)
		if(p[x][i]!=p[y][i])
			x=p[x][i],y=p[y][i];
	return p[x][0];
}
int main(){
	n=read();
	for(int i=1;i<=n;i++)a[i]=read();
	init();
	Build(1,1,n);
	for(int i=1;i<=n;i++){
		while(tp1&&a[stkmx[tp1]]<=a[i])update(1,stkmx[tp1-1]+1,stkmx[tp1],-a[stkmx[tp1]]),tp1--;
		while(tp2&&a[stkmi[tp2]]>=a[i])update(1,stkmi[tp2-1]+1,stkmi[tp2],a[stkmi[tp2]]),tp2--;
		update(1,stkmx[tp1]+1,i,a[i]);
		update(1,stkmi[tp2]+1,i,-a[i]);
		stkmx[++tp1]=i;stkmi[++tp2]=i;
		int tl=query(1);idf[i]=++cnt;L[cnt]=R[cnt]=i;
		int u=cnt;
		while(tp&&L[stk[tp]]>=tl){
			if(ishe[stk[tp]]&&pd(lim[stk[tp]],i)){
				R[stk[tp]]=i;
				Ins(stk[tp],u);
				u=stk[tp--];
			}else if(pd(L[stk[tp]],i)){
				ishe[++cnt]=1;
				Ins(cnt,stk[tp]);Ins(cnt,u);
				L[cnt]=L[stk[tp]];R[cnt]=i;tp--;
				lim[cnt]=L[u];u=cnt;
			}else{
				Ins(++cnt,u);
				do{
					Ins(cnt,stk[tp--]);
				}while(tp&&!pd(L[stk[tp]],i));
				L[cnt]=L[stk[tp]];R[cnt]=i;Ins(cnt,stk[tp--]);
				u=cnt;
			}
		}
		stk[++tp]=u;
		update(1,1,i,-1);
	}
	dfs(stk[1]);
	int m=read();
	while(m--){
		int x=read(),y=read();
		x=idf[x];y=idf[y];
		int lc=lca(x,y);
		if(ishe[lc]){
			x=get(x,dep[x]-dep[lc]-1);y=get(y,dep[y]-dep[lc]-1);
			printf("%d %d\n",L[x],R[y]);
		}else printf("%d %d\n",L[lc],R[lc]);
	}	
	return 0;
}

网络流

最大流

#include<bits/stdc++.h>
using namespace std;
const int N=5e4+10;
struct Edge{
	int to,nxt,val;
}e[N<<1];
int h[N],idx=1;
void Ins(int a,int b,int c){
	e[++idx].to=b;e[idx].nxt=h[a];
	h[a]=idx;e[idx].val=c;
}
int s,t;
int d[N],now[N],q[N],hh,tt;
bool bfs(){
	memset(d,-1,sizeof(d));
	d[s]=1;hh=tt=0;q[0]=s;now[s]=h[s];
	while(hh<=tt){
		int u=q[hh++];
		for(int i=h[u];i;i=e[i].nxt){
			int v=e[i].to;
			if(e[i].val&&d[v]==-1){
				d[v]=d[u]+1;
				now[v]=h[v];
				if(v==t)return 1;
				q[++tt]=v;
			}
		}
	}
	return 0;
}
int dinic(int u,int f){
	if(u==t)return f;
	int re=f;
	for(int i=now[u];i&&re;i=e[i].nxt){
		int v=e[i].to;now[u]=i;
		if(e[i].val&&d[v]==d[u]+1){
			int k=dinic(v,min(re,e[i].val));
			if(!k)d[v]=-1;
			re-=k;e[i].val-=k;e[i^1].val+=k;
		}
	}
	return f-re;
}
int main(){
	int n,m;
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=1;i<=m;i++){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		Ins(a,b,c);Ins(b,a,0);
	}
	long long flow=0;
	while(bfs())flow+=dinic(s,0x7fffffff);
	printf("%lld\n",flow);
	return 0;
}

最小割树

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=2e3+10;
struct Edge{
	int to,nxt,val;
}e[N<<1],e2[N<<1];
int h[N],idx=1;
void Ins(int a,int b,int c){
	e[++idx].to=b;e[idx].nxt=h[a];
	h[a]=idx;e[idx].val=c;
}
int h2[N],idx2=1;
void Ins2(int a,int b,int c){
	e2[++idx2].to=b;e2[idx2].nxt=h2[a];
	h2[a]=idx2;e2[idx2].val=c;
}
int a[N],q[N],d[N],hh,tt;
int st,ed,now[N];
bool bfs(){
	memset(d,-1,sizeof(d));
	hh=0;tt=-1;q[++tt]=st;
	now[st]=h[st];d[st]=1;
	while(hh<=tt){
		int u=q[hh++];
		for(int i=h[u];i;i=e[i].nxt){
			int v=e[i].to;
			if(d[v]==-1&&e[i].val){
				d[v]=d[u]+1;
				now[v]=h[v];
				if(v==ed)return 1;
				q[++tt]=v;
			}
		}
	}
	return 0;
}
int dinic(int u,int f){
	if(u==ed)return f;
	int re=f;
	for(int i=now[u];i&&re;i=e[i].nxt){
		int v=e[i].to;
		now[u]=i;
		if(d[v]==d[u]+1&&e[i].val){
			int k=dinic(v,min(e[i].val,re));
			if(!k)d[v]=-1;
			re-=k;e[i].val-=k;e[i^1].val+=k;
		}
	}
	return f-re;
}
void get(int s,int t){
	st=s;ed=t;
	for(int i=2;i<=idx;i+=2){
		int tmp=e[i].val+e[i^1].val>>1;
		e[i].val=e[i^1].val=tmp;
	}
	int Max=0;
	while(bfs())Max+=dinic(st,1e9);
	Ins2(s,t,Max);Ins2(t,s,Max);
}
int tmp1[N],tmp2[N];
void dfs(int l,int r){
	if(l==r)return ;
	get(a[l],a[r]);
	int cl=0,cr=0;
	for(int i=l+1;i<r;i++){
		if(d[a[i]]!=-1)tmp1[++cl]=a[i];
		else tmp2[++cr]=a[i];
	}
	int now=l;
	for(int i=1;i<=cl;i++)
		a[++now]=tmp1[i];
	for(int i=1;i<=cr;i++)
		a[++now]=tmp2[i];
	dfs(l,l+cl);dfs(l+cl+1,r);
}
int dep[N],p[N][25],Max[N][25];
void init(int u){
	for(int i=0;p[u][i];i++){
		p[u][i+1]=p[p[u][i]][i];
		Max[u][i+1]=min(Max[u][i],Max[p[u][i]][i]);
	}
	for(int i=h2[u];i;i=e2[i].nxt){
		int v=e2[i].to;
		if(v==p[u][0])continue;
		dep[v]=dep[u]+1;
		p[v][0]=u;
		Max[v][0]=e2[i].val;
		init(v);
	}
}
int lca(int x,int y){
	int res=1e9;
	if(dep[x]<dep[y])swap(x,y);
	for(int d=dep[x]-dep[y],i=0;d;d>>=1,i++)
		if(d&1)res=min(res,Max[x][i]),x=p[x][i];
	if(x==y)return res;
	for(int i=20;~i;i--)
		if(p[x][i]!=p[y][i]){
			res=min(res,Max[x][i]);
			res=min(res,Max[y][i]);
			x=p[x][i];
			y=p[y][i];
		}
	res=min(res,Max[x][0]);res=min(res,Max[y][0]);
	return res;
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	while(m--){
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		Ins(a,b,c);Ins(b,a,c);
	}
	for(int i=0;i<=n;i++)
		a[i]=i;
	dfs(0,n);
	init(1);
	int ask;
	scanf("%d",&ask);
	while(ask--){
		int u,v;
		scanf("%d%d",&u,&v);
		printf("%d\n",lca(u,v));
	}
	return 0;
}

spfa 费用流

#include<bits/stdc++.h>
using namespace std;
const int N=5e3+10;
const int M=5e4+10;
struct Edge{
	int to,nxt,val,cost;
}e[M<<1];
int h[N],idx=1;
void Ins(int a,int b,int c,int d){
	e[++idx].to=b;e[idx].nxt=h[a];
	h[a]=idx;e[idx].val=c;e[idx].cost=d;
}
int n,m,st,ed,dis[N],pre[N];
deque<int> q;
bool inq[N];
bool spfa(){
	memset(dis,0x3f,sizeof(dis));
	memset(inq,0,sizeof(inq));
	q.push_back(st);dis[st]=0;
	while(!q.empty()){
		int u=q.front();q.pop_front();inq[u]=0;
		for(int i=h[u];i;i=e[i].nxt){
			int v=e[i].to;
			if(dis[v]>dis[u]+e[i].cost&&e[i].val){
				pre[v]=i;
				dis[v]=dis[u]+e[i].cost;
				if(!inq[v]){
					if(q.empty()||dis[v]<=dis[q.front()])q.push_front(v);
					else q.push_back(v);
					inq[v]=1;
				}
			}
		}
	}
	return dis[ed]!=0x3f3f3f3f;
}
int main(){
	scanf("%d%d%d%d",&n,&m,&st,&ed);
	for(int i=1;i<=m;i++){
		int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		Ins(a,b,c,d);
		Ins(b,a,0,-d);
	}
	int res1=0,res2=0;
	while(spfa()){
		int Min=0x3f3f3f3f;
		for(int now=ed;now!=st;now=e[pre[now]^1].to)
			Min=min(Min,e[pre[now]].val);
		for(int now=ed;now!=st;now=e[pre[now]^1].to){
			e[pre[now]].val-=Min;
			e[pre[now]^1].val+=Min;
		}
		res1+=Min;
		res2+=dis[ed]*Min;
	}
	printf("%d %d\n",res1,res2);
	return 0;
}

dinic 费用流

#include<bits/stdc++.h>
using namespace std;
const int N=5e4+10;
const int INF=0x3f3f3f3f;
#define rint register int
struct Edge{
	int to,nxt,val,cost;
}e[N<<1];
int h[N],idx=1;
void Ins(rint a,rint b,rint c,rint d){
	e[++idx].to=b;e[idx].nxt=h[a];h[a]=idx;
	e[idx].val=c;e[idx].cost=d;
}
int s,t,dis[N];
queue<int> q;
bool vis[N];
bool spfa(){
	memset(dis,0x3f,sizeof(dis));
	dis[t]=0;q.push(t);
	while(!q.empty()){
		rint u=q.front();q.pop();vis[u]=0;
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(e[i^1].val&&dis[v]>dis[u]+e[i^1].cost){
				dis[v]=dis[u]+e[i^1].cost;
				if(!vis[v]){
					vis[v]=1;
					q.push(v);
				}
			}
		}
	}
	return dis[s]!=INF;
}
int dinic(rint u,rint f){
	if(u==t)return f;
	rint re=f;
	vis[u]=1;
	for(rint i=h[u];i&&re;i=e[i].nxt){
		rint v=e[i].to;
		if(!vis[v]&&e[i].val&&dis[v]+e[i].cost==dis[u]){
			rint k=dinic(v,min(re,e[i].val));
			if(!k)dis[v]=INF;
			re-=k;e[i].val-=k;e[i^1].val+=k;
		}
	}
	vis[u]=0;
	return f-re;
}
int main(){
	rint n,m;
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(rint i=1;i<=m;i++){
		rint x,y,w,c;
		scanf("%d%d%d%d",&x,&y,&w,&c);
		Ins(x,y,w,c);Ins(y,x,0,-c);
	}
	rint res1=0,res2=0;
	while(spfa()){
		rint now=dinic(s,INF);
		res1+=now;
		res2+=now*dis[s];
	}
	printf("%d %d\n",res1,res2);
	return 0;
}

有源汇上下界最大流

构造方案，每个边的流量加上下界

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=5e5+10;
#define rint register int
#define rll register long long
#define ll long long
const int INF=0x3f3f3f3f;
struct Edge{
	int to,nxt;
	ll val;
}e[N<<1];
int h[N],idx;
void Ins(rint a,rint b,rint c){
	e[++idx].to=b;e[idx].nxt=h[a];
	h[a]=idx;e[idx].val=c;
}
int st,ed,cnt,s,t;
ll rd[N],cd[N];
int q[N],hh,tt,now[N],d[N];
bool bfs(){
	hh=0;tt=-1;q[++tt]=s;
	memset(d,-1,sizeof(d));
	d[s]=1;now[s]=h[s];
	while(hh<=tt){
		rint u=q[hh++];
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(d[v]==-1&&e[i].val){
				d[v]=d[u]+1;
				if(v==t)return 1;
				now[v]=h[v];
				q[++tt]=v;
			}
		}
	}
	return 0;
}
ll dinic(rint u,rll f){
	if(u==t)return f;
	rll re=f;
	for(rint i=now[u];i&&re;i=e[i].nxt){
		rint v=e[i].to;now[u]=i;
		if(d[v]==d[u]+1&&e[i].val){
			rll k=dinic(v,min(e[i].val,re));
			if(!k)d[v]=-1;
			re-=k;e[i].val-=k;e[i^1].val+=k;
		}
	}
	return f-re;
}
int main(){
	rint n,m;
	while(~scanf("%d%d",&m,&n)){
		memset(h,0,sizeof(h));idx=1;cnt=n;
		memset(rd,0,sizeof(rd));memset(cd,0,sizeof(cd));
		st=++cnt,ed=++cnt;
		rll res=0;
		for(rint i=1;i<=n;i++){
			rint g;
			scanf("%d",&g);
			Ins(i,ed,INF-g);Ins(ed,i,0);
			rd[ed]+=g;cd[i]+=g;
		}
		for(rint i=1;i<=m;i++){
			rint c,d;
			scanf("%d%d",&c,&d);
			++cnt;Ins(st,cnt,d);Ins(cnt,st,0);
			while(c--){
				rint x,l,r;
				scanf("%d%d%d",&x,&l,&r);
				++x;
				Ins(cnt,x,r-l);Ins(x,cnt,0);
				rd[x]+=l;cd[cnt]+=l;
			}
		}
		Ins(ed,st,INF);Ins(st,ed,0);
		s=++cnt;t=++cnt;
		rll sd=0;
		for(rint i=1;i<=cnt;i++){
			if(rd[i]>cd[i])Ins(s,i,rd[i]-cd[i]),Ins(i,s,0),sd+=rd[i]-cd[i];
			if(cd[i]>rd[i])Ins(i,t,cd[i]-rd[i]),Ins(t,i,0);
		}
		rll tot=0;
		while(bfs())tot+=dinic(s,INF);
		if(tot==sd){
			for(rint i=h[ed];i;i=e[i].nxt){
				rint v=e[i].to;
				if(v==st){
					res+=e[i^1].val;
					e[i].val=e[i^1].val=0;
					break;
				}
			}
			s=st;t=ed;
			while(bfs())res+=dinic(s,INF);
			printf("%lld\n",res);
		}else puts("-1");
		puts("");
	}
	return 0;
}

无源汇上下界可行流

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=5e5+10;
#define rint register int
#define rll register long long
#define ll long long
const int INF=0x3f3f3f3f;
struct Edge{
	int to,nxt;
	ll val;
}e[N<<1];
int h[N],idx;
void Ins(rint a,rint b,rint c){
	e[++idx].to=b;e[idx].nxt=h[a];
	h[a]=idx;e[idx].val=c;
}
int st,ed,cnt,s,t;
ll rd[N],cd[N];
int q[N],hh,tt,now[N],d[N];
bool bfs(){
	hh=0;tt=-1;q[++tt]=s;
	memset(d,-1,sizeof(d));
	d[s]=1;now[s]=h[s];
	while(hh<=tt){
		rint u=q[hh++];
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(d[v]==-1&&e[i].val){
				d[v]=d[u]+1;
				if(v==t)return 1;
				now[v]=h[v];
				q[++tt]=v;
			}
		}
	}
	return 0;
}
ll dinic(rint u,rll f){
	if(u==t)return f;
	rll re=f;
	for(rint i=now[u];i&&re;i=e[i].nxt){
		rint v=e[i].to;now[u]=i;
		if(d[v]==d[u]+1&&e[i].val){
			rll k=dinic(v,min(e[i].val,re));
			if(!k)d[v]=-1;
			re-=k;e[i].val-=k;e[i^1].val+=k;
		}
	}
	return f-re;
}
int main(){
	rint n;
	scanf("%d",&n);
	idx=1;cnt=n;
	st=++cnt,ed=++cnt;
	rll res=0;
	for(rint i=1;i<=n;i++){
		Ins(st,i,INF);Ins(i,st,0);
		Ins(i,ed,INF);Ins(ed,i,0);
		rint c;
		scanf("%d",&c);
		while(c--){
			rint x;
			scanf("%d",&x);
			Ins(i,x,INF-1);Ins(x,i,0);
			cd[i]++;rd[x]++;
		}
	}
	Ins(ed,st,INF);Ins(st,ed,0);
	s=++cnt;t=++cnt;
	rll sd=0;
	for(rint i=1;i<=cnt;i++){
		if(rd[i]>cd[i])Ins(s,i,rd[i]-cd[i]),Ins(i,s,0),sd+=rd[i]-cd[i];
		if(cd[i]>rd[i])Ins(i,t,cd[i]-rd[i]),Ins(t,i,0);
	}
	rll tot=0;
	while(bfs())tot+=dinic(s,INF);
	for(rint i=h[ed];i;i=e[i].nxt){
		rint v=e[i].to;
		if(v==st){
			res+=e[i^1].val;
			e[i].val=e[i^1].val=0;
			break;
		}
	}
	s=ed;t=st;
	while(bfs())res-=dinic(s,INF);
	printf("%lld\n",res);
	return 0;
}

有源汇最小费用可行流

#include<queue>
#include<cstdio>
#include<cassert>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+10;
const int INF=0x3f3f3f3f;
#define rint register int
#define rll register long long
#define ll long long
struct Edge{
	int to,nxt,val;
	ll cost;
}e[N<<1];
int h[N],idx=1;
void Ins(rint a,rint b,rint c,rll d){
	e[++idx].to=b;e[idx].nxt=h[a];
	h[a]=idx;e[idx].val=c;e[idx].cost=d;
}
int st,ed,mp[N],cnt;
queue<int> q;
bool inq[N];
ll dis[N];
int pre[N];
bool spfa(){
	memset(dis,0x3f,sizeof(dis));
	q.push(st);dis[st]=0;
	while(!q.empty()){
		int u=q.front();q.pop();inq[u]=0;
		for(rint i=h[u];i;i=e[i].nxt){
			rint v=e[i].to;
			if(e[i].val&&dis[v]>dis[u]+e[i].cost){
				dis[v]=dis[u]+e[i].cost;
				pre[v]=i;
				if(!inq[v]){
					q.push(v);
					inq[v]=1;
				}
			}
		}
	}
	return dis[ed]!=dis[0];
}
ll res;
int rd[N],cd[N];
int main(){
	rint n;
	scanf("%d",&n);cnt=n;
	st=1;ed=++cnt;
	for(rint i=1;i<=n;i++){
		Ins(i,ed,INF,0);Ins(ed,i,0,0);
		rint k;
		scanf("%d",&k);
		while(k--){
			rint y,c;
			scanf("%d%d",&y,&c);
			Ins(i,y,INF-1,c);Ins(y,i,0,-c);
			cd[i]++;rd[y]++;
			res+=c;
		}
	}
	Ins(ed,st,INF,0);Ins(st,ed,0,0);
	st=++cnt;
	ed=++cnt;
	for(rint i=1;i<=n;i++){
		if(rd[i]>cd[i])Ins(st,i,rd[i]-cd[i],0),Ins(i,st,0,0);
		if(cd[i]>rd[i])Ins(i,ed,cd[i]-rd[i],0),Ins(ed,i,0,0);
	}
	while(spfa()){
		rint Min=INF;
		for(rint now=ed;now!=st;now=e[pre[now]^1].to)
			Min=min(Min,e[pre[now]].val);
		for(rint now=ed;now!=st;now=e[pre[now]^1].to)
			e[pre[now]].val-=Min,e[pre[now]^1].val+=Min;
		res+=Min*dis[ed];
	}
	printf("%lld\n",res);
	return 0;
}

有源汇上下界可行流

只需要在s和t之间连边然后跑无源汇

有源汇最小流

求ss->tt最大流
连边t->s,inf
求ss->tt最大流
答案即为边t->s,inf的实际流量

最大独立集/最小点覆盖

# include <cstdio>
# include <algorithm>
# include <cmath>
# include <cstring>
# include <vector>

using namespace std;

const int N_MAX = 1000, E_MAX = 1000000;

struct Edge
{
	int to, next;
};

int n, m, e;
vector <int> g[N_MAX + 10];

int opp[N_MAX + 10];
bool vis[N_MAX + 10];

bool vx[N_MAX + 10], vy[N_MAX + 10];

void addEdge(int x, int y)
{
	g[x].push_back(y);
}

bool find(int x)
{
	if (vis[x]) return false;
	vis[x] = true;
	for (int i = 0; i < (int) g[x].size(); i++) {
		int y = g[x][i];
		if (opp[y] == 0 || find(opp[y])) {
			opp[y] = x;
			return true;
		}
	}
	return false;
}

void mark(int x)
{
	if (vx[x]) return;
	vx[x] = true;
	for (int i = 0; i < (int) g[x].size(); i++) {
		int y = g[x][i];
		if (opp[y] && !vy[y]) {
			vy[y] = true;
			mark(opp[y]);
		}
	}
}

int hungary()
{
	memset(opp, 0, sizeof(opp));
	int ans = 0;
	for (int i = 1; i <= n; i++) {
		memset(vis, false, sizeof(vis));
		ans += find(i);
	}
	return ans;
}

int maxIndSet()
{
	int ans = hungary();
	memset(vis, false, sizeof(vis));
	for (int i = 1; i <= m; i++)
		vis[opp[i]] = true;
	memset(vx, false, sizeof(vx));
	memset(vy, false, sizeof(vy));
	for (int i = 1; i <= n; i++)
		if (!vis[i]) mark(i);
	return n + m - ans;
}

int main()
{
	scanf("%d%d%d", &n, &m, &e);
	for (int i = 1; i <= e; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		addEdge(x, y);
	}
	printf("%d\n", maxIndSet());
	for (int i = 1; i <= n; i++)
		if (vx[i]) printf("%d ", i);
	puts("");
	for (int i = 1; i <= m; i++)
		if (!vy[i]) printf("%d ", i);
	puts("");
	return 0;
}

最小割相关

求最小边的最小割

设总边数为 $E$ ，跑最大流之前所有的边权都乘 $E+1$ 然后再 $+1$
得到的结果应该是 $mincut∗(E+1)+$ 割边数量（这个比较显然吧）
由于割边数越小，跑出来结果越小，所以就自动选了割边数量小的边（但相同不能保证字典序）

输出任意一种最小割的方案

跑过一次最大流之后，在残量网络上，s和t之间不连通了
进行一次dfs/bfs，求出从s出发能到达的点集S，和不能到达的点集T
所有从S跨越到T的满流边（残留网络为0）构成了一组最小割

判断一条边是否满流

运行一次最大流算法，得到一个残量网络
取残量网络上的一条满流边(u, v)，判断这条边是否一定满流
对残量网络运行Tarjan算法，求出所有SCC
当u和v不属于同一个SCC的时候，这条边一定满流
否则，我们可以在SCC中找到一个包含这条边的反向边的环，沿着环增广一次，仍然不破坏流量平衡，但是这条边已经不满流了

判断某一条边是否可能为最小割中的一条

所有一定满流的边都可能为最小割

判断某条边是否一定出现在最小割中

首先还是对残量网络求SCC
考虑一条满流边(u, v)，判断她是否一定出现在最小割中
当u和s属于同一个SCC，并且v和t属于同一个SCC的时候，这条边一定出现在最小割中

多项式

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<18;
const int p=998244353;
const int inv2=p+1>>1;
#define rint register int
#define rll register long long 
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline int Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
inline int fpow(rint a,rint b){
	rint res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int w[N],r[N],pd;
unsigned long long A[N],t;
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd^n){
		for(rint i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
		pd=n;
	}
	for(rint i=0;i<n;i++)A[i]=a[r[i]];
	for(rint mid=1;mid<n;mid<<=1)
	for(rint R=mid<<1,i=0;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=w[mid+j]*A[i+j+mid]%p,A[i+j+mid]=A[i+j]+p-t,A[i+j]+=t;
	for(rint i=0;i<n;i++)a[i]=A[i]%p;
	if(typ==1)return;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)a[i]=Mul(a[i],inv);
}
void dfs_inv(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=fpow(g[0],p-2);return;}
	dfs_inv(f,g,n+1>>1);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);f.resize(n);
}
void dfs_sqrt(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=1;return;}
	dfs_sqrt(f,g,n+1>>1);
	vector<int> a;
	dfs_inv(a,f,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,a,1);ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(Add(Mul(f[i],f[i]),g[i]),Mul(inv2,a[i]));
	ntt(Max,f,-1);f.resize(n);
}
void Ln(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	dfs_inv(f,g,n);
	for(rint i=1;i<n;i++)g[i-1]=Mul(g[i],i);
	g[n-1]=0;
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);
	f.resize(n);
	for(rint i=n-1;i;i--)f[i]=Mul(f[i-1],fpow(i,p-2));
	f[0]=0;
}
void Exp(vector<int> &f,vector<int> g,int n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=1;return;}
	Exp(f,g,n+1>>1);
	vector<int> a;
	Ln(a,f,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	g.resize(Max);a.resize(Max);
	for(rint i=0;i<Max;i++)a[i]=Del(g[i],a[i]);a[0]++;
	ntt(Max,a,1);ntt(Max,f,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],a[i]);
	ntt(Max,f,-1);f.resize(n);
}
struct Eval{
	vector<int> vec[N<<2],A,res,tmp;
	vector<int> Mult(vector<int> a,vector<int> b,rint n){
		rint Max=a.size();
		for(int i=0;i<Max;i++)
			a[i]=Mul(a[i],b[i]);
		ntt(Max,a,1);a.resize(n);
		return a;
	}
	void dfs1(rint x,rint l,rint r){
		if(l==r){
			vec[x].resize(2);
			vec[x][0]=1;
			vec[x][1]=p-tmp[l];
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1;
		dfs1(L,l,mid);dfs1(R,mid+1,r);
		rint Max=1,n=r-l+2;
		while(Max<n)Max<<=1;vec[x].resize(Max);
		ntt(Max,vec[L],1);ntt(Max,vec[R],1);
		for(rint i=0;i<Max;i++)
			vec[x][i]=Mul(vec[L][i],vec[R][i]);
		ntt(Max,vec[x],-1);vec[x].resize(n);
	}
	void dfs2(rint x,rint l,rint r,vector<int> v){
		rint n=r-l+2;
		if(l==r){
			res[l]=v[0];
			return;
		}
		rint Max=1,mid=l+r>>1;
		while(Max<n)Max<<=1;
		ntt(Max,v,-1);
		dfs2(x<<1,l,mid,Mult(v,vec[x<<1|1],mid-l+1));
		dfs2(x<<1|1,mid+1,r,Mult(v,vec[x<<1],r-mid));
	}
	vector<int> query(vector<int> A,vector<int> a){
		rint n=A.size(),m=a.size()-1;res.resize(m+1);tmp=a;
		dfs1(1,1,m);
		dfs_inv(vec[1],vec[1],n);
		rint Max=1;
		while(Max<n+n)Max<<=1;
		ntt(Max,vec[1],1);
		ntt(Max,A,-1);
		dfs2(1,1,m,Mult(A,vec[1],n));
		return res;
	}
}T;
struct Fin{
	vector<int> vec[N<<2],vec2[N<<2],tmp,tmp2;
	void dfs1(rint x,rint l,rint r){
		if(l==r){
			vec[x].resize(2);
			vec[x][0]=p-tmp[l];
			vec[x][1]=1;
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1,n=r-l+2;
		dfs1(L,l,mid);dfs1(R,mid+1,r);
		rint Max=1;
		while(Max<n)Max<<=1;
		ntt(Max,vec[L],1);ntt(Max,vec[R],1);
		vec[x].resize(Max);
		for(rint i=0;i<Max;i++)
			vec[x][i]=Mul(vec[L][i],vec[R][i]);
		ntt(Max,vec[x],-1);
		vec[x].resize(n);
	}
	void dfs2(rint x,rint l,rint r){
		if(l==r){
			vec2[x].resize(1);
			vec2[x][0]=Mul(tmp2[l],fpow(tmp[l],p-2));
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1,n=r-l+2;
		dfs2(x<<1,l,mid);dfs2(x<<1|1,mid+1,r);
		rint Max=1;
		while(Max<n)Max<<=1;
		vec2[x].resize(Max);
		ntt(Max,vec2[L],1);ntt(Max,vec2[R],1);
		for(rint i=0;i<Max;i++)
			vec2[x][i]=Add(Mul(vec2[L][i],vec[R][i]),Mul(vec2[R][i],vec[L][i]));
		ntt(Max,vec2[x],-1);vec2[x].resize(n-1);
	}
	vector<int> query(vector<int> x,vector<int> y){
		int n=x.size()-1;tmp=x;tmp2=y;
		dfs1(1,1,n);
		for(rint i=1;i<=n;i++)
			vec[1][i-1]=Mul(vec[1][i],i);
		vec[1][n]=0;
		tmp=T.query(vec[1],x);
		dfs2(1,1,n);
		return vec2[1];
	}
}F;
int main(){
	rint n,m;
	w[N>>1]=1;w[N/2+1]=n=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],n);
	for(rint i=N/2-1;~i;i--)w[i]=w[i<<1];
	scanf("%d",&n);
	vector<int> f,g;++n;
	f.resize(n);g.resize(n);
	for(rint i=1;i<n;i++)
		scanf("%d%d",&f[i],&g[i]);
	f=F.query(f,g);
	for(rint v:f)printf("%d ",v);
	return 0;
}

多项式求逆

给定多项式 $F$ ,求一个 $G$ 使得 $F*G\equiv1 \pmod {x^n}$

多项式的模意义是指次数大于等于 $n$ 的项都当成 $0$ ，考虑倍增，设有多项式 $A$ 满足 $F*A\equiv 1 \pmod {x^{\lceil{\frac{n}2}\rceil}}$ ，上取整的意思为了让它平方后大于等于 $n$ 的项都当成 $0$ 。

显然有 $F*G\equiv1 \pmod {x^{\lceil{\frac{n}2}\rceil}}$ ，根据模的定义可知。

所以有 $F*(G-A)\equiv 0 \pmod{x^{\lceil{\frac{n}2}\rceil}}$ ， $F$ 不可能为 $0$ ，于是 $(G-A)^2\equiv0 \pmod{x^n}$ ，继续推一发，得到 $G^2-2AG+A^2\equiv 0 \pmod{x^n}$ ，两边同时乘 $F$ 得到 $G-2A+A^2F\equiv 0 \pmod{x^n}$ ，所以推出 $G\equiv 2A-A^2F$ ，只有常数项的时候求逆元即可，分治即可求出。

void dfs_inv(rint n,vector<int> &f,vector<int> g){
	g.resize(n);
	if(n==1){
		f.resize(1);
		f[0]=fpow(g[0],p-2);
		return ;
	}
	dfs_inv(n+1>>1,f,g);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);
	f.resize(n);
}

多项式开根

求一个多项式 $F$ 满足 $F^2=G$ ，若 $F$ 有两个，一般取正的。

仍然使用倍增，设 $F^2\equiv G\pmod {x^{2n}}$ ，假设已经求出了 $A^2\equiv G\pmod {x^n}$ ，那么有 $F^2-A^2\equiv0 \pmod {x^n}$ 。

易得

(F+A)(F-A)\equiv0 \pmod {x^n}

(F-A)\equiv0 \pmod {x^n}

(F-A)^2\equiv0 \pmod {x^{2n}}

F^2+A^2-2AF\equiv0 \pmod {x^{2n}}

G+A^2-2AF\equiv0 \pmod {x^{2n}}

F\equiv \frac{G+A^2}{2A}\pmod {x^{2n}}

在只有常数项的时候可以直接求出。

若常数项为 $1$ 或 $0$ ，开根后较为显然，但是有些毒瘤出题人并不喜欢简单，所以对于模意义下开根就比较重要。

void dfs_sqr(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){
		f.resize(1);f[0]=1;
		return;
	}
	dfs_sqr(f,g,n+1>>1);
	vector<int> a;
	dfs_inv(a,f,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);ntt(Max,a,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(Add(Mul(f[i],f[i]),g[i]),Mul(inv2,a[i]));
	ntt(Max,f,-1);f.resize(n);
}

二次剩余

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
#define rint register int
#define rll register long long
ll I,n,p;
inline ll Add(rll x,rll y){return (x+y)%p;}
inline ll Del(rll x,rll y){return ((x-y)%p+p)%p;}
inline ll Mul(rll x,rll y){x*=y;return x>=p?x%p:x;}
struct cp{
	ll x,y;
	cp operator * (const cp&A)const{return (cp){Add(Mul(x,A.x),Mul(Mul(I,y),A.y)),Add(Mul(x,A.y),Mul(y,A.x))};}
};
ll fpow(rll a,rll b){
	rll res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
cp fpow(cp a,rll b){
	cp res=(cp){1,0};
	for(;b;b>>=1,a=a*a)
		if(b&1)res=res*a;
	return res;
}
inline bool check(rll x){
	if(fpow(x,p-1>>1)==1)return 1;
	return 0;
}
int main(){
	int T;
	scanf("%d",&T);
	while(T--){
		scanf("%lld%lld",&n,&p);
		if(n==0)printf("0\n");
		else if(!check(n))printf("Hola!\n");
		else {
			rll a;I=0;
			while(I==0||check(I))a=rand()%p,I=Del(Mul(a,a),n);
			rll res1=fpow((cp){a,1},p+1>>1).x;
			rll res2=p-res1;
			if(res1==res2)printf("%lld ",res1);
			else printf("%lld %lld\n",min(res1,res2),max(res1,res2));
		}
	}
	return 0;
}

多项式除法

给定多项式 $F,G$ ，求两个多项式 $A,B$ 满足 $F=A*G+B$ 。

其中 $F,G,A$ 的项数分别为 $n,m,n-m$ ，要求 $B$ 的项数小于 $m$

定义一种变换， $F_{R(x)}=x^n F(\frac{1}x)$ ，容易发现这种变换就是将 $F$ 的系数reverse了一下。

F=A*G+B

F(\frac{1}x)=A(\frac{1}x)G(\frac{1}x)+B(\frac{1}x)

x^nF(\frac{1}x)=x^nA(\frac{1}x)G(\frac{1}x)+x^nB(\frac{1}x)

x^nF(\frac{1}x)=x^{n-m}A(\frac{1}x)x^mG(\frac{1}x)+x^{n-m+1}x^{m-1}B(\frac{1}x)

F_{R(x)}=A_{R(x)}G_{R(x)}+x^{n-m+1}B_{R(x)}

模一下 $x^{n-m+1}$ 就能得到

F_{R(x)}\equiv A_{R(x)}G_{R(x)} \pmod {x^{n-m+1}}

A_{R(x)}\equiv F_{R(x)}G^{-1}_{R(x)} \pmod {x^{n-m+1}}

多项式求逆即可。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=3e5+10;
const int p=998244353;
typedef long long ll;
ll fpow(ll a,ll b){
	ll res;
	for(res=1;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
ll tmp[2][N],F[N],G[N],A[N],B[N];
int r[N];
void ntt(int n,ll *a,int typ){
	for(int i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(int mid=1;mid<n;mid<<=1){
		ll Wn=fpow(3,(p-1)/(mid*2));
		if(typ==-1)Wn=fpow(Wn,p-2);
		for(int i=0,R=mid<<1;i<n;i+=R){
			ll w=1;
			for(int j=0;j<mid;j++,w=w*Wn%p){
				const ll x=a[i+j],y=w*a[i+j+mid]%p;
				a[i+j]=(x+y)%p;
				a[i+j+mid]=(x-y+p)%p;
			}
		}
	}
}
void dfs(int n){
	if(n==1){B[0]=fpow(G[0],p-2);return;}
	dfs(n+1>>1);
	int Max=1,l=0;
	while(Max<n+n)Max<<=1,l++;
	for(int i=0;i<Max;i++){
		r[i]=(r[i>>1]>>1)|((i&1)<<l-1);
		A[i]=i<n?G[i]:0;
	}
	ntt(Max,A,1);ntt(Max,B,1);
	for(int i=0;i<Max;i++)
		B[i]=(2ll*B[i]%p-A[i]*B[i]%p*B[i]%p+p)%p;
	ntt(Max,B,-1);
	ll inv=fpow(Max,p-2);
	for(int i=0;i<Max;i++)
		B[i]=i<n?B[i]*inv%p:0;
}
int main(){
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i<=n;i++)scanf("%lld",&tmp[0][i]),F[n-i]=tmp[0][i];
	for(int i=0;i<=m;i++)scanf("%lld",&tmp[1][i]),G[m-i]=tmp[1][i];
	for(int i=n-m+2;i<=m;i++)G[i]=0;
	for(int i=n-m+2;i<=n;i++)F[i]=0;
	dfs(n-m+1);
	int Max=1,l=0;
	while(Max<=2*n-2*m+4)Max<<=1,l++;
	for(int i=0;i<Max;i++)r[i]=(r[i>>1]>>1)|((i&1)<<l-1);
	ntt(Max,F,1);ntt(Max,B,1);
	for(int i=0;i<Max;i++)
		F[i]=F[i]*B[i]%p;
	ntt(Max,F,-1);
	ll inv=fpow(Max,p-2);
	for(int i=0;i<Max;i++)
		A[i]=(i<=n-m)?F[i]*inv%p:0;
	reverse(A,A+n-m+1);
	memcpy(B,A,sizeof(A));
	memcpy(F,tmp[0],sizeof(tmp[0]));
	memcpy(G,tmp[1],sizeof(tmp[1]));
	Max=1,l=0;
	while(Max<=n)Max<<=1,l++;
	for(int i=0;i<Max;i++)r[i]=(r[i>>1]>>1)|((i&1)<<l-1);
	ntt(Max,B,1);ntt(Max,G,1);
	for(int i=0;i<Max;i++)
		B[i]=B[i]*G[i]%p;
	ntt(Max,B,-1);
	inv=fpow(Max,p-2);
	for(int i=0;i<Max;i++)
		B[i]=B[i]*inv%p;
	for(int i=0;i<=m;i++)
		B[i]=(F[i]-B[i]+p)%p;
	for(int i=0;i<=n-m;i++)
		printf("%lld ",A[i]);
	puts("");
	for(int i=0;i<m;i++)
		printf("%lld ",B[i]);
	return 0;
}

多项式求导和积分

求导， $[x^{i-1}]F'=[x^i]F\times i$

积分，积分是求导的逆运算，对一个多项式积分的结果表示哪个多项式求导得到的是这个多项式

$[x^i]\int F'=\frac{[x^{i-1}]F'}i$ 。

多项式ln

求一个多项式 $F$ 满足 $G=\ln F$

两边求导得到 $G'=\ln'F F'$

由 $\ln$ 的导数可以知道 $\ln'F=\frac{1}F$

所以 $G'=\frac{F'}F$

于是 $G=\int\frac{F'}F$

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=1<<18;
const int p=998244353;
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline ll Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
ll fpow(rll a,rll b){
	rll res=1;
	for(;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
int r[N],w[N],t,pd;
void ntt(rint n,vector<int> &a,rint typ){
	if(pd!=n){
		for(rint i=0;i<n;i++)
			r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
		pd=n;
	}
	a.resize(n);
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint i=0;i<n;i+=mid<<1)
	for(rint j=0;j<mid;j++)
		t=Mul(w[mid+j],a[i+j+mid]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(~typ)return ;
	t=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)
		a[i]=Mul(a[i],t);
}
void dfs_inv(rint n,vector<int> &f,vector<int> g){
	g.resize(n);
	if(n==1){
		f.resize(1);
		f[0]=fpow(g[0],p-2);
		return ;
	}
	dfs_inv(n+1>>1,f,g);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);
	f.resize(n);
}
void Ln(rint n,vector<int> &f,vector<int> g){
	g.resize(n);
	f=g;
	dfs_inv(n,g,g);
	for(rint i=1;i<n;i++)
		f[i-1]=Mul(f[i],i);
	f[n-1]=0;
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);
	for(rint i=n-1;i;i--)
		f[i]=Mul(f[i-1],fpow(i,p-2));
	f[0]=0;
	f.resize(n);
}
int main(){
	w[N/2]=1;w[N/2+1]=t=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],t);
	for(rint i=N/2-1;i;i--)w[i]=w[i<<1];
	rint n;
	scanf("%d",&n);
	vector<int> f;
	f.resize(n);
	for(rint i=0;i<n;i++)
		scanf("%d",&f[i]);
	Ln(n,f,f);
	for(rint i=0;i<n;i++)
		printf("%d ",f[i]);
	return 0;
}

多项式exp

令 $G(x)\equiv e^{f(x)}\pmod {x^n},G_1(x)\equiv e^{f(x)} \pmod {x^{\frac{n}2}}$

有 $G\equiv G_1(1-lnG_1+F) \pmod {x^n}$

证明略

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<18;
const int p=998244353;
#define rint register int
#define rll register long long 
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline int Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
inline int fpow(rint a,rint b){
	rint res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int w[N],r[N],pd,t;
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd^n){
		for(rint i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
		pd=n;
	}
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint R=mid<<1,i=0;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=Mul(w[mid+j],a[i+j+mid]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(typ==1)return;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)a[i]=Mul(a[i],inv);
}
void dfs_inv(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=fpow(g[0],p-2);return;}
	dfs_inv(f,g,n+1>>1);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);f.resize(n);
}
void Ln(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	dfs_inv(f,g,n);
	for(rint i=1;i<n;i++)g[i-1]=Mul(g[i],i);
	g[n-1]=0;
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);
	f.resize(n);
	for(rint i=n-1;i;i--)f[i]=Mul(f[i-1],fpow(i,p-2));
	f[0]=0;
}
void Exp(vector<int> &f,vector<int> g,int n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=1;return;}
	Exp(f,g,n+1>>1);
	vector<int> a;
	Ln(a,f,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	g.resize(Max);a.resize(Max);
	for(rint i=0;i<Max;i++)a[i]=Del(g[i],a[i]);a[0]++;
	ntt(Max,a,1);ntt(Max,f,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],a[i]);
	ntt(Max,f,-1);f.resize(n);
}
int main(){
	rint n;
	w[N>>1]=1;w[N/2+1]=n=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],n);
	for(rint i=N/2-1;~i;i--)w[i]=w[i<<1];
	vector<int> f;
	scanf("%d",&n);f.resize(n);
	for(rint i=0;i<n;i++)
		scanf("%d",&f[i]);
	Exp(f,f,n);
	for(rint v:f)printf("%d ",v);
}

#include<algorithm>
#include<cstring>
#include<cstdio>
#define ll long long
#define ull unsigned long long
#define clr(f,n) memset(f,0,sizeof(int)*(n))
#define cpy(f,g,n) memcpy(f,g,sizeof(int)*(n))
using namespace std;
const int _G=3,mod=998244353,Maxn=1050000;
inline int read(){
  int X=0;char ch=0;
  while(ch<48||ch>57)ch=getchar();
  while(ch>=48&&ch<=57)X=X*10+(ch^48),ch=getchar();
  return X;
}
ll powM(ll a,int t=mod-2){
  ll ans=1;
  while(t){
    if(t&1)ans=ans*a%mod;
    a=a*a%mod;t>>=1;
  }return ans;
}
const int invG=powM(_G);
int tr[Maxn<<1],tf;
void tpre(int n){
  if (tf==n)return ;tf=n;
  for(int i=0;i<n;i++)
    tr[i]=(tr[i>>1]>>1)|((i&1)?n>>1:0);
}
void NTT(int *g,bool op,int n)
{
  tpre(n);
  static ull f[Maxn<<1],w[Maxn<<1];w[0]=1;
  for (int i=0;i<n;i++)f[i]=g[tr[i]];
  for(int l=1;l<n;l<<=1){
    ull tG=powM(op?_G:invG,(mod-1)/(l+l));
    for (int i=1;i<l;i++)w[i]=w[i-1]*tG%mod;
    for(int k=0;k<n;k+=l+l)
      for(int p=0;p<l;p++){
        int tt=w[p]*f[k|l|p]%mod;
        f[k|l|p]=f[k|p]+mod-tt;
        f[k|p]+=tt;
      }      
    if (l==(1<<17))
      for (int i=0;i<n;i++)f[i]%=mod;
  }if (!op){
    ull invn=powM(n);
    for(int i=0;i<n;++i)
      g[i]=f[i]%mod*invn%mod;
  }else for(int i=0;i<n;++i)g[i]=f[i]%mod;
}
void px(int *f,int *g,int n)
{for(int i=0;i<n;++i)f[i]=1ll*f[i]*g[i]%mod;}
int F[Maxn],G[Maxn],s1[Maxn],s2[Maxn];
void cdq(int l,int r)
{
  if (l+1==r){
    if (l>0)G[l]=powM(l)*G[l]%mod;
    return ;
  }int mid=(l+r)>>1,n=r-l;
  cdq(l,mid);
  cpy(s1,G+l,n/2);clr(s1+(n/2),n/2);NTT(s1,1,n);
  cpy(s2,F,n);NTT(s2,1,n);
  px(s1,s2,n);NTT(s1,0,n);
  for (int i=n/2;i<n;i++)
    G[l+i]=(G[l+i]+s1[i])%mod;
  cdq(mid,r);
}
int n,m;
int main()
{
  m=read();G[0]=1;
  for (int i=0;i<m;i++)F[i]=1ll*read()*i%mod;
  for (n=1;n<m;n<<=1);
  cdq(0,n);
  for (int i=0;i<m;i++)
    printf("%d ",G[i]);
  return 0;
}

泰勒展开

对于一个函数 $F$ ， $F(x)$ 的值可以近似看作在 $x_0$ 处泰勒展开的值，即

F(x)=\sum_{i=0}^{+\infty} \frac{F^{(i)}_{x_0}(x-x_0)^i}{i!}

这个式子看上去貌似到正无穷没有办法求，但是由于它递减的很快所以基本上只需要求前几项即可。

牛顿迭代

上边的式子虽然易懂但是比较繁琐，因为对于不同的东西需要用到不同的式子，考虑更加通用的办法——牛顿迭代。

考虑如果求出了 $F(x) \equiv 0 \pmod {x^{\lceil\frac{n}2\rceil}}$ 的一个解 $x_0$ ，如何求出另一个解 $x$ 满足 $F(x) \equiv 0 \pmod {x^{n}}$ 。

将 $F(x)$ 在 $x_0$ 处泰勒展开，容易得到

F(x)=F_{x_0}+F'_{x_0}(x-x_0)+....

注意到后边写了省略号，~~并不是我懒得写了~~，而是因为没有用了，因为 $x$ 和 $x_0$ 在模 ${x^n}$ 的意义下，前一半项是一样的，所以 $x-x_0$ 的最低次项也是 $\frac{n}2$ 即大于 $2$ 的次幂全是 $0$ 。

于是有

F_{x_0}+F'_{x_0}(x-x_0) \equiv 0 \pmod {x^n}

F'_{x_0}x\equiv F'_{x_0}x_0-F_{x_0} \pmod {x^n}

x\equiv x_0-\frac{F_{x_0}}{F'_{x_0}} \pmod {x^n}

这样推式子就简单的多了，更具体的，只需要构造一个函数 $F(x)\equiv0 \pmod {x^n}$ 即可。

例如多项式乘法逆，构造函数 $F(x)=\frac{1}x-G$ ，其中 $G$ 为给定的多项式。

那么我们要求的实际上就是 $F(x)\equiv0 \pmod {x^n}$ 的一个解。

直接代牛顿迭代，有

x\equiv x_0 - \frac{\frac{1}{x_0}-G}{-\frac{1}{x_0^2}} \pmod {x^n}

x\equiv 2x_0-Gx_0^2 \pmod {x^n}

多项式开根,构造函数 $F(x)=x^2-G$ ，其中 $G$ 为给定多项式。

代牛顿迭代，得到

x\equiv x_0-\frac{x_0^2-G}{2x_0} \pmod {x^n}

x\equiv \frac{x_0^2+G}{2x_0} \pmod {x^n}

~~简单易懂~~

任意模数多项式乘法

总有一些毒瘤出题人喜欢把模数搞得奇怪，让你没办法做 $NTT$ ，这时候可以用 $CRT$ ，但是常数过于大，就不大好玩。

考虑一个古老的算法，其实也不是很古老，只是一段时间没怎么用过， $FFT$ 。

因为 $FFT$ 是不用取模的，所以可以直接乘然后最后一起取模，不过模数很大容易炸精，所以要拆系数，不妨令 $[x^i]F=a_0\times T +b_0$ ，然后分别求出 $a,b$ 最后合并即可，即分别求出 $a_0a_1,a_0b_1+a_1b_0,b_0b_1$ ，这样做大概需要七次 $FFT$ ，还是太慢了。

于是使用魔法，使得能够快速求出要求的几项。

令 $P=a+bi,Q=a-bi$ ，显然他们的系数是共轭的，有一个结论

$Q_0=P_0$ ，其余的 $Q_i=P_{n-i}$ ，即 $Q_i=P_{(n-1)\&(n-i)}$ ，所以只需要求出 $P$ ，然后就有了 $Q$ ，这样就把两次 $NTT$ 合并成了一次。此时再暴力求出剩下的几项只要五次，考虑继续优化。

令 $A=a_0a_1+b_0b_1i,B=a_0b_1+a_1b_0$ ，这样只需要做两次 $IDFT$ 即可，所以就可以写出四次的 $MTT$ 了。

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=4e5+10;
const int M=(1<<15)-1;
typedef double db;
const db pi=acos(-1.0);
struct complex{
	double x,y;
	complex operator + (const complex&A)const{return (complex){x+A.x,y+A.y};}
	complex operator - (const complex&A)const{return (complex){x-A.x,y-A.y};}
	complex operator * (const complex&A)const{return (complex){x*A.x-y*A.y,x*A.y+y*A.x};}
	friend complex operator ~ (const complex&A){return (complex){A.x,-A.y};}
}A[N],B[N],C[N],D[N],a0[N],a1[N],b0[N],b1[N],w[N],tmp;
int r[N];
void fft(int n,complex *a,int typ=1){
	for(int i=0;i<n;i++)if(i<r[i])swap(a[i],a[r[i]]);
	for(int mid=1,t0=n>>1;mid<n;mid<<=1,t0>>=1)
	for(int i=0,R=mid<<1;i<n;i+=R)
	for(int j=0,t=0;j<mid;j++,t+=t0)
		tmp=w[t]*a[i+j+mid],a[i+j+mid]=a[i+j]-tmp,a[i+j]=a[i+j]+tmp;
	if(typ)return;
	reverse(a+1,a+n);
	for(int i=0;i<n;i++)a[i]=a[i]*(complex){1.0/n,0.0};
}
int main(){
	int n,m,p;
	scanf("%d%d%d",&n,&m,&p);
	for(int i=0;i<=n;i++){
		int x;
		scanf("%d",&x);
		a0[i].x=x>>15;
		a1[i].x=x&M;
	}
	for(int i=0;i<=m;i++){
		int x;
		scanf("%d",&x);
		b0[i].x=x>>15;
		b1[i].x=x&M;
	}
	for(int i=0;i<=n;i++)A[i].x=a0[i].x,A[i].y=a1[i].x;
	for(int i=0;i<=m;i++)B[i].x=b0[i].x,B[i].y=b1[i].x;
	int Max=1,l=0;
	while(Max<=n+m)Max<<=1,l++;
	for(int i=0;i<Max;i++){
		r[i]=(r[i>>1]>>1)|((i&1)<<l-1);
		w[i]=(complex){cos(2*i*pi/Max),sin(2*i*pi/Max)};
	}
	fft(Max,A);fft(Max,B);
	for(int i=0;i<Max;i++){
		tmp=~A[(Max-1)&(Max-i)];
		a0[i]=(A[i]+tmp)*(complex){0.5,0.0};
		a1[i]=(tmp-A[i])*(complex){0.0,0.5};
		tmp=~B[(Max-1)&(Max-i)];
		b0[i]=(B[i]+tmp)*(complex){0.5,0.0};
		b1[i]=(tmp-B[i])*(complex){0.0,0.5};
	}
	for(int i=0;i<Max;i++){
		A[i]=a0[i]*b0[i]+a1[i]*b1[i]*(complex){0.0,1.0};
		B[i]=a1[i]*b0[i]+a0[i]*b1[i];
	}
	fft(Max,A,0);fft(Max,B,0);
	for(int i=0;i<=n+m;i++){
		long long w1=(long long)(A[i].x+0.5)%p;
		long long w2=(long long)(B[i].x+0.5)%p;
		long long w3=(long long)(A[i].y+0.5)%p;
		printf("%lld ",((w1<<30)+(w2<<15)+w3)%p);
	}
	return 0;
}

#include <algorithm>
#include <cstdio>
#include <cstring>
int mod;
namespace Math {
	inline int pw(int base, int p, const int mod) {
		static int res;
		for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
		return res;
	}
	inline int inv(int x, const int mod) { return pw(x, mod - 2, mod); }
}

const int mod1 = 998244353, mod2 = 1004535809, mod3 = 469762049, G = 3;
const long long mod_1_2 = static_cast<long long> (mod1) * mod2;
const int inv_1 = Math::inv(mod1, mod2), inv_2 = Math::inv(mod_1_2 % mod3, mod3);
struct Int {
	int A, B, C;
	explicit inline Int() { }
	explicit inline Int(int __num) : A(__num), B(__num), C(__num) { }
	explicit inline Int(int __A, int __B, int __C) : A(__A), B(__B), C(__C) { }
	static inline Int reduce(const Int &x) {
		return Int(x.A + (x.A >> 31 & mod1), x.B + (x.B >> 31 & mod2), x.C + (x.C >> 31 & mod3));
	}
	inline friend Int operator + (const Int &lhs, const Int &rhs) {
		return reduce(Int(lhs.A + rhs.A - mod1, lhs.B + rhs.B - mod2, lhs.C + rhs.C - mod3));
	}
	inline friend Int operator - (const Int &lhs, const Int &rhs) {
		return reduce(Int(lhs.A - rhs.A, lhs.B - rhs.B, lhs.C - rhs.C));
	}
	inline friend Int operator * (const Int &lhs, const Int &rhs) {
		return Int(static_cast<long long> (lhs.A) * rhs.A % mod1, static_cast<long long> (lhs.B) * rhs.B % mod2, static_cast<long long> (lhs.C) * rhs.C % mod3);
	}
	inline int get() {
		long long x = static_cast<long long> (B - A + mod2) % mod2 * inv_1 % mod2 * mod1 + A;
		return (static_cast<long long> (C - x % mod3 + mod3) % mod3 * inv_2 % mod3 * (mod_1_2 % mod) % mod + x) % mod;
	}
} ;

#define maxn 131072

namespace Poly {
#define N (maxn << 1)
	int lim, s, rev[N];
	Int Wn[N | 1];
	inline void init(int n) {
		s = -1, lim = 1; while (lim < n) lim <<= 1, ++s;
		for (register int i = 1; i < lim; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
		const Int t(Math::pw(G, (mod1 - 1) / lim, mod1), Math::pw(G, (mod2 - 1) / lim, mod2), Math::pw(G, (mod3 - 1) / lim, mod3));
		*Wn = Int(1); for (register Int *i = Wn; i != Wn + lim; ++i) *(i + 1) = *i * t;
	}
	inline void NTT(Int *A, const int op = 1) {
		for (register int i = 1; i < lim; ++i) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
		for (register int mid = 1; mid < lim; mid <<= 1) {
			const int t = lim / mid >> 1;
			for (register int i = 0; i < lim; i += mid << 1) {
				for (register int j = 0; j < mid; ++j) {
					const Int W = op ? Wn[t * j] : Wn[lim - t * j];
					const Int X = A[i + j], Y = A[i + j + mid] * W;
					A[i + j] = X + Y, A[i + j + mid] = X - Y;
				}
			}
		}
		if (!op) {
			const Int ilim(Math::inv(lim, mod1), Math::inv(lim, mod2), Math::inv(lim, mod3));
			for (register Int *i = A; i != A + lim; ++i) *i = (*i) * ilim;
		}
	}
#undef N
}

int n, m;
Int A[maxn << 1], B[maxn << 1];
int main() {
	scanf("%d%d%d", &n, &m, &mod); ++n, ++m;
	for (int i = 0, x; i < n; ++i) scanf("%d", &x), A[i] = Int(x % mod);
	for (int i = 0, x; i < m; ++i) scanf("%d", &x), B[i] = Int(x % mod);
	Poly::init(n + m);
	Poly::NTT(A), Poly::NTT(B);
	for (int i = 0; i < Poly::lim; ++i) A[i] = A[i] * B[i];
	Poly::NTT(A, 0);
	for (int i = 0; i < n + m - 1; ++i) {
		printf("%d", A[i].get());
		putchar(i == n + m - 2 ? '\n' : ' ');
	}
	return 0;
}

多项式快速幂

给定多项式 $G$ ，求一个 $F$ ，使得 $F=G^k$

比较好推， $\ln F=k\ln G$ , $F=e^{k\ln G}$

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=4e5+10;
const int p=998244353;
typedef long long ll;
int cnt[N];
ll A[N],B[N],inv[N],w[25][N];
ll Add(ll x,ll y){
	x+=y;return x>=p?x-p:x;
}
ll Del(ll x,ll y){
	x-=y;return x>=0?x:x+p;
}
ll Mul(ll x,ll y){
	x*=y;return x>=p?x%p:x;
}
ll fpow(ll a,ll b){
	ll res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int r[N];
ll t;
void ntt(int n,ll *a,int typ){
	for(int i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(int mid=1,t0=0;mid<n;t0++,mid<<=1)
	for(int i=0,R=mid<<1;i<n;i+=R)
	for(int j=0;j<mid;j++){
		t=Mul(a[i+j+mid],w[t0][j]);
		a[i+j+mid]=Del(a[i+j],t);
		a[i+j]=Add(a[i+j],t);
	}
	if(typ==1)return;
	ll inv=fpow(n,p-2);
	reverse(a+1,a+n);
	for(int i=0;i<n;i++)a[i]=Mul(a[i],inv);
}
struct Poly{
	ll A[N],B[N];
	void dfs_inv(ll *a,ll *b,int n){
		if(n==1){a[0]=fpow(b[0],p-2);return;}
		dfs_inv(a,b,n+1>>1);
		int Max=1,l=0;
		while(Max<n+n)Max<<=1,l++;
		for(int i=0;i<Max;i++){
			r[i]=(r[i>>1]>>1)|((i&1)<<l-1);
			A[i]=i<n?b[i]:0;
		}
		ntt(Max,A,1);ntt(Max,a,1);
		for(int i=0;i<Max;i++)a[i]=Del(Mul(2,a[i]),Mul(Mul(a[i],a[i]),A[i]));
		ntt(Max,a,-1);
		for(int i=n;i<Max;i++)a[i]=0;
	}
	void Ln(ll *a,ll *b,int n){
		dfs_inv(B,b,n);
		memcpy(A,b,sizeof(A));
		for(int i=1;i<n;i++)
			A[i-1]=Mul(i,A[i]);
		A[n-1]=0;
		int Max=1,l=0;
		while(Max<n+n)Max<<=1,l++;
		for(int i=0;i<Max;i++)
			r[i]=(r[i>>1]>>1)|((i&1)<<l-1),a[i]=0;
		ntt(Max,A,1);ntt(Max,B,1);
		for(int i=0;i<Max;i++)
			A[i]=Mul(A[i],B[i]),B[i]=0;
		ntt(Max,A,-1);
		for(int i=n-1;i;i--)
			A[i]=Mul(A[i-1],fpow(i,p-2));
		A[0]=0;
		for(int i=0;i<n;i++)
			a[i]=A[i];
	}
}T;
ll C[N];
void Exp(ll *a,ll *b,int n){
	if(n==1){a[0]=1;return;}
	Exp(a,b,n+1>>1);
	T.Ln(C,a,n);
	int Max=1,l=0;
	while(Max<n+n)Max<<=1,l++;
	for(int i=0;i<Max;i++){
		r[i]=(r[i>>1]>>1)|((i&1)<<l-1);
		if(i<n)C[i]=Del(b[i],C[i]);
	}
	C[0]++;
	ntt(Max,a,1);ntt(Max,C,1);
	for(int i=0;i<Max;i++)a[i]=Mul(a[i],C[i]);
	ntt(Max,a,-1);
	for(int i=n;i<Max;i++)a[i]=0;
}
ll D[N];
int main(){
	int n;
	scanf("%d",&n);
	char ch=getchar();int k=0;
	while(ch<'0'||ch>'9')ch=getchar();
	while(ch<='9'&&ch>='0')k=Add(ch-'0',Mul(k,10)),ch=getchar();
	for(int i=0;i<n;i++)scanf("%lld",&A[i]);
	for(int i=0,k=1;k<=n+n;k<<=1,i++){
		w[i][0]=1;w[i][1]=fpow(3,(p-1)/(k*2));
		for(int j=2;j<k;j++)
			w[i][j]=Mul(w[i][j-1],w[i][1]);
	}
	T.Ln(B,A,n);
	for(int i=0;i<n;i++)B[i]=Mul(B[i],k);
	Exp(D,B,n);
	for(int i=0;i<n;i++)
		printf("%lld ",D[i]);
	return 0;
}

多项式复合

给定两个多项式 $F,G$ ，求一个多项式 $H(x)\equiv F(G(x)) \pmod {x^n}$ 。

考虑暴力做，依次考虑 $G^i$ 的贡献，与 $F$ 的系数相乘后加入到答案里边，时间复杂度 $O(n^2logn)$ ，常数略大。

事实上没有必要把他们都乘出来

H=\sum_{i=0}^{n-1}[x^i]FG^i

H=\sum_{i=0}^{L}\sum_{j=0}^{L-1}[x^{iL+j}]FG^{iL+j}

H=\sum_{i=0}^{L}G^{iL}\sum_{j=0}^{L-1}[x^{iL+j}]FG^{j}

令 $L=\sqrt n$ ，这样只需要预处理出 $G^{iL},G^{i}$ 即可，时间复杂度 $O(n^2+n\sqrt nlogn)$ ，实际上 $n^2$ 的部分常数比较小所以跑的飞快。

#include<cstdio>
#include<cmath>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<18;
const int p=998244353;
const int inv2=p+1>>1;
#define rint register int
#define rll register long long 
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline int Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
inline int fpow(rint a,rint b){
	rint res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int w[N],r[N],pd,t;
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd^n){
		for(rint i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
		pd=n;
	}
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint R=mid<<1,i=0;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=Mul(w[mid+j],a[i+j+mid]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(typ==1)return;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)a[i]=Mul(a[i],inv);
}
vector<int> Mul(vector<int> a,vector<int> b){
	rint n=a.size(),m=b.size(),Max=1;
	while(Max<n+m)Max<<=1;
	ntt(Max,a,1);ntt(Max,b,1);
	for(rint i=0;i<Max;i++)a[i]=Mul(a[i],b[i]);
	ntt(Max,a,-1);
	return a;
}
vector<int> vec[210],vec2[210],ans;
int main(){
	rint n,m;
	w[N>>1]=1;w[N/2+1]=n=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],n);
	for(rint i=N/2-1;~i;i--)w[i]=w[i<<1];
	scanf("%d%d",&n,&m);++n;++m;
	vector<int> f,g;ans.resize(n);
	f.resize(n);g.resize(m);
	for(rint i=0;i<n;i++)scanf("%d",&f[i]);
	for(rint i=0;i<m;i++)scanf("%d",&g[i]);
	int sqr=sqrt(n)+1;
	vec[0].resize(n);vec[0][0]=1;
	for(int i=1;i<=sqr;i++){
		vec[i]=Mul(vec[i-1],g);
		vec[i].resize(n);
	}
	vec2[0].resize(n);vec2[0][0]=1;
	for(int i=1;i<=sqr;i++){
		vec2[i]=Mul(vec2[i-1],vec[sqr]);
		vec2[i].resize(n);
	}
	for(rint i=0;i<sqr;i++){
		vector<int> now;
		now.clear();now.resize(n);
		for(rint j=0;j<sqr;j++){
			if(i*sqr+j>=n)break;
			rint val=f[i*sqr+j];
			for(rint k=0;k<n;k++){
				now[k]=Add(now[k],Mul(val,vec[j][k]));
			}
		}
		rint Max=1;
		while(Max<n+n)Max<<=1;
		ntt(Max,now,1);ntt(Max,vec2[i],1);
		for(rint j=0;j<Max;j++)now[j]=Mul(vec2[i][j],now[j]);
		ntt(Max,now,-1);now.resize(n);
		for(rint i=0;i<n;i++)
			ans[i]=Add(ans[i],now[i]);
	}
	for(rint i=0;i<n;i++)
		printf("%d ",ans[i]);
	return 0;
}

多项式复合逆

给定多项式 $G$ ，求一个多项式 $F$ ，使得 $G(F(x))=x$ 。大概考虑一下这个东西有什么意义，比如有一个函数 $H(G(x))=P(x)$ ，那么代入复合逆就能够得到， $H(x)=P(F(x))$ ，只需要多项式复合就能够求出 $H$ 。

#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<15;
const int p=998244353;
#define rint register int
#define rll register long long 
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline int Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
int fpow(rint a,rint b){
	rint res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int r[N],w[N],t,pd;
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd^n){
		for(rint i=0;i<n;i++)
			r[i]=(r[i>>1]>>1)|((i&1)?(n>>1):0);
		pd=n;
	}
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint i=0,R=mid<<1;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=Mul(a[i+j+mid],w[mid+j]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(~typ)return ;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)
		a[i]=Mul(a[i],inv);
}
vector<int> vec[N],vec2[N];
void dfs_inv(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=fpow(g[0],p-2);return;}
	dfs_inv(f,g,n+1>>1);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);f.resize(n);
}
vector<int> Mul(vector<int> a,vector<int> b){
	rint n=a.size(),m=b.size();
	rint Max=1;
	while(Max<n+m)Max<<=1;
	ntt(Max,a,1);ntt(Max,b,1);
	for(rint i=0;i<Max;i++)
		a[i]=Mul(a[i],b[i]);
	ntt(Max,a,-1);
	return a;
}
int main(){
	w[N/2]=1;w[N/2+1]=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],w[N/2+1]);
	for(rint i=N/2-1;i;i--)w[i]=w[i<<1];
	rint n;
	vector<int> a;
	scanf("%d",&n);
	a.resize(n);
	rint sqr=sqrt(n)+1;
	scanf("%*d");--n;
	for(rint i=0;i<n;i++)
		scanf("%d",&a[i]);
	dfs_inv(a,a,n);
	//for(rint v:a)printf("xx%d\n",v);
	vec[0].resize(n);vec[0][0]=1;
	vec2[0].resize(n);vec2[0][0]=1;
	for(rint i=1;i<=sqr;i++){
		vec[i]=Mul(vec[i-1],a);
		vec[i].resize(n);
	}
	for(rint i=1;i<=sqr;i++){
		vec2[i]=Mul(vec2[i-1],vec[sqr]);
		vec2[i].resize(n);
	}
	printf("0 ");
	for(rint i=0;i<=sqr;i++){
		for(rint j=1;j<=sqr;j++){
			rint x=i*sqr+j-1;
			if(x>=n)return 0;
			rint res=0;
			for(rint k=0;k<=x;k++)
				res=Add(res,Mul(vec[j][k],vec2[i][x-k]));
			printf("%d ",Mul(res,fpow(x+1,p-2)));
		}
	}
	return 0;
}

拉格朗日反演

如果有 $F(G(x))=x$ ，即 $F,G$ 互为复合逆，同时一定有 $G(F(x))=x$ ，可以称 $G(x)=F^{-1}(x)$ , $F(x)=G^{-1}(x)$ 。

在这种情况下，有这样的式子：

[x^n]F(x)=\frac{1}n[x^{n-1}](\frac{x}{G(x)})^n

[x^n]H(F(x))=\frac{1}n[x^{n-1}]H'(x)(\frac{x}{G(x)})^n

可以 $nlogn$ 的求出一项系数。

考虑

F=\sum_{i=1}^{n}\frac{1}i[x^{i-1}](\frac{x}{G(x)})^i

F=\sum_{i=0}^{L}\sum_{j=1}^{L}\frac{1}{iL+j}[x^{iL+j-1}](\frac{x}{G(x)})^{iL+j}

F=\sum_{i=0}^{L}\sum_{j=1}^{L}\frac{1}{iL+j}[x^{iL+j-1}](\frac{x}{G(x)})^{iL}(\frac{x}{G(x)})^{j}

令 $L=\sqrt n$ ，然后预处理一下后边的式子，可以 $O(n)$ 求出两个多项式乘积的某一项，时间复杂度同上。

有标号连通图计数

考虑一个简单无向图的个数，显然是 $2^{\binom{i}{2}}$ ，设其 $EGF$ 为 $F$

一个无向图是由若干连通图拼在一起的，设连通图的 $EGF$ 为 $G$ ，枚举联通块个数，有

F=\sum_{i=0}^{+\infty}\frac{G^i}{i!}

除掉阶乘是因为连通块之间没有区别。

所以 $F=e^G$ ， $G=\ln F$

边双连通图计数

求大小为 $n$ 的边双连通图的数量。

由于边双联通图没有一个固定的点，所以不大好做，考虑将其转化为有根的边双，其实就是加入了一个特殊点方便计数而已。

容易发现，设无根边双的 $EGF$ 为 $P$ ，有根边双为 $D$ ，那么 $D_i=iP_i$ ，原因是每个点都成为过根。

考虑有根边双怎么求，一张有根连通图一定是由若干个边双联通分量构成的，考虑根所在的边双联通分量，除去这个边双联通分量后，原图一定会分为若干个有根联通图，枚举根所在的连通分量的大小和划分成的有根连通图的个数，设有根联通图的 $EGF$ 为 $F$ ，有

F=\sum_{i=1}^{+\infty}\frac{D_ix^i\sum_{j=0}^{+\infty}\frac{F^ji^j}{j!}}{i!}

由

\frac{F^ji^j}{j!}=\frac{(Fi)^j}{j!}=e^{Fi}

所以

F=D(xe^F)

$F$ 显然比较好求，考虑 $D$ 怎么求，令 $H=xe^F$ ，有 $F(x)=D(H(x))$ ，两边同时带入 $H$ 的复合逆，有， $F(H^{-1}(x))=D(x)$ 。

所以只需要求出 $F(H^{-1})$ ，由拉格朗日反演

[x^i]F(H^{-1})=\frac{1}i[x^{i-1}]F'(\frac{x}{H})^i

代入 $H=xe^F$ ，有

[x^i]F(H^{-1})=\frac{1}i[x^{i-1}]F'e^{-iF}

于是可以 $nlogn$ 求出 $D$ 的一项。

#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<18;
const int p=998244353;
#define rint register int
#define rll register long long
#define ll long long
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline ll Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
inline ll fpow(rll a,rll b){
	rll res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int w[N],r[N],t,pd,fac[N],inv[N];
unsigned long long A[N];
vector<int> G;
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd!=n){
		pd=n;
		for(rint i=0;i<n;i++)
			r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
	}
	for(rint i=0;i<n;i++)A[i]=a[r[i]];
	for(rint mid=1;mid<n;mid<<=1)
	for(rint i=0,R=mid<<1;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=A[i+j+mid]*w[mid+j]%p,A[i+j+mid]=A[i+j]-t+p,A[i+j]+=t;
	for(rint i=0;i<n;i++)a[i]=A[i]%p;
	if(typ==1)return ;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)a[i]=Mul(inv,a[i]);
}
void dfs_inv(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){
		f.resize(1);f[0]=fpow(g[0],p-2);
		return ;
	}
	dfs_inv(f,g,n+1>>1);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);f.resize(n);
}
void Ln(vector<int> &f,vector<int> g,rint n){
	f.resize(n);g.resize(n);
	for(rint i=1;i<n;i++)
		f[i-1]=Mul(i,g[i]);
	f[n-1]=0;
	dfs_inv(g,g,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);
	for(rint i=n-1;i;i--)
		f[i]=Mul(f[i-1],fpow(i,p-2));
	f[0]=0;f.resize(n);
}
void Exp(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){
		f.resize(1);f[0]=1;return ;
	}
	Exp(f,g,n+1>>1);
	vector<int> c=f;
	Ln(c,c,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	g.resize(Max);c.resize(Max);
	for(rint i=0;i<Max;i++)
		g[i]=Del(g[i],c[i]);
	g[0]++;
	ntt(Max,g,1);ntt(Max,f,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);f.resize(n);
}
vector<int> D;
int solve(rint n){
	vector<int> f=D,g=G;
	for(rint i=0;i<n;i++)
		g[i]=p-Mul(g[i],n);
	Exp(g,g,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);
	return Mul(fac[n-1],Mul(f[n-1],fpow(n,p-2)));
}
int main(){
	w[N/2]=1;w[N/2+1]=t=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],t);
	for(rint i=N/2-1;i;i--)w[i]=w[i<<1];
	fac[0]=fac[1]=inv[0]=inv[1]=1;
	for(rint i=2;i<N;i++){
		fac[i]=Mul(fac[i-1],i);
		inv[i]=p-Mul(p/i,inv[p%i]);
	}
	for(rint i=2;i<N;i++)inv[i]=Mul(inv[i-1],inv[i]);
	G.resize(1e5+1);
	for(rint i=0;i<=1e5;i++)
		G[i]=Mul(fpow(2,1ll*(i-1)*i/2%(p-1)),inv[i]);
	Ln(G,G,1e5+1);
	D.resize(1e5+1);
	for(rint i=0;i<=1e5;i++)
		G[i]=Mul(i,G[i]);
	for(rint i=1;i<=1e5;i++){
		D[i-1]=Mul(i,G[i]);
	}
	for(rint i=1;i<=5;i++){
		rint x;
		scanf("%d",&x);
		printf("%d\n",solve(x));
	}
	return 0;
}

多项式多点求值

给定一个 $n$ 次多项式和 $m$ 个值，求出 $F(a_1),F(a_2),……F(a_m)$ 。

构造两个多项式 $P=\prod_{i=1}^{\frac{n}2}(x-a_i),Q=\prod_{i=\frac{n}2+1}^n(x-a_i)$ ，那么如果令 $F=P*A+B$ ，对于任意 $[1,\frac{n}2]$ 中的点， $F(a_i)=B(a_i)$ ，原因是代入后 $P$ 为 $0$ ，所以就有了理论上是 $nlogn$ 实际上有大常数的多项式取模写法。

实际上取模的做法貌似已经成了时代的眼泪，我们现在已经有了更快的转置做法。

考虑一种减法卷积，定义 $c=Mult(a,b)$ 即 $c_i=\sum_{j=0}^{+\infty}a_jb_{j-i}$ 。

它有一个性质是， $Mult(Mult(a,b),c)=Mult(a,b*c)$ ，可以利用这个性质进行多点求值。

将 $F(a_j)$ 展开，得到

F(a_j)=\sum_{i=0}^{n}[x_i]Fa_j^i

观察发现，如果构造一个多项式 $G_j=\sum_{i=0}^{n}a_j^ix^i$ ，那么 $ F(a_j) $ 实际上是 $Mult(F,G_j)$ 的常数项，证明很简单，代入即可。

$G$ 是一个 $n$ 次的多项式，暴力做的话是 $nmlogn$ ，不如直接乘，考虑优化，将 $G$ 写成封闭形式， $G_j=\frac{1}{1-a_jx}$ ，这样就可以 $O(1)$ 的算出每一个 $G_j$ 的分母，但是还是没啥用，因为乘起来还是要 $O(nlogn)$ ，最后我们要的是每一个 $Mult(F,G_j)$ 的常数项，考虑求出 $\prod_{j=1}^mG_j$ ，可以用分治乘法和求逆求出，直接令 $F=Mult(F,\prod_{j=1}^mG_j)$ ，再一次进行分治，如果进去左区间就把右区间的贡献去掉，具体的，设 $T=\prod_{j=\frac{n}2+1}^{n}\frac{1}{G_j}$ ，只需要令 $F=Mult(F,T)$ ，进入右区间时同理，原因是暴力做是对于右区间的每一个 $j$ ，进行 $Mult(Mult(F,\frac{1}{G_j}),\frac{1}{G_j})$ ，由 $Mult(Mult(a,b),c)=Mult(a,b*c)$ 知，可以先对右区间卷积起来一起做减法卷积。

然后就有了小常数的多点求值。

多项式快速插值

给定 $n+1$ 个点值，要求一个 $n$ 次多项式。

考虑拉格朗日插值法，有

F=\sum_{i=0}^ny_i\prod_{i!=j}\frac{x-x_j}{x_i-x_j}

这个式子为什么是对的，其实很简单，发现带入任意 $x_i$ 都成立，然后就没了。

但是直接做是 $O(n^2)$ 的，将其进行简单变形。

F=\sum_{i=0}^n\frac{y_i}{\prod_{i!=j}x_i-x_j}\prod_{i!=j}{x-x_j}

考虑带有 $y_i$ 的那一项，容易发现只需要求出下边的即可，有点难求于是构造一个多项式 $G=\prod_{i=0}^nx-x_i$ ，只需要求出 $\frac{G(x)}{x-x_i}$ ，这个东西看起来好求但是实际上不是很好玩，因为分子分母在 $x=x_i$ 的时候都是 $0$ 。

这时候就要用到洛必达法则了。

众所周知，两个无穷小之比或两个无穷大之比的极限可能存在，也可能不存在。因此，求这类极限时往往需要适当的变形，转化成可利用极限运算法则或重要极限的形式进行计算。洛必达法则便是应用于这类极限计算的通用方法。

洛必达法则：对于 $\frac{f(x)}{g(x)}$ ，若 $\lim\limits_{x→a}f(x)=0,\lim\limits_{x→a}g(x)=0$ ，那么 $\lim\limits_{x→a}\frac{f(x)}{g(x)}=\lim\limits_{x→a}\frac{f'(x)}{g'(x)}$ 。

也就是说，当 $x=x_i$ 时，$\frac{G(x)}{x-x_i}=\frac{G’(x)}{(x-x_i)‘}=G’(x) $

于是有，

F=\sum_{i=0}^n\frac{y_i}{G'(x_i)}\prod_{i!=j}{x-x_j}

使用多项式多点求值即可求出所有的 $G'$ ，不妨令 $y_i=\frac{y_i}{G'(x_i)}$ 。

那么实际上要求的是

F=\sum_{i=0}^ny_i\prod_{i!=j}{x-x_j}

令

F_{l,r}=\sum_{i=l}^ry_i\prod_{i!=j}{x-x_j}

直接暴力拆开，

F_{l,r}=\sum_{i=l}^{mid}y_i\prod_{i!=j}{x-x_j}+\sum_{i=mid+1}^ry_i\prod_{i!=j}{x-x_j}

F_{l,r}=\sum_{i=l}^{mid}y_i\prod_{i=l,i!=j}^{mid}{x-x_j}\prod_{i=mid+1}^nx-x_j+\sum_{i=mid+1}^ry_i\prod_{i=mid+1,i!=j}^n{x-x_j}\prod_{i=l}^{mid}x-x_j

F_{l,r}=F_{l,mid}\prod_{i=mid+1}^nx-x_j+F_{mid+1,r}\prod_{i=l}^{mid}x-x_j

自底向上乘即可。

下降幂多项式乘法

给定下降幂多项式 $F,G$ ，求 $F*G$ 。

首先考虑一个下降幂单项式 $x^{\underline n}$ 的点值 $EGF$

\sum_{i=n}^{+\infty}\frac{i^\underline n}{i!}x^i

\sum_{i=n}^{+\infty}\frac{1}{(i-n)!}x^i

\sum_{i=0}^{+\infty}\frac{1}{i!}x^{i+n}

即 $e^xx^{n}$ ，那么只需要将 $F,G$ 乘上 $e^x$ 即可转化为点值的 $EGF$ ，之后直接用点值相乘即可，注意得到的是点值的 $EGF$ 所以相乘的时候要乘上一个阶乘。

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<19;
const int p=998244353;
const int inv2=p+1>>1;
#define rint register int
#define rll register long long 
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline int Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
inline int fpow(rint a,rint b){
	rint res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int w[N],r[N],pd,t,inv[N],fac[N];
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd^n){
		for(rint i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
		pd=n;
	}
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint R=mid<<1,i=0;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=Mul(w[mid+j],a[i+j+mid]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(typ==1)return;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)a[i]=Mul(a[i],inv);
}
int main(){
	rint n,m;
	w[N>>1]=1;w[N/2+1]=n=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],n);
	for(rint i=N/2-1;~i;i--)w[i]=w[i<<1];
	vector<int> a,b,e,inve;
	scanf("%d%d",&n,&m);++n;++m;
	a.resize(n);b.resize(m);
	for(rint i=0;i<n;i++)
		scanf("%d",&a[i]);
	for(rint i=0;i<m;i++)
		scanf("%d",&b[i]);
	e.resize(n+m);inve.resize(n+m);
	inv[0]=inv[1]=fac[0]=fac[1]=1;
	for(rint i=2;i<N;i++){
		inv[i]=p-1ll*p/i*inv[p%i]%p;
		fac[i]=Mul(fac[i-1],i);
	}
	for(rint i=2;i<N;i++)
		inv[i]=Mul(inv[i],inv[i-1]);
	for(rint i=0;i<n+m;i++)
		e[i]=inv[i],inve[i]=(i&1)?p-inv[i]:inv[i];
	rint Max=1;
	while(Max<n*2+m*2)Max<<=1;
	ntt(Max,a,1);ntt(Max,b,1);
	ntt(Max,e,1);ntt(Max,inve,1);
	for(rint i=0;i<Max;i++)
		a[i]=Mul(a[i],e[i]),b[i]=Mul(b[i],e[i]);
	ntt(Max,a,-1);ntt(Max,b,-1);
	for(rint i=0;i<Max;i++)a[i]=Mul(fac[i],Mul(a[i],b[i]));
	for(rint i=n+m-1;i<Max;i++)a[i]=0;
	ntt(Max,a,1);
	for(rint i=0;i<Max;i++)a[i]=Mul(a[i],inve[i]);
	ntt(Max,a,-1);
	for(rint i=0;i<n+m-1;i++)
		printf("%d ",a[i]);
	return 0;
}

下降幂多项式转普通多项式

给定一个下降幂多项式，求一个与其相等的普通多项式。

直接将下降幂转化成点值的形式然后快速插值即可。

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<19;
const int p=998244353;
const int inv2=p+1>>1;
#define rint register int
#define rll register long long 
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline int Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
inline int fpow(rint a,rint b){
	rint res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int w[N],r[N],pd,t;
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd^n){
		for(rint i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
		pd=n;
	}
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint R=mid<<1,i=0;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=Mul(w[mid+j],a[i+j+mid]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(typ==1)return;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)a[i]=Mul(a[i],inv);
}
void dfs_inv(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=fpow(g[0],p-2);return;}
	dfs_inv(f,g,n+1>>1);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);f.resize(n);
}
void dfs_sqrt(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=1;return;}
	dfs_sqrt(f,g,n+1>>1);
	vector<int> a;
	dfs_inv(a,f,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,a,1);ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(Add(Mul(f[i],f[i]),g[i]),Mul(inv2,a[i]));
	ntt(Max,f,-1);f.resize(n);
}
void Ln(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	dfs_inv(f,g,n);
	for(rint i=1;i<n;i++)g[i-1]=Mul(g[i],i);
	g[n-1]=0;
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);
	f.resize(n);
	for(rint i=n-1;i;i--)f[i]=Mul(f[i-1],fpow(i,p-2));
	f[0]=0;
}
void Exp(vector<int> &f,vector<int> g,int n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=1;return;}
	Exp(f,g,n+1>>1);
	vector<int> a;
	Ln(a,f,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	g.resize(Max);a.resize(Max);
	for(rint i=0;i<Max;i++)a[i]=Del(g[i],a[i]);a[0]++;
	ntt(Max,a,1);ntt(Max,f,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],a[i]);
	ntt(Max,f,-1);f.resize(n);
}
struct Eval{
	vector<int> vec[N<<2],A,res,tmp;
	vector<int> Mult(vector<int> a,vector<int> b,rint n){
		rint Max=a.size();
		for(int i=0;i<Max;i++)
			a[i]=Mul(a[i],b[i]);
		ntt(Max,a,1);a.resize(n);
		return a;
	}
	void dfs1(rint x,rint l,rint r){
		if(l==r){
			vec[x].resize(2);
			vec[x][0]=1;
			vec[x][1]=p-tmp[l];
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1;
		dfs1(L,l,mid);dfs1(R,mid+1,r);
		rint Max=1,n=r-l+2;
		while(Max<n)Max<<=1;vec[x].resize(Max);
		ntt(Max,vec[L],1);ntt(Max,vec[R],1);
		for(rint i=0;i<Max;i++)
			vec[x][i]=Mul(vec[L][i],vec[R][i]);
		ntt(Max,vec[x],-1);vec[x].resize(n);
	}
	void dfs2(rint x,rint l,rint r,vector<int> v){
		rint n=r-l+2;
		if(l==r){
			res[l]=v[0];
			return;
		}
		rint Max=1,mid=l+r>>1;
		while(Max<n)Max<<=1;
		ntt(Max,v,-1);
		dfs2(x<<1,l,mid,Mult(v,vec[x<<1|1],mid-l+1));
		dfs2(x<<1|1,mid+1,r,Mult(v,vec[x<<1],r-mid));
	}
	vector<int> query(vector<int> A,vector<int> a){
		rint n=A.size(),m=a.size()-1;res.resize(m+1);tmp=a;
		dfs1(1,1,m);
		dfs_inv(vec[1],vec[1],n);
		rint Max=1;
		while(Max<n+n)Max<<=1;
		ntt(Max,vec[1],1);
		ntt(Max,A,-1);
		dfs2(1,1,m,Mult(A,vec[1],n));
		return res;
	}
}T;
struct Fin{
	vector<int> vec[N<<2],vec2[N<<2],tmp,tmp2;
	void dfs1(rint x,rint l,rint r){
		if(l==r){
			vec[x].resize(2);
			vec[x][0]=p-tmp[l];
			vec[x][1]=1;
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1,n=r-l+2;
		dfs1(L,l,mid);dfs1(R,mid+1,r);
		rint Max=1;
		while(Max<n)Max<<=1;
		ntt(Max,vec[L],1);ntt(Max,vec[R],1);
		vec[x].resize(Max);
		for(rint i=0;i<Max;i++)
			vec[x][i]=Mul(vec[L][i],vec[R][i]);
		ntt(Max,vec[x],-1);
		vec[x].resize(n);
	}
	void dfs2(rint x,rint l,rint r){
		if(l==r){
			vec2[x].resize(1);
			vec2[x][0]=Mul(tmp2[l-1],fpow(tmp[l],p-2));
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1,n=r-l+2;
		dfs2(x<<1,l,mid);dfs2(x<<1|1,mid+1,r);
		rint Max=1;
		while(Max<n)Max<<=1;
		vec2[x].resize(Max);
		ntt(Max,vec2[L],1);ntt(Max,vec2[R],1);
		for(rint i=0;i<Max;i++)
			vec2[x][i]=Add(Mul(vec2[L][i],vec[R][i]),Mul(vec2[R][i],vec[L][i]));
		ntt(Max,vec2[x],-1);vec2[x].resize(n-1);
	}
	vector<int> query(vector<int> x,vector<int> y){
		int n=x.size()-1;tmp=x;tmp2=y;
		dfs1(1,1,n);
		for(rint i=1;i<=n;i++)
			vec[1][i-1]=Mul(vec[1][i],i);
		vec[1][n]=0;
		tmp=T.query(vec[1],x);
		dfs2(1,1,n);
		return vec2[1];
	}
}F;
int inv[N],fac[N];
int main(){
	rint n,m;
	w[N>>1]=1;w[N/2+1]=n=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],n);
	for(rint i=N/2-1;~i;i--)w[i]=w[i<<1];
	scanf("%d",&n);
	vector<int> f,g;
	f.resize(n);g.resize(n);
	for(rint i=0;i<n;i++)scanf("%d",&f[i]);
	inv[0]=inv[1]=fac[1]=fac[0]=1;
	for(rint i=2;i<=n;i++)inv[i]=p-1ll*p/i*inv[p%i]%p;
	for(rint i=2;i<=n;i++){
		inv[i]=Mul(inv[i],inv[i-1]);
		fac[i]=Mul(fac[i-1],i);
	}
	for(rint i=0;i<n;i++)g[i]=inv[i];
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);f.resize(n);
	for(rint i=0;i<n;i++)f[i]=Mul(f[i],fac[i]);
	g.resize(n+1);
	for(rint i=1;i<=n;i++)g[i]=i-1;
	f=F.query(g,f);
	for(rint i=0;i<n;i++)
		printf("%d ",f[i]);
	return 0;
}

普通多项式转下降幂多项式

给定一个普通多项式，求一个与其相等的下降幂多项式。

使用多点求值求出下降幂的点值 $EGF$ ，然后乘上 $e^{-x}$ 即可。

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1<<18;
const int p=998244353;
const int inv2=p+1>>1;
#define rint register int
#define rll register long long 
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline int Mul(rll x,rint y){x*=y;return x>=p?x%p:x;}
inline int fpow(rint a,rint b){
	rint res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int w[N],r[N],pd,t;
void ntt(rint n,vector<int> &a,rint typ){
	a.resize(n);
	if(pd^n){
		for(rint i=0;i<n;i++)r[i]=(r[i>>1]>>1)|((i&1)?n>>1:0);
		pd=n;
	}
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint R=mid<<1,i=0;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=Mul(w[mid+j],a[i+j+mid]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(typ==1)return;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)a[i]=Mul(a[i],inv);
}
void dfs_inv(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=fpow(g[0],p-2);return;}
	dfs_inv(f,g,n+1>>1);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],Del(2,Mul(f[i],g[i])));
	ntt(Max,f,-1);f.resize(n);
}
void dfs_sqrt(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=1;return;}
	dfs_sqrt(f,g,n+1>>1);
	vector<int> a;
	dfs_inv(a,f,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,a,1);ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(Add(Mul(f[i],f[i]),g[i]),Mul(inv2,a[i]));
	ntt(Max,f,-1);f.resize(n);
}
void Ln(vector<int> &f,vector<int> g,rint n){
	g.resize(n);
	dfs_inv(f,g,n);
	for(rint i=1;i<n;i++)g[i-1]=Mul(g[i],i);
	g[n-1]=0;
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);
	f.resize(n);
	for(rint i=n-1;i;i--)f[i]=Mul(f[i-1],fpow(i,p-2));
	f[0]=0;
}
void Exp(vector<int> &f,vector<int> g,int n){
	g.resize(n);
	if(n==1){f.resize(1);f[0]=1;return;}
	Exp(f,g,n+1>>1);
	vector<int> a;
	Ln(a,f,n);
	rint Max=1;
	while(Max<n+n)Max<<=1;
	g.resize(Max);a.resize(Max);
	for(rint i=0;i<Max;i++)a[i]=Del(g[i],a[i]);a[0]++;
	ntt(Max,a,1);ntt(Max,f,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],a[i]);
	ntt(Max,f,-1);f.resize(n);
}
struct Eval{
	vector<int> vec[N<<2],A,res,tmp;
	vector<int> Mult(vector<int> a,vector<int> b,rint n){
		rint Max=a.size();
		for(int i=0;i<Max;i++)
			a[i]=Mul(a[i],b[i]);
		ntt(Max,a,1);a.resize(n);
		return a;
	}
	void dfs1(rint x,rint l,rint r){
		if(l==r){
			vec[x].resize(2);
			vec[x][0]=1;
			vec[x][1]=p-tmp[l];
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1;
		dfs1(L,l,mid);dfs1(R,mid+1,r);
		rint Max=1,n=r-l+2;
		while(Max<n)Max<<=1;vec[x].resize(Max);
		ntt(Max,vec[L],1);ntt(Max,vec[R],1);
		for(rint i=0;i<Max;i++)
			vec[x][i]=Mul(vec[L][i],vec[R][i]);
		ntt(Max,vec[x],-1);vec[x].resize(n);
	}
	void dfs2(rint x,rint l,rint r,vector<int> v){
		rint n=r-l+2;
		if(l==r){
			res[l]=v[0];
			return;
		}
		rint Max=1,mid=l+r>>1;
		while(Max<n)Max<<=1;
		ntt(Max,v,-1);
		dfs2(x<<1,l,mid,Mult(v,vec[x<<1|1],mid-l+1));
		dfs2(x<<1|1,mid+1,r,Mult(v,vec[x<<1],r-mid));
	}
	vector<int> query(vector<int> A,vector<int> a){
		rint n=A.size(),m=a.size()-1;res.resize(m+1);tmp=a;
		dfs1(1,1,m);
		dfs_inv(vec[1],vec[1],n);
		rint Max=1;
		while(Max<n+n)Max<<=1;
		ntt(Max,vec[1],1);
		ntt(Max,A,-1);
		dfs2(1,1,m,Mult(A,vec[1],n));
		return res;
	}
}T;
struct Fin{
	vector<int> vec[N<<2],vec2[N<<2],tmp,tmp2;
	void dfs1(rint x,rint l,rint r){
		if(l==r){
			vec[x].resize(2);
			vec[x][0]=p-tmp[l];
			vec[x][1]=1;
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1,n=r-l+2;
		dfs1(L,l,mid);dfs1(R,mid+1,r);
		rint Max=1;
		while(Max<n)Max<<=1;
		ntt(Max,vec[L],1);ntt(Max,vec[R],1);
		vec[x].resize(Max);
		for(rint i=0;i<Max;i++)
			vec[x][i]=Mul(vec[L][i],vec[R][i]);
		ntt(Max,vec[x],-1);
		vec[x].resize(n);
	}
	void dfs2(rint x,rint l,rint r){
		if(l==r){
			vec2[x].resize(1);
			vec2[x][0]=Mul(tmp2[l],fpow(tmp[l],p-2));
			return;
		}
		rint mid=l+r>>1,L=x<<1,R=x<<1|1,n=r-l+2;
		dfs2(x<<1,l,mid);dfs2(x<<1|1,mid+1,r);
		rint Max=1;
		while(Max<n)Max<<=1;
		vec2[x].resize(Max);
		ntt(Max,vec2[L],1);ntt(Max,vec2[R],1);
		for(rint i=0;i<Max;i++)
			vec2[x][i]=Add(Mul(vec2[L][i],vec[R][i]),Mul(vec2[R][i],vec[L][i]));
		ntt(Max,vec2[x],-1);vec2[x].resize(n-1);
	}
	vector<int> query(vector<int> x,vector<int> y){
		int n=x.size()-1;tmp=x;tmp2=y;
		dfs1(1,1,n);
		for(rint i=1;i<=n;i++)
			vec[1][i-1]=Mul(vec[1][i],i);
		vec[1][n]=0;
		tmp=T.query(vec[1],x);
		dfs2(1,1,n);
		return vec2[1];
	}
}F;
int inv[N];
int main(){
	rint n,m;
	w[N>>1]=1;w[N/2+1]=n=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],n);
	for(rint i=N/2-1;~i;i--)w[i]=w[i<<1];
	scanf("%d",&n);
	vector<int> f,g;
	f.resize(n);g.resize(n+1);
	for(rint i=0;i<n;i++)scanf("%d",&f[i]);
	inv[0]=inv[1]=1;
	for(rint i=1;i<=n;i++){
		g[i]=i-1;
		if(i!=1)inv[i]=p-1ll*p/i*inv[p%i]%p;
	}
	for(rint i=2;i<=n;i++)inv[i]=Mul(inv[i-1],inv[i]);
	f=T.query(f,g);
	for(rint i=0;i<f.size()-1;i++)f[i]=Mul(f[i+1],inv[i]);
	f.resize(n);
	g.resize(n);
	for(rint i=0;i<n;i++)g[i]=(i&1)?p-inv[i]:inv[i];
	rint Max=1;
	while(Max<n+n)Max<<=1;
	ntt(Max,f,1);ntt(Max,g,1);
	for(rint i=0;i<Max;i++)f[i]=Mul(f[i],g[i]);
	ntt(Max,f,-1);f.resize(n);
	for(rint i=0;i<n;i++)
		printf("%d ",f[i]);
	return 0;
}

Chirp Z-Transform

给定一个 $n$ 项多项式 $P(x)$ 以及 $c, m$ ，请计算 $P(c^0),P(c^1),\dots,P(c^{m-1})$ 。所有答案都对 $998244353$ 取模。

#include<vector>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define db double
#define ll long long
#define rint register int
#define rll register long long
#define ull unsigned long long
const int N=1<<22;
const int p=998244353;
const int sqr=1<<15;
inline int read(){
    rint x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline int Add(rint x,rint y){x+=y;return x>=p?x-p:x;}
inline int Del(rint x,rint y){x-=y;return x>=0?x:x+p;}
inline ll Mul(rll x,rll y){x*=y;return x>=p?x%p:x;}
inline ll fpow(rll a,rll b){
	rll res=1;
	for(;b;b>>=1,a=Mul(a,a))
		if(b&1)res=Mul(res,a);
	return res;
}
int w[N],r[N],t,pd;
void ntt(rint n,vector<int>&a,rint typ){
	a.resize(n);
	for(rint i=0;i<n;i++)
		if(i<r[i])swap(a[i],a[r[i]]);
	for(rint mid=1;mid<n;mid<<=1)
	for(rint i=0,R=mid<<1;i<n;i+=R)
	for(rint j=0;j<mid;j++)
		t=Mul(a[i+j+mid],w[mid+j]),a[i+j+mid]=Del(a[i+j],t),a[i+j]=Add(a[i+j],t);
	if(~typ)return ;
	rint inv=fpow(n,p-2);
	reverse(a.begin()+1,a.end());
	for(rint i=0;i<n;i++)
		a[i]=Mul(a[i],inv);
}
int pw1[sqr+1],pw2[sqr+1];
inline int pw(rint k){
	return Mul(pw1[k%sqr],pw2[k/sqr]);
}
int main(){
	w[N/2]=1;w[N/2+1]=t=fpow(3,(p-1)/N);
	for(rint i=N/2+2;i<N;i++)w[i]=Mul(w[i-1],t);
	for(rint i=N/2-1;i;i--)w[i]=w[i<<1];
	rint n=read(),c=read(),m=read();
	pw1[0]=pw2[0]=1;
	for(rint i=1;i<=sqr;i++)pw1[i]=Mul(pw1[i-1],c);
	for(rint i=1;i<=sqr;i++)pw2[i]=Mul(pw2[i-1],pw1[sqr]);
	vector<int> f,g;f.resize(n),g.resize(n+m);
	for(rint i=0;i<n;i++){
		f[i]=read();
		f[i]=Mul(f[i],pw(p-1-1ll*(i-1)*i/2%(p-1)));
	}
	for(rint i=0;i<n+m;i++)
		g[i]=pw(1ll*i*(i-1)/2%(p-1));
	rint Max=1;
	while(Max<n+m)Max<<=1;
	for(rint i=0;i<Max;i++)
		r[i]=(r[i>>1]>>1)|((i&1)?Max>>1:0);
	ntt(Max,g,-1);ntt(Max,f,1);
	for(rint i=0;i<Max;i++)
		f[i]=Mul(f[i],g[i]);
	ntt(Max,f,1);
	for(rint i=0;i<m;i++)
		printf("%d ",Mul(pw(p-1-1ll*(i-1)*i/2%(p-1)),f[i]));
	return 0;
}

常系数齐次线性递推

#include<cstdio>
#include<algorithm>
using namespace std;const int N=65536+10;typedef long long ll;const ll mod=998244353;
int n;int k;int rv[20][N];ll rt[20][20];int Len;ll tr1[N];ll tr2[N];long long st[N];long long xs[N];
ll sg[N];ll a[N];ll res[N];ll irg[N];ll q[N];ll rf[N];int DL=-1;ll ans=0;ll ret[N];
inline ll po(ll a,ll p){ll r=1;for(;p;p>>=1,a=a*a%mod)if(p&1)r=r*a%mod;return r;}
inline void ntt(ll* a,int o,int len,int d)//ntt
{
    for(int i=0;i<len;i++)if(i<rv[d][i])swap(a[i],a[rv[d][i]]);
    for(int k=1,j=1;k<len;k<<=1,j++)
        for(int s=0;s<len;s+=(k<<1))
            for(int i=s,w=1;i<s+k;i++,w=w*rt[o][j]%mod)
            {ll a0=a[i];ll a1=a[i+k]*w%mod;a[i]=(a0+a1)%mod,a[i+k]=(a0+mod-a1)%mod;}
    if(o==1){ll inv=po(len,mod-2);for(int i=0;i<len;i++)(a[i]*=inv)%=mod;}
}
inline void poly_inv(ll* a,ll* b,int len)//求逆
{
    b[0]=po(a[0],mod-2);
    for(int k=1,j=0;k<=len;k<<=1,j++)
    {
        for(int i=0;i<k;i++)tr1[i]=a[i];for(int i=0;i<k;i++)tr2[i]=b[i];
        ntt(tr1,0,k<<1,j);ntt(tr2,0,k<<1,j);
        for(int i=0;i<(k<<1);i++)b[i]=tr2[i]*(2+mod-tr1[i]*tr2[i]%mod)%mod;
        ntt(b,1,k<<1,j);for(int i=k;i<(k<<1);i++)b[i]=0;
    }
}
inline void poly_mod(ll* a)//取模
{
    int mi=(k<<1);while(a[--mi]==0);if(mi<k)return;
    for(int i=0;i<(Len<<1);i++)rf[i]=0;for(int i=0;i<=mi;i++)rf[i]=a[i];
    reverse(rf,rf+mi+1);for(int i=mi-k+1;i<=mi;i++)rf[i]=0;ntt(rf,0,Len<<1,DL+1);
    for(int i=0;i<(Len<<1);i++)q[i]=(rf[i]*irg[i])%mod;ntt(q,1,(Len<<1),DL+1);
    for(int i=mi-k+1;i<=(Len<<1);i++)q[i]=0;reverse(q,q+mi-k+1);ntt(q,0,(Len<<1),DL+1);
    for(int i=0;i<(Len<<1);i++)(q[i]*=sg[i])%=mod;ntt(q,1,(Len<<1),DL+1);
    for(int i=0;i<k;i++)(a[i]+=mod-q[i])%=mod;for(int i=k;i<=mi;i++)a[i]=0;
}
int main()
{
    for(int i=0;i<=15;i++)
        for(int j=0;j<(1<<(i+1));j++)rv[i][j]=(rv[i][j>>1]>>1)|((j&1)<<i);
    for(int t=2,j=1;j<=18;t<<=1,j++)rt[0][j]=po(3,(mod-1)/t);
    for(int t=2,j=1;j<=18;t<<=1,j++)rt[1][j]=po(332748118,(mod-1)/t);
    scanf("%d%d",&n,&k);
    for(Len=1;Len<=k;Len<<=1,DL++); //预处理
    for(int i=1;i<=k;i++){scanf("%lld",&xs[i]);xs[i]=xs[i]<0?xs[i]+mod:xs[i];}
    for(int i=0;i<k;i++){scanf("%lld",&st[i]);st[i]=st[i]<0?st[i]+mod:st[i];}
    for(int i=1;i<=k;i++)sg[k-i]=mod-xs[i];sg[k]=1;for(int i=0;i<=k;i++)ret[i]=sg[i];
    for(int i=0;i<=k;i++)rf[i]=sg[i];reverse(rf,rf+k+1);poly_inv(rf,irg,Len);
    for(int i=0;i<=k;i++)rf[i]=0;ntt(sg,0,Len<<1,DL+1);ntt(irg,0,Len<<1,DL+1);a[1]=1;res[0]=1;
    while(n)//快速幂
    {
        if(n&1)
        {
            ntt(res,0,Len<<1,DL+1);ntt(a,0,Len<<1,DL+1);
            for(int i=0;i<(Len<<1);i++)(res[i]*=a[i])%=mod;
            ntt(res,1,Len<<1,DL+1);ntt(a,1,Len<<1,DL+1);poly_mod(res);
        }ntt(a,0,Len<<1,DL+1);for(int i=0;i<(Len<<1);i++)(a[i]*=a[i])%=mod;
        ntt(a,1,Len<<1,DL+1);poly_mod(a);n>>=1;
    }for(int i=0;i<k;i++)(ans+=res[i]*st[i])%=mod;printf("%lld",ans);return 0;
}

常系数非齐次线性递推

#pragma GCC optimize ("unroll-loops")
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#define N 131077
#define p 998244353
#define ll long long
#define reg register
#define add(x,y) (x+y>=p?x+y-p:x+y)
#define dec(x,y) (x<y?x-y+p:x-y)
using namespace std;

inline void read(int &x){
	x = 0;
	char c = getchar();
	while(c<'0'||c>'9') c = getchar();
	while(c>='0'&&c<='9'){
		x = (x<<3)+(x<<1)+(c^48);
		c = getchar();
	}
}

void print(int x){
	if(x>9) print(x/10);
	putchar(x%10+'0');	
}

inline int power(int a,int t){
	int res = 1;
	while(t){
		if(t&1) res = (ll)res*a%p;
		a = (ll)a*a%p;
		t >>= 1; 
	}
	return res;
}

int rt[N],rev[N],inv[N],fac[N],ifac[N];
int siz;

inline int binom(int x,int y){
    if(x<y) return 0;
    return (ll)fac[x]*ifac[y]%p*ifac[x-y]%p;
}

void init(int n){
	int w,lim = 1;
	while(lim<=n) lim <<= 1,++siz;
	for(reg int i=1;i!=lim;++i) rev[i] = (rev[i>>1]>>1)|((i&1)<<(siz-1));
	w = power(3,(p-1)>>siz);
	fac[0] = fac[1] = ifac[0] = ifac[1] = inv[1] = rt[lim>>1] = 1;
	for(reg int i=(lim>>1)+1;i!=lim;++i) rt[i] = (ll)rt[i-1]*w%p;
	for(reg int i=(lim>>1)-1;i;--i) rt[i] = rt[i<<1];	
	for(reg int i=2;i<=lim;++i) inv[i] = (ll)(p-p/i)*inv[p%i]%p;
    for(reg int i=2;i<=lim;++i) ifac[i] = fac[i] = (ll)fac[i-1]*i%p;
    ifac[lim] = power(fac[lim],p-2);
    for(reg int i=lim-1;i;--i) ifac[i] = (ll)ifac[i+1]*(i+1)%p;
}

inline int getlen(int n){
	return 1<<(32-__builtin_clz(n));	
}

inline void NTT(int *f,int type,int lim){
	if(type==-1) reverse(f+1,f+lim);
	reg int x,shift = siz-__builtin_ctz(lim);
	static int a[N];
	for(reg int i=0;i!=lim;++i) a[rev[i]>>shift] = f[i];
	for(reg int mid=1;mid!=lim;mid<<=1){
		for(reg int j=0;j!=lim;j+=(mid<<1)){
			for(reg int k=0;k!=mid;++k){
				x = (ll)a[j|k|mid]*rt[mid|k]%p;
				a[j|k|mid] = dec(a[j|k],x);
				a[j|k] = add(a[j|k],x);
			}
		}
	}
	memcpy(f,a,lim<<2);
	if(type==1) return;
	x = inv[lim];
	for(reg int i=0;i!=lim;++i) f[i] = (ll)f[i]*x%p;
}

void inverse(const int *f,int *R,int n){
	static int g[N],h[N],q[N];
	memset(g,0,getlen(n<<1)+2<<2);
	int lim = 1,top = 0;
	int s[30];
	while(n){
		s[++top] = n;
		n >>= 1;
	}
	g[0] = power(f[0],p-2);
	while(top--){
		n = s[top+1];
		while(lim<=(n<<1)) lim <<= 1;
		memcpy(q,g,(n+1)<<2);
		memcpy(h,f,(n+1)<<2);
		memset(h+n+1,0,(lim-n)<<2);
		NTT(g,1,lim),NTT(h,1,lim);
		for(reg int i=0;i!=lim;++i) g[i] = (ll)g[i]*g[i]%p*h[i]%p;
		NTT(g,-1,lim);
		for(reg int i=0;i<=n;++i) g[i] = dec(add(q[i],q[i]),g[i]);
		memset(g+n+1,0,(lim-n)<<2);
	}
	memcpy(R,g,(n+1)<<2);
}

void divide(const int *f,const int *g,int n,int m,int *R){
	static int A[N],B[N];
    memcpy(A,f,(n+1)<<2);
	memcpy(B,g,(m+1)<<2);
    reverse(A,A+n+1);
    reverse(B,B+m+1);
    int tt = n-m,lim = getlen((n-m)<<1);
    for(reg int i=tt+1;i!=lim;++i) A[i] = 0;
    for(reg int i=min(m,tt)+1;i!=lim;++i) B[i] = 0;
    inverse(B,B,tt);
    NTT(A,1,lim),NTT(B,1,lim);
    for(reg int i=0;i!=lim;++i) A[i] = (ll)A[i]*B[i]%p;
    NTT(A,-1,lim);
    reverse(A,A+tt+1);
    memcpy(R,A,(tt+1)<<2);
}

void mod(const int *f,const int *g,int n,int m,int *R){
    if(n<m){
        memcpy(R,f,(n+1)<<2);
        return;
    }   
    static int A[N],B[N];
    memcpy(B,f,(n+1)<<2);
    int lim = getlen(n);
	divide(f,g,n,m,R);
	for(int i=0;i<=m;++i) A[i] = g[i];
	for(int i=m+1;i!=lim;++i) A[i] = 0;
	for(int i=n-m+1;i!=lim;++i) R[i] = 0;
	NTT(A,1,lim),NTT(R,1,lim);
    for(reg int i=0;i!=lim;++i) R[i] = (ll)A[i]*R[i]%p;
    NTT(R,-1,lim);
    for(reg int i=0;i!=m;++i) R[i] = dec(B[i],R[i]);
    for(int i=m;i!=lim;++i) R[i] = 0;
}

#define mid ((l+r)>>1)
#define ls (u<<1)
#define rs (u<<1|1)
int bflim;
int *P[N],len[N];

void prepare(int l,int r,int u,const int *a){ //分治乘,多点求值用
	if(l==r){
		len[u] = 1;
		P[u] = new int[2];
		P[u][0] = p-a[l],P[u][1] = 1;
		return;
	}
	prepare(l,mid,ls,a);
	prepare(mid+1,r,rs,a);
	len[u] = r-l+1;
	int lim = getlen(len[u]);
	P[u] = new int[len[u]+1];
	int F[lim+1],G[lim+1];
	memcpy(F,P[ls],(len[ls]+1)<<2);
	memcpy(G,P[rs],(len[rs]+1)<<2);
	if(r-l>bflim){
		memset(F+len[ls]+1,0,(lim-len[ls]+1)<<2);
		memset(G+len[rs]+1,0,(lim-len[rs]+1)<<2);
		NTT(F,1,lim),NTT(G,1,lim);
		for(reg int i=0;i!=lim;++i) F[i] = (ll)F[i]*G[i]%p;
		NTT(F,-1,lim);
		memcpy(P[u],F,(len[u]+1)<<2);
	}else{
		memset(P[u],0,(len[u]+1)<<2);
		for(reg int i=0;i<=len[ls];++i)
		for(reg int j=0;j<=len[rs];++j)
			P[u][i+j] = (P[u][i+j]+(ll)F[i]*G[j])%p;
	}
}

void solve(const int *F,int l,int r,const int *a,int u,int n,int *R){ //多点求值
	if(r-l<=bflim){ //小范围暴力
		ll pw[17];
		int res,x;
		ll s1,s2,s3,s4;
		pw[0] = 1;
		for(reg int j=l;j<=r;++j){
			res = F[n],x = a[j];
			reg int i = 1;
			for(;i<=16;++i) pw[i] = pw[i-1]*x%p;
			i = n-1;
			while(i>=15){
				s1 = res*pw[16]+F[i]*pw[15]+F[i-1]*pw[14]+F[i-2]*pw[13];
				s2 = F[i-3]*pw[12]+F[i-4]*pw[11]+F[i-5]*pw[10]+F[i-6]*pw[9];
				s3 = F[i-7]*pw[8]+F[i-8]*pw[7]+F[i-9]*pw[6]+F[i-10]*pw[5];
				s4 = F[i-11]*pw[4]+F[i-12]*pw[3]+F[i-13]*pw[2]+F[i-14]*x;
				res = ((F[i-15]+s1+s2)%p+s3+s4)%p;
				i -= 16;
			}
			i = (n&15)-1;
			for(;~i;--i) res = ((ll)res*x+F[i])%p;
			R[j] = res;
		}
		return;
	}
	int G[getlen(n<<1)+1];
	memset(G,0,sizeof(G));
	mod(F,P[ls],n,len[ls],G);
	solve(G,l,mid,a,ls,len[ls]-1,R);
	memset(G,0,sizeof(G));
	mod(F,P[rs],n,len[rs],G);
	solve(G,mid+1,r,a,rs,len[rs]-1,R);
}

void evaluation(const int *F,int *a,int n,int m,int *R){
	bflim = log2(m);
	prepare(1,m,1,a);
	solve(F,1,m,a,1,n,R);
}

#undef ls
#undef rs
#undef mid

void multiply(const int *F,const int *G,int n,int m,int len,int *R){
    static int A[N],B[N];
    memcpy(A,F,(n+1)<<2);
    memcpy(B,G,(m+1)<<2);
    int lim = getlen(n+m);
    memset(A+n+1,0,(lim-n+2)<<2);
    memset(B+m+1,0,(lim-m+2)<<2);
    NTT(A,1,lim),NTT(B,1,lim);
    for(reg int i=0;i!=lim;++i) R[i] = (ll)A[i]*B[i]%p;
    NTT(R,-1,lim);
    memset(R+len+1,0,(lim-len+2)<<2);
}

void mod_power(const int *G,int k,int t,int *R){ //多项式快速幂 (模G(x))
    int f[N],g[N];
    memset(f,0,sizeof(f));
    memset(g,0,sizeof(g));
    int n = 1,m = 0;
    f[1] = g[0] = 1;
    while(t){
        if(t&1){
            multiply(f,g,n,m,n+m,g);
            mod(g,G,n+m,k,g);
            m = min(n+m,k-1);
        }
        int lim = getlen(n<<1);
        NTT(f,1,lim);
        for(reg int i=0;i!=lim;++i) f[i] = (ll)f[i]*f[i]%p;
        NTT(f,-1,lim);
        mod(f,G,n<<1,k,f);
        n = min(n<<1,k-1);
        t >>= 1;
    }
    memcpy(R,g,k<<2);
}

int n,m,k,ans,T;
int F[N],G[N],B[N],A[N],a[N],f[N],d[N];

int main(){
    init(100000);
	read(n),read(m),read(k);
	for(reg int i=0;i!=k;++i) read(a[i]);
	for(reg int i=1;i<=k;++i) read(f[i]);
    for(reg int i=0;i<=m;++i) read(G[i]);
    for(reg int i=0;i<=m;++i) d[i+1] = i+k;
	evaluation(G,d,m,m+1,B); //构造B的后半部分
    for(reg int i=k+m;i>=k;--i) B[i] = B[i-k+1];
    for(reg int i=0;i!=k;++i) B[i] = 0;
    multiply(f,a,k,k,k-1,d); //一波卷积求出B的前半部分
    for(reg int i=0;i!=k;++i) B[i] = dec(a[i],d[i]);
    f[0] = p-1;
    for(reg int i=0;i<=k;++i) F[i] = p-f[i];
    inverse(F,F,m+k);
    multiply(F,B,m+k,m+k,m+k,A); //求逆算出A
    for(reg int i=0;i<=m+k;++i) a[i] = A[i];
    memset(F,0,sizeof(F));
    for(reg int i=0;i<=m+1;++i) F[i] = (m+1-i)&1?p-binom(m+1,i):binom(m+1,i); //高阶差分系数
    T = m+k+1;
    multiply(F,f,m+1,k,T,f); //化为齐次递推
    memset(G,0,sizeof(G));
    for(reg int i=0;i<=T;++i) G[T-i] = p-f[i];
    mod_power(G,T,n,F);
    for(reg int i=0;i!=T;++i) ans = (ans+(ll)F[i]*a[i])%p;
    print(ans);
	return 0;	
}

字符串

最小表示法

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=3e5+10;
int a[N];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
		scanf("%d",&a[i]);
	int i=0,j=1,k=0;
	while(i<n&&j<n&&k<n){
		int p=a[(i+k)%n]-a[(j+k)%n];
		if(!p)k++;
		else{
			if(p>0)i+=k+1;
			else j+=k+1;
			if(i==j)j++;
			k=0;
		}
	}
	k=min(i,j);
	for(int i=0;i<n;i++)
		printf("%d ",a[(i+k)%n]);
	return 0;
}

kmp

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=1e6+10;
int nxt[N];
char s1[N],s2[N];
int main(){
    scanf("%s%s",s1+1,s2+1);
    rint n=strlen(s1+1),m=strlen(s2+1);
    for(rint i=2,j=0;i<=m;i++){
        while(j&&s2[j+1]!=s2[i])j=nxt[j];
        if(s2[i]==s2[j+1])j++;
        nxt[i]=j;
    }
    for(rint i=1,j=0;i<=n;i++){
        while(j&&s2[j+1]!=s1[i])j=nxt[j];
        if(s2[j+1]==s1[i])j++;
        if(j==m){
            printf("%d\n",i-m+1);
            j=nxt[j];
        }
    }
    for(rint i=1;i<=m;i++)
        printf("%d ",nxt[i]);
    return 0;
}

exkmp

#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N = 2e7 + 10;
ll nxt[N], ext[N];
void qnxt(char *c)
{
	int len = strlen(c);
	int p = 0, k = 1, l; //我们会在后面先逐位比较出 nxt[1] 的值，这里先设 k 为 1
	//如果 k = 0，p 就会锁定在 |c| 不会被更改，无法达成算法优化的效果啦
	nxt[0] = len; //以 c[0] 开始的后缀就是 c 本身，最长公共前缀自然为 |c|
	while(p + 1 < len && c[p] == c[p + 1]) p++;
	nxt[1] = p; //先逐位比较出 nxt[1] 的值
	for(int i = 2; i < len; i++)
	{
		p = k + nxt[k] - 1; //定义
		l = nxt[i - k]; //定义
		if(i + l <= p) nxt[i] = l; //如果灰方框小于初始的绿方框,直接确定 nxt[i] 的值
		else
		{
			int j = max(0, p - i + 1);
			while(i + j < len && c[i + j] == c[j]) j++; //否则进行逐位比较
			nxt[i] = j;
			k = i; //此时的 x + nxt[x] - 1 一定刷新了最大值，于是我们要重新赋值 k
		}
	}
}
void exkmp(char *a, char *b)
{
	int la = strlen(a), lb = strlen(b);
	int p = 0, k = 0, l;
	while(p < la && p < lb && a[p] == b[p]) p++; //先算出初值用于递推
	ext[0] = p;
	for(int i = 1; i < la; i++) //下面都是一样的逻辑啦
	{
		p = k + ext[k] - 1;
		l = nxt[i - k];
		if(i + l <= p) ext[i] = l;
		else
		{
			int j = max(0, p - i + 1);
			while(i + j < la && j < lb && a[i + j] == b[j]) j++;
			ext[i] = j;
			k = i;
		}
	}
}
int la, lb;
char a[N], b[N];
ll ans;
int main()
{
	cin.tie(0); cout.tie(0);
	ios::sync_with_stdio(0);
	cin >> a >> b;
	qnxt(b);
	exkmp(a, b);
	la = strlen(a), lb = strlen(b), ans = 0;
	for(int i = 0; i < lb; i++) //要注意下标从 0 开始
	{
		ans ^= (i + 1) * (nxt[i] + 1);
	}
	cout << ans << "\n";
	ans = 0;
	for(int i = 0; i < la; i++)
	{
		ans ^= (i + 1) * (ext[i] + 1);
	}
	cout << ans;
	return 0;
}

manacher

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=3e7+10;
char s1[N],s2[N];
int n,m,f[N];
int main(){
	scanf("%s",s1+1);
	n=strlen(s1+1);
	s2[m]='~';s2[++m]='#';
	for(rint i=1;i<=n;i++)
		s2[++m]=s1[i],s2[++m]='#';
	rint res=0;
	for(rint i=1,mid=0,r=0;i<=m;i++){
		if(i<=r)f[i]=min(r-i+1,f[2*mid-i]);
		while(s2[i+f[i]]==s2[i-f[i]])f[i]++;
		if(i+f[i]-1>r)r=i+f[i]-1,mid=i;
		res=max(res,f[i]-1);
	}
	printf("%d\n",res);
	return 0;
}

ac自动机

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e6+10;
int Trie[N][26],cnt,val[N],fail[N];
char s[N];
void Ins(char s[]){
	int l=strlen(s),now=0;
	for(int i=0;i<l;i++){
		int t=s[i]-'a';
		if(!Trie[now][t])Trie[now][t]=++cnt;
		now=Trie[now][t];
	}
	val[now]++;
}
void bfs(){
	queue<int > q;
	for(int i=0;i<26;i++)
		if(Trie[0][i])q.push(Trie[0][i]);
	while(!q.empty()){
		int u=q.front();q.pop();
		for(int i=0;i<26;i++){
			if(Trie[u][i])fail[Trie[u][i]]=Trie[fail[u]][i],q.push(Trie[u][i]);
			else Trie[u][i]=Trie[fail[u]][i];
		}
	}
}
int query(char s[]){
	int l=strlen(s),now=0,tot=0;
	for(int i=0;i<l;i++){
		int t=s[i]-'a';
		now=Trie[now][t];
		for(int j=now;j&&val[j]!=-1;j=fail[j]){
			tot+=val[j];
			val[j]=-1;
		}
	}
	return tot;
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%s",s);
		Ins(s);
	}
	bfs();
	scanf("%s",s);
	printf("%d\n",query(s));
	return 0;
}

回文自动机

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=5e5+10;
char s[N];
int ch[N][26],siz[N],fail[N],len[N],lst,cnt;
int get(int x,int n){
	while(s[n]!=s[n-len[x]-1])x=fail[x];
	return x;
}
void extend(int n,int c){
	int p=get(lst,n);
	if(!ch[p][c]){
		++cnt;
		fail[cnt]=ch[get(fail[p],n)][c];
		len[cnt]=len[p]+2;ch[p][c]=cnt;
		siz[cnt]=siz[fail[cnt]]+1;
	}
	lst=ch[p][c];
}
int main(){
	scanf("%s",s+1);
	int n=strlen(s+1);
	cnt++;len[1]=-1;fail[0]=1;s[0]=-1;
	for(int i=1,res=0;i<=n;i++){
		s[i]=(s[i]-97+res)%26;
		extend(i,s[i]);
		printf("%d ",res=siz[lst]);
	}
}

后缀数组

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=6e5+10;
typedef long long ll;
char s[N];
int rk[N],tp[N],sa[N],n,m,t[N],ht[N];
void suf(){
	for(int i=1;i<=m;i++)t[i]=0;
	for(int i=1;i<=n;i++)t[rk[i]]++;
	for(int i=1;i<=m;i++)t[i]+=t[i-1];
	for(int i=n;i;i--)sa[t[rk[tp[i]]]--]=tp[i];
}
ll a[N],stmx[N][25],stmi[N][25];
int b[N],stk[N],top,L[N],R[N];
ll qmax(int l,int r){
	int len=b[r-l+1];
	return max(stmx[l][len],stmx[r-(1<<len)+1][len]);
}
ll qmin(int l,int r){
	int len=b[r-l+1];
	return min(stmi[l][len],stmi[r-(1<<len)+1][len]);
}
ll ans1[N],ans2[N];
int main(){
	scanf("%d%s",&n,s+1);
	for(int i=1;i<=n;i++){
		scanf("%lld",&a[i]);
		rk[i]=s[i]-'a'+1;tp[i]=i;
	}
	m=26;
	suf();
	for(int w=1,p=0;p!=n;m=p,w<<=1){
		p=0;
		for(int i=n-w+1;i<=n;i++)tp[++p]=i;
		for(int i=1;i<=n;i++)
			if(sa[i]>w)tp[++p]=sa[i]-w;
		suf();
		memcpy(tp,rk,sizeof(rk));
		rk[sa[1]]=p=1;
		for(int i=2;i<=n;i++)
			rk[sa[i]]=(tp[sa[i]]==tp[sa[i-1]]&&tp[sa[i]+w]==tp[sa[i-1]+w])?p:++p;
	}
	for(int i=1,j,k=0;i<=n;i++){
		if(k)--k;
		j=sa[rk[i]-1];
		while(s[i+k]==s[j+k])k++;
		ht[rk[i]]=k;
		stmx[i][0]=stmi[i][0]=a[sa[i]];
	}
	for(int j=1;(1<<j)<=n;j++){
		b[1<<j]=j;
		for(int i=1;i+(1<<j)-1<=n;i++){
			stmx[i][j]=max(stmx[i][j-1],stmx[i+(1<<j-1)][j-1]);
			stmi[i][j]=min(stmi[i][j-1],stmi[i+(1<<j-1)][j-1]);
		}
	}
	for(int i=1;i<=n;i++)if(!b[i])b[i]=b[i-1];
	for(int i=2;i<=n;i++){
		while(top&&ht[stk[top]]>ht[i])R[stk[top]]=i-1,top--;
		stk[++top]=i;
	}
	while(top)R[stk[top]]=n,top--;
	for(int i=n;i>1;i--){
		while(top&&ht[stk[top]]>=ht[i])L[stk[top]]=i+1,top--;
		stk[++top]=i;
	}
	memset(ans2,-0x3f,sizeof(ans2));
	while(top)L[stk[top]]=2,top--;
	for(int i=2;i<=n;i++){
		ll ls=i-L[i]+1,rs=R[i]-i+1;
		ans1[0]+=ls*rs;
		ans1[ht[i]+1]-=ls*rs;
		ls=qmax(L[i]-1,i-1);rs=qmax(i,R[i]);
		ans2[ht[i]]=max(ans2[ht[i]],ls*rs);
		ls=qmin(L[i]-1,i-1);rs=qmin(i,R[i]);
		ans2[ht[i]]=max(ans2[ht[i]],ls*rs);
	}
	for(int i=1;i<=n;i++)ans1[i]+=ans1[i-1];
	for(int i=n;~i;i--){
		if(!ans1[i])ans2[i]=0;
		if(ans1[i+1])ans2[i]=max(ans2[i],ans2[i+1]);
	}
	for(int i=0;i<n;i++)printf("%lld %lld\n",ans1[i],ans2[i]);
	return 0;
}

#include<cstdio>
#include<cstring>
#include<algorithm>
const int MAX_LEN = 1e6 + 5;
const int SIGMA = 128;
struct SA{
	int str[MAX_LEN << 1];
	int type[MAX_LEN << 1];
	int LMS[MAX_LEN << 1];
	int sa[MAX_LEN];
	int rank[MAX_LEN];
	int cnt[MAX_LEN];
	int cur[MAX_LEN];
	void inducedsort(int *str, int len, int sigma, int *LMS, int LMSC, int *type) {
		#define PUSH_S(x) (sa[cur[str[x]]--] = x)
		#define PUSH_L(x) (sa[cur[str[x]]++] = x)
		std::fill_n(sa, len, -1), std::fill_n(cnt, sigma, 0);
		for(register int i = 0; i < len; ++i) ++cnt[str[i]];
		for(register int i = 1; i < sigma; ++i) cnt[i] += cnt[i - 1];
		for(register int i = 0; i < sigma; ++i) cur[i] = cnt[i] - 1;
		for(register int i = LMSC - 1; i >= 0; --i) PUSH_S(LMS[i]);
		for(register int i = 1; i < sigma; ++i) cur[i] = cnt[i - 1];
		for(register int i = 0; i < len; ++i) sa[i] > 0 && type[sa[i] - 1] == 0 && PUSH_L(sa[i] - 1);
		for(register int i = 0; i < sigma; ++i) cur[i] = cnt[i] - 1;
		for(register int i = len - 1; i >= 0; --i) sa[i] > 0 && type[sa[i] - 1] && PUSH_S(sa[i] - 1);
		#undef PUSH_S
		#undef PUSH_L
	}
	void sais(int *str, int len, int sigma, int *LMS, int *type) {
		type[len - 1] = 1;
		for(register int i = len - 2; i >= 0; --i) type[i] = (str[i] == str[i + 1] ? type[i + 1] : str[i] < str[i + 1]);
		int LMSC = 0;
		rank[0] = -1;
		for(register int i = 1; i < len; ++i) rank[i] = (type[i] && !type[i - 1]) ? (LMS[LMSC] = i, LMSC++) : -1;
		inducedsort(str, len, sigma, LMS, LMSC, type);
		int tot = -1;
		int *s1 = str + len;
		for(register int i = 0, now, last; i < len; ++i) {
		    if(-1 == (now = rank[sa[i]])) continue;
		    if(tot < 1 || LMS[now + 1] - LMS[now] != LMS[last + 1] - LMS[last]) ++tot;
		    else for(register int j = LMS[now], k = LMS[last]; j < LMS[now + 1]; ++j, ++k) if((str[j] << 1 | type[j]) != (str[k] << 1 | type[k])) {
		        ++tot;
		        break;
		    }
		    s1[last = now] = tot;
		}
		if(tot + 1 < LMSC) sais(s1, LMSC, tot + 1, LMS + LMSC, type + len);
		else for(register int i = 0; i < LMSC; ++i) sa[s1[i]] = i;
		for(register int i = 0; i < LMSC; ++i) s1[i] = LMS[sa[i]];
		inducedsort(str, len, sigma, s1, LMSC, type);
	}
	void Getsa(char *s,int *ans){
		int len = strlen(s);
		for(register int i = 0; i < len; ++i) {
		    str[i] = s[i];
		}
		sais(str, len + 1, SIGMA, LMS, type);
		for(register int i = 0; i < len; ++i) ans[i+1]=sa[i + 1] + 1;
	}
}S;
const int N=1e6+10;
char s[N];
int sa[N];
int main(){
	scanf("%s",s+1);
	int n=strlen(s+1);
	S.Getsa(s+1,sa);
	for(int i=1;i<=n;i++)
		printf("%d ",sa[i]);
}

树上后缀数组

#include<iostream>
#include<vector>
#include<utility>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N=5e5+5,base1=20041001,base2=20040607,md1=167772161,md2=104857601;
int n,F[19][N],head[N],cnt,dep[N],x[N],y[N],c[N],sa[N],hx1[N],hx2[N];
char s[N];
vector<int>vec[N],vd[N];
vector<pair<int,int>>pd[N];
struct edge{
	int to,nxt;
}e[N];
void dfs(int now){
	hx1[now]=((LL)hx1[F[0][now]]*base1+s[now])%md1,hx2[now]=((LL)hx2[F[0][now]]*base2+s[now])%md2;
	for(int i=head[now];i;i=e[i].nxt)dep[e[i].to]=dep[now]+1,F[0][e[i].to]=now,dfs(e[i].to);
}
void ssort(){
	int m=256;
	for(int i=1;i<=m;++i)c[i]=0;
	for(int i=1;i<=n;++i)++c[x[i]=s[i]];
	for(int i=1;i<=m;++i)c[i]+=c[i-1];
	for(int i=n;i;--i)sa[c[x[i]]--]=i;
	for(int k=0;1<<k<=n;++k){
		for(int i=1;i<=n;++i)vec[i].clear();
		for(int i=1;i<=n;++i)if(F[k][i])vec[F[k][i]].push_back(i);
		int p=0;
		for(int i=1;i<=1<<k;++i)
		for(int j:vd[i])y[++p]=j;
		for(int i=1;i<=n;++i)
		for(int j:vec[sa[i]])y[++p]=j;
		for(int i=1;i<=m;++i)c[i]=0;
		for(int i=1;i<=n;++i)++c[x[i]];
		for(int i=1;i<=m;++i)c[i]+=c[i-1];
		for(int i=n;i;--i)sa[c[x[y[i]]]--]=y[i];
		std::swap(x,y);
		x[sa[1]]=p=1;
		for(int i=2;i<=n;++i)
		x[sa[i]]=(y[sa[i]]==y[sa[i-1]]&&y[F[k][sa[i]]]==y[F[k][sa[i-1]]]?p:++p);
		if(p==n)break;
		m=p;
	}
}
inline int cmp(int a,int b){return F[0][a]!=F[0][b]?x[F[0][a]]<x[F[0][b]]:a<b;}
int main(){
	ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	cin>>n;
	for(int i=2;i<=n;++i){
		int&f=F[0][i];
		cin>>f;
		e[++cnt]=(edge){i,head[f]},head[f]=cnt;
	}
	cin>>(s+1);
	dfs(dep[1]=1);
	for(int i=1;i<=n;++i)vd[dep[i]].push_back(i);
	for(int i=1;i<19;++i)
	for(int j=1;j<=n;++j)
	F[i][j]=F[i-1][F[i-1][j]];
	ssort();
	int lst=1;
	for(int i=2;i<=n+1;++i)
	if(hx1[sa[i]]!=hx1[sa[i-1]]||hx2[sa[i]]!=hx2[sa[i-1]])pd[dep[sa[i-1]]].emplace_back(lst,i-1),lst=i;
	for(int i=1;i<=n;++i)
	for(auto j:pd[i]){
		sort(sa+j.first,sa+j.second+1,cmp);
		for(int k=j.first;k<=j.second;++k)
		x[sa[k]]=k;
	}
	for(int i=1;i<=n;++i)cout<<sa[i]<<' ';
	return 0;
}

后缀自动机

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define rint register int
const int N=2e6+10;
char s[N];
int ch[N][26],len[N],fa[N],lst,cnt;
long long siz[N];
void extend(rint c){
	rint p=lst;rint np=lst=++cnt;
	len[np]=len[p]+1;siz[np]=1;
	for(;p&&!ch[p][c];p=fa[p])ch[p][c]=np;
	if(!p)fa[np]=1;
	else {
		rint t=ch[p][c];
		if(len[t]==len[p]+1)fa[np]=t;
		else {
			rint ct=++cnt;
			memcpy(ch[ct],ch[t],sizeof(ch[t]));
			fa[ct]=fa[t];fa[t]=fa[np]=ct;
			len[ct]=len[p]+1;
			for(;p&&ch[p][c]==t;p=fa[p])ch[p][c]=ct;
		}
	}
}
int a[N],t[N];
int main(){
	rint n;
	scanf("%s",s+1);
	n=strlen(s+1);lst=++cnt;
	for(rint i=1;i<=n;i++)extend(s[i]-'a');
	for(rint i=1;i<=cnt;i++)t[len[i]]++;
	for(rint i=1;i<=n;i++)t[i]+=t[i-1];
	for(rint i=1;i<=cnt;i++)a[t[len[i]]--]=i;
	long long res=0;
	for(rint i=cnt;i;i--){
		siz[fa[a[i]]]+=siz[a[i]];
		if(siz[a[i]]>1)res=max(res,siz[a[i]]*len[a[i]]);
	}
	printf("%lld\n",res);
	return 0;
}

后缀平衡树

#include<bits/stdc++.h>
using namespace std;
const int N=8e5+5,M=3e6+6;
char buf[M+5],*p1,*p2,c;
#define gc (p1==p2&&(p2=(p1=buf)+fread(buf,1,M,stdin),p1==p2)?EOF:*p1++)
inline void read(int &x){
	for(c=gc;c<'0'||c>'9';c=gc);
	for(x=0;c>='0'&&c<='9';x=x*10+c-'0',c=gc);
}
char s[N],st[M];
int t[N][2],Q,n,tot,mask,ans,rt,*nrb,sz[N],val,rb[N],bt;
typedef long double ld;
ld d[N][3],lv,rv;
inline void read(char *s){
	while(!isalpha(c=gc));n=0;
	do s[++n]=c;while(isalpha(c=gc));
}
inline void decode(char *st,int mask){
	for(int i=0;i<n;i++)
		mask=(mask*131+i)%n,swap(st[i],st[mask]);
}
void Rem(int x){if(t[x][0])Rem(t[x][0]);rb[++bt]=x;if(t[x][1])Rem(t[x][1]);}
int Mak(int l,int r,ld L,ld R){
	if(l>r)return 0;int md=l+r>>1,x=rb[md];ld Md=(L+R)/2;
	t[x][0]=Mak(l,md-1,L,Md),t[x][1]=Mak(md+1,r,Md,R);
	d[x][0]=L,d[x][1]=Md,d[x][2]=R,sz[x]=sz[t[x][0]]+sz[t[x][1]]+1;
	return x;
}
void Reb(int &x){bt=0,Rem(x),x=Mak(1,bt,d[x][0],d[x][2]);}
void Ins(int &x,ld L,ld R){
	if(!x){
		x=val,t[x][0]=t[x][1]=0,sz[x]=1;
		d[x][0]=L,d[x][1]=(L+R)/2,d[x][2]=R;
	}else{
		int k=s[x]==s[val]?d[x-1][1]<d[val-1][1]:s[x]<s[val];
		Ins(t[x][k],d[x][k],d[x][k+1]),++sz[x];
		if((rand()+rand())%sz[x]==0)nrb=&x;
	}
}
void Del(int &x){
	int k=d[val][1]>=d[x][1],&y=t[x][k],&z=t[x][!k];
	if(y){
		Del(y);if(k&&val){
			t[val][0]=t[x][0],t[val][1]=t[x][1];
			d[val][0]=d[x][0],d[val][1]=d[x][1];
			d[val][2]=d[x][2],x=val,val=0;
		}sz[x]=sz[y]+sz[z]+1;
	}else {val=(x==val)?0:x;x=z;}
	return;
}
void Ins(int x){val=x,nrb=NULL,Ins(rt,1,1e18);if(nrb)Reb(*nrb);}
int lwb(int x){
	if(!x)return 0;int i,k=0;
	for(i=0;i<n;++i)
		if(i==x){k=1;break;}
		else if(s[x-i]!=st[n-i]){k=s[x-i]<st[n-i];break;}
	return k?sz[t[x][0]]+1+lwb(t[x][1]):lwb(t[x][0]);
}
int main(){
    srand(1919810u^time(0));
	read(Q),read(s);int i,j,k;
	for(i=1;i<=n;++i)Ins(i);tot=n;
	while(Q--){
		read(st);
		if(st[1]=='A'){
			read(st),decode(st+1,mask);
			for(i=1;i<=n;++i)s[++tot]=st[i],Ins(tot);
		}else if(st[1]=='Q'){
			read(st),decode(st+1,mask),ans=lwb(rt),++st[1];
			ans=lwb(rt)-ans;printf("%d\n",ans),mask^=ans;
		}else for(i=1,read(n);i<=n;++i)val=tot--,Del(rt);
	}
	return 0;
}

杂项

LGV引理

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=410;
const int p=998244353;
typedef long long ll;
ll a[N][N],fac[2100000],inv[2100000];
ll fpow(ll a,ll b){
	ll res;
	for(res=1;b;b>>=1,a=a*a%p)
		if(b&1)res=res*a%p;
	return res;
}
ll C(int n,int m){
	if(n<m)return 0;
	return fac[n]*inv[m]%p*inv[n-m]%p;
}
int x[N],y[N];
ll calc(int n){
	ll res=1;
	for(int i=1;i<=n;i++){
		if(a[i][i]==0){
			for(int j=i+1;j<=n;j++){
				if(a[j][i]){
					swap(a[i],a[j]);
					res=-res;
					goto L;
				}
			}
			return 0;
		}L:;
		for(int j=i+1;j<=n;j++){
			ll r=a[j][i]*fpow(a[i][i],p-2)%p;
			for(int k=i;k<=n;k++)
				a[j][k]=(a[j][k]-a[i][k]*r%p+p)%p;
		}
	}
	for(int i=1;i<=n;i++)res=res*a[i][i]%p;
	return (res+p)%p;
}
int main(){
	fac[0]=fac[1]=inv[0]=inv[1]=1;
	for(int i=2;i<=2e6;i++){
		fac[i]=(ll)fac[i-1]*i%p;
		inv[i]=p-(ll)p/i*inv[p%i]%p;
	}
	for(int i=2;i<=2e6;i++)
		inv[i]=inv[i-1]*inv[i]%p;
	int T;
	scanf("%d",&T);
	while(T--){
		int n,m;
		scanf("%d%d",&n,&m);
		for(int i=1;i<=m;i++)scanf("%d%d",&x[i],&y[i]);
		for(int i=1;i<=m;i++)
			for(int j=1;j<=m;j++)
				a[i][j]=C(y[j]-x[i]+n-1,n-1);
		printf("%lld\n",calc(m));
	}
	return 0;
}

树哈希

h(x)=1+\sum_{v\in son(x)}f(h(v))

ll h(ll x) {
    return x * x * x * 1237123 + 19260817;
}
ll f(ll x) {
    ll cur = h(x & ((1 << 31) - 1)) + h(x >> 31);
    return cur;
}

spfa优化

#include<ctime>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define reg register
#define uni unsigned
using namespace std;
const uni N=100010,W=5,WT=10,INF=0x7fffffff;    //W为取首位若干项数，WT为随机次数
uni n,m,s,d[N<<4],a[N],b[N<<2][3],t[N],seed=time(0),w[N],OnF,st[10];
bool vis[N];
char BuF[1<<24],*InF=BuF,WuF[1<<23];
uni read(){
    reg uni x=0;
    for(;47>*InF||*InF>58;++InF);
    for(;47<*InF&&*InF<58;x=x*10+(*InF++^48));
    return(x);
}
void write(uni x){
    if(!x){
        WuF[OnF++]=48;
    }
    for(;x;x/=10){
        st[++st[0]]=x%10+48;
    }
    for(;st[0];WuF[OnF++]=st[st[0]--]);
    WuF[OnF++]=' ';
}
uni rnd(uni l,uni r){
    seed^=seed<<17;seed^=seed>>15;seed^=seed<<3;
    return(seed%(r-l+1)+l);
}
void spfa(uni s){
    memset(w,127,sizeof(w));
    w[d[0]=s]=0;
    for(reg uni h=0,t=0;h<=t;++h){
        for(reg uni i=0;i<=WT&&h+W+W+1<t;++i){
            reg uni x=rnd(h+W+1,t-W-1);
            if(w[d[h]]>w[d[x]]){
                swap(d[h],d[x]);    //随机判断交换
            }
        }
        for(reg uni i=0;i<=W&&h+i+i<t;++i){
            if(w[d[h]]>w[d[h+i+1]]){
                swap(d[h],d[h+i+1]);    //与队首判断
            }
            if(w[d[h]]>w[d[t-i]]){
                swap(d[h],d[t-i]);    //与队尾判断
            }
        }
        for(reg uni i=a[d[h]],mi=INF;i;i=b[i][0]){
            reg uni nxt=b[i][1];
            if(w[nxt]>w[d[h]]+b[i][2]){
                w[nxt]=w[d[h]]+b[i][2];
                if(!vis[nxt]){
                    d[++t]=nxt;
                    if(mi>w[nxt]){
                        if(mi<INF){
                            swap(d[t],d[t-1]);    //保持单次松弛最小值在队尾
                        }
                        mi=w[nxt];
                    }
                    vis[nxt]=1;
                }
            }
        }
        vis[d[h]]=0;
    }
}
int main(){
    fread(BuF,1,1<<24,stdin);
    n=read();m=read();s=read();
    for(reg uni last=0;m;--m){
        reg uni x=read(),y=read(),z=read();
        b[++last][0]=a[x];
        b[a[x]=last][1]=y;
        b[last][2]=z;
    }
    spfa(s);
    for(reg uni *i=w+1,*r=w+n+1;i!=r;++i){
        write(*i);
    }
    fwrite(WuF,1,OnF,stdout);
    fclose(stdin);
    fclose(stdout);
    return(0);
}

2-sat

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#define reset(a) memset((a),0,sizeof((a)))
using namespace std;
static char buf[100000],*pa,*pd;
#define gc pa==pd&&(pd=(pa=buf)+fread(buf,1,100000,stdin),pa==pd)?EOF:*pa++
inline int read(){
	register int x(0);register char c(gc);
	while(c>'9'||c<'0')c=gc;
	while(c>='0'&&c<='9')x=x*10+c-48,c=gc;
	return x;
}
const int N=4001000;
struct edge{
	int to,next;
}e[N];
int head[N],tot,HEAD[N];
int n,m,cnt,turn[N],belong[N],vis[N];
void add(int x,int y){
	e[++tot].to=y;e[tot].next=head[x];head[x]=tot;
	e[++tot].to=x;e[tot].next=HEAD[y];HEAD[y]=tot;
}
void dfs1(int u){
	int i;
	vis[u]=1;
	for(i=head[u];i;i=e[i].next)
		if(!vis[e[i].to])
			dfs1(e[i].to);
	turn[++cnt]=u;
}
void dfs2(int u){
	belong[u]=cnt;vis[u]=1;
	for(int i=HEAD[u];i;i=e[i].next)
		if(!vis[e[i].to])
			dfs2(e[i].to);
}
void kosaraju(){
	for(int i=1;i<=2*n;i++)
		if(!vis[i])dfs1(i);
	reset(vis);cnt=0;
	for(int i=2*n;i>=1;i--)
		if(!vis[turn[i]])
			cnt++,dfs2(turn[i]);
}
int main(){
	n=read();m=read();
	register int i,x,y,f1,f2;
	for(i=1;i<=m;i++){
		x=read();f1=read();y=read();f2=read();
		add(x+n*(f1&1),y+n*(f2^1));
		add(y+n*(f2&1),x+n*(f1^1));
	}
	kosaraju();
	for(i=1;i<=n;i++)
		if(belong[i]==belong[i+n]){
			cout<<"IMPOSSIBLE";return 0;
		}
	cout<<"POSSIBLE\n";
	for(i=1;i<=n;i++){
		cout<<(belong[i]>belong[i+n])<<' ';
	}	
	return 0;
}

差分约束

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
const int N=5e3+10;
struct Edge{
    int to,nxt,val;
}e[N<<1];
int h[N],idx;
void Ins(rint x,rint y,rint z){
    e[++idx].to=y;e[idx].val=z;
    e[idx].nxt=h[x];h[x]=idx;
}
int dis[N],cnt[N],n,m;
bool inq[N];
queue<int> q;
bool spfa(rint s){
    memset(dis,0x3f,sizeof(dis));
    dis[s]=0;
    q.push(s);inq[s]=0;
    while(!q.empty()){
        rint u=q.front();q.pop();inq[u]=0;
        for(rint i=h[u];i;i=e[i].nxt){
            rint v=e[i].to;
            if(dis[v]>dis[u]+e[i].val){
                dis[v]=dis[u]+e[i].val;
                if(!inq[v]){
                    inq[v]=1;
                    q.push(v);
                    if(++cnt[v]>n+1)return 0;
                }
            }
        }
    }
    return 1;
}
int main(){
    scanf("%d%d",&n,&m);
    for(rint i=1;i<=m;i++){
        rint x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        Ins(y,x,z);
    }
    for(rint i=1;i<=n;i++)
        Ins(0,i,0);
    if(spfa(0)){
        for(rint i=1;i<=n;i++)
            printf("%d ",dis[i]);
    }else {
        printf("NO");
    }
    return 0;
}

prufer 序列

#include<bits/stdc++.h>
using namespace std;
const int maxn=5000005;
int read(){//快读
	int res,f=1;
	char c;
	while((c=getchar())<'0'||c>'9')
		if(c=='-')f=-1;
	res=c-48;
	while((c=getchar())>='0'&&c<='9')
		res=res*10+c-48;
	return res*f;
}
int n,type;
int f[maxn],d[maxn],topss[maxn],x[maxn],top,maxv[maxn],p[maxn];
long long make_prufer(int n){
	int headss=n;long long ans=0;
    for(int i=1;i<n;++i)f[i]=read(),++d[f[i]];
    for (int i=1,j=1;i<=n-2;++i,++j){
		while (d[j])++j;
	    p[i]=f[j];
		while (i<=n-2&&!--d[p[i]]&&p[i]<j)p[i+1]=f[p[i]],++i;//代替朴素算法中的if
	}
	for(int i=1;i<=n-2;++i)ans^=(long long)i*p[i];
	return ans;
}
long long make_fa(int n){
	long long ans=0;
	for(int i=1;i<=n-2;++i)p[i]=read(),++d[p[i]];
	p[n-1]=n;
	for(int i=1,j=1;i<n;++i,++j){
		while(d[j])++j;
		f[j]=p[i];
		while(i<n&&!--d[p[i]]&&p[i]<j)f[p[i]]=p[i+1],++i;
	}
	for(int i=1;i<n;++i)ans^=(long long)i*f[i];
    return ans;
}
int main(){
	n=read(),type=read();
	if(type==1)printf("%lld",make_prufer(n));
	if(type==2)printf("%lld",make_fa(n));
	return 0;
}

每一个prufer序列和一个有标号无根树对应

竞赛图相关

兰道定理，用于判断竞赛图

其中 $s_i$ 是每个点出度从小到大排序，且 $k=n$ 时必须取等号

\forall 1 \le k \le n,\sum_{i=1}^{k}s_i\ge \binom{k}{2}

若竞赛图有环，则一定存在三元环

一般图最大匹配

带花树算法

#include<bits/stdc++.h>
using namespace std;
#define I inline int
#define V inline void
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define REP(u) for(int i=h[u],v;v=e[i].t,i;i=e[i].n)
const int N=1e3+1,M=1e5+1;
queue<int>q;
int n,m,tot,qwq,ans;
int h[N],lk[N],tag[N],fa[N],pre[N],dfn[N];
struct edge{int t,n;}e[M];
V link(int x,int y){lk[x]=y,lk[y]=x;}
V add_edge(int x,int y){
	if(!lk[x]&&!lk[y])link(x,y),ans++;
	e[++tot]=(edge){y,h[x]},h[x]=tot;
	e[++tot]=(edge){x,h[y]},h[y]=tot;
}
V rev(int x){if(x)rev(x[pre][lk]),link(x,pre[x]);}
I find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
I lca(int x,int y){
	for(qwq++;;x=x[lk][pre],swap(x,y))
		if(dfn[x=find(x)]==qwq)return x;
		else if(x)dfn[x]=qwq;
}
V shrink(int x,int y,int p){
	for(;find(x)!=p;x=pre[y]){
		pre[x]=y,y=lk[x],fa[x]=fa[y]=p;
		if(tag[y]==2)tag[y]=1,q.push(y);
	}
}
I blossom(int u){
	FOR(i,1,n)tag[i]=pre[i]=0,fa[i]=i;
	tag[u]=1,q=queue<int>(),q.push(u);
	for(int p;!q.empty();q.pop())REP(u=q.front())
		if(tag[v]==1)
			p=lca(u,v),shrink(u,v,p),shrink(v,u,p);
		else if(!tag[v]){
			pre[v]=u,tag[v]=2;
			if(!lk[v])return rev(v),1;
			else tag[lk[v]]=1,q.push(lk[v]);
		}
	return 0;
}
int main(){
	scanf("%d%d",&n,&m);
	for(int x,y;m--;add_edge(x,y))scanf("%d%d",&x,&y);
	FOR(i,1,n)ans+=!lk[i]&&blossom(i);
	cout<<ans<<'\n';
	FOR(i,1,n)cout<<lk[i]<<' ';
	return 0;
}

一般图最大权匹配



#include<bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
using namespace std;
inline int read()
{
	char c=getchar();int x=0;bool f=0;
	for(;!isdigit(c);c=getchar())f^=!(c^45);
	for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
	if(f)x=-x;return x;
}

#define fi first
#define se second
#define pb push_back
#define mkp make_pair
typedef pair<int,int>pii;
typedef vector<int>vi; 

#define maxn 4000005
#define inf 0x3f3f3f3f

#define N 805
int n,m;
struct edge{
	int u,v,w;
}e[N][N];
int lab[N],sl[N],q[N],hd,tl,rt[N],col[N],mat[N],frm[N][N],fa[N];
int vis[N],tim;
vi p[N];
int d(edge x){return lab[x.u]+lab[x.v]-e[x.u][x.v].w*2;}
void upd(int u,int v){if(!sl[v]||d(e[u][v])<d(e[sl[v]][v]))sl[v]=u;}
void updall(int u){
	sl[u]=0;
	For(i,1,n)if(e[i][u].w>0 && rt[i]!=u && col[rt[i]]==0) upd(i,u);
}
void push(int u){
	if(u<=n) q[++tl]=u;
	else for(int x:p[u]) push(x);
}
void setrt(int u,int r){
	rt[u]=r;
	if(u>n) for(int x:p[u]) setrt(x,r);
}
int rot(int b,int x){
	int pos=find(p[b].begin(),p[b].end(),x)-p[b].begin();
	if(pos%2==0)return pos;
	reverse(p[b].begin()+1,p[b].end());
	return p[b].size()-pos;
}
void match(int u,int v){
	mat[u]=e[u][v].v;
	if(u<=n)return;
	auto w=e[u][v];
	int x=frm[u][w.u],pos=rot(u,x);
	For(i,0,pos-1) match(p[u][i],p[u][i^1]);
	match(x,v);
	rotate(p[u].begin(),p[u].begin()+pos,p[u].end());
}
void aug(int u,int v){
	assert(u&&v);
	int t=rt[mat[u]];
	match(u,v);
	if(t) match(t,rt[fa[t]]),aug(rt[fa[t]],t);
}

int lca(int u,int v){
	++tim;
	while(u||v){
		if(vis[u]==tim)return u;
		vis[u]=tim;
		u=rt[mat[u]];
		if(u)u=rt[fa[u]];
		if(!u)swap(u,v);
	}return 0;
}

int newn(){
	int x=n+1;
	while(x<=m&&rt[x])++x; if(x>m)++m;
	p[x].clear(),lab[x]=mat[x]=rt[x]=col[x]=0;
	return x;
}
void blossom(int u,int v,int r){
	int x=newn();
	mat[x]=mat[r];
	p[x].pb(r);
	for(int i=u;i!=r;i=rt[fa[i]]){
		p[x].pb(i),p[x].pb(i=rt[mat[i]]);
		push(i);
	}
	reverse(p[x].begin()+1,p[x].end());
	for(int i=v;i!=r;i=rt[fa[i]]){
		p[x].pb(i),p[x].pb(i=rt[mat[i]]);
		push(i); 
	}
	setrt(x,x);
	For(i,1,m)e[x][i].w=e[i][x].w=0,frm[x][i]=0;
	for(int t:p[x]){
		For(i,1,m){
			if(!e[x][i].w||d(e[t][i])<d(e[x][i]))
				e[x][i]=e[t][i],e[i][x]=e[i][t];
		}
		For(i,1,n)if(frm[t][i])frm[x][i]=t;
	}
	updall(x);
}
void expand(int x){
	for(int u:p[x]) setrt(u,u);
	int u=frm[x][e[x][fa[x]].u],pos=rot(x,u);
	for(int i=0;i<pos;i+=2){
		int a=p[x][i],b=p[x][i+1];
		fa[a]=e[b][a].u;
		col[a]=1,col[b]=0;
		sl[a]=0;
		updall(b),push(b);
	}
	col[u]=1;
	fa[u]=fa[x];
	for(int i=pos+1;i<p[x].size();++i)
		col[p[x][i]]=-1,updall(p[x][i]);
	rt[x]=0;
}

bool path(edge e){
	int u=rt[e.u],v=rt[e.v];
	assert(!d(e));
	if(col[v]==-1){
		fa[v]=e.u;
		col[v]=1;
		int t=rt[mat[v]];
		sl[v]=sl[t]=col[t]=0;
		push(t);
	}
	else if(!col[v]){
		int t=lca(u,v);
		if(!t)return aug(u,v),aug(v,u),1;
		else blossom(u,v,t);
	}
	return 0;
}
bool bfs()
{
	For(i,0,m) col[i]=-1,sl[i]=0;
	hd=1,tl=0;
	For(i,1,m)if(rt[i]==i&&!mat[i])fa[i]=col[i]=0,push(i);
	if(!tl)return 0;
	while(1){
		while(hd<=tl){
			int u=q[hd++];
			For(v,1,n)
				if(e[u][v].w>0&&rt[u]!=rt[v]){
					if(d(e[u][v])) upd(u,rt[v]);
					else if(path(e[u][v])) return 1; 
				}
		}
		int tmp=inf;
		For(i,1,n) if(col[rt[i]]==0) tmp=min(tmp,lab[i]);
		For(i,n+1,m) if(rt[i]==i && col[i]==1) tmp=min(tmp,lab[i]>>1);
		For(i,1,m) if(rt[i]==i && sl[i] && col[i]!=1) tmp=min(tmp,d(e[sl[i]][i])>>(col[i]+1));
		For(i,1,n){
			if(col[rt[i]]==0) lab[i]-=tmp;
			if(col[rt[i]]==1) lab[i]+=tmp;
			if(lab[i]<=0) return 0;
		}
		For(i,n+1,m){
			if(rt[i]==i){
				if(col[i]==0) lab[i]+=tmp*2;
				if(col[i]==1) lab[i]-=tmp*2;
			}
		}
		hd=1,tl=0;
		For(i,1,m)
			if(rt[i]==i && sl[i] && rt[sl[i]]!=i && !d(e[sl[i]][i]) && path(e[sl[i]][i])) return 1;
		For(i,n+1,m)
			if(rt[i]==i && col[i]==1 && !lab[i]) expand(i);
	}
}

signed main()
{
	n=read(),m=n;
	int M=read(),mx=0;
	For(i,1,n)For(j,1,n)e[i][j]=(edge){i,j,0},frm[i][j]=0;
	For(i,1,M){
		int u=read(),v=read(),w=read();
		mx=max(mx,w);
		e[u][v].w=e[v][u].w=max(e[u][v].w,w);
	}
	For(i,1,n)frm[i][i]=i;
	For(i,1,n)lab[i]=mx,rt[i]=i,p[i].clear();
	while(bfs());
	long long res=0;
	For(i,1,n)if(mat[i]&&i<mat[i])res+=e[i][mat[i]].w;
	cout<<res<<"\n";
	For(i,1,n)cout<<mat[i]<<" ";
	return 0;
}

最小树形图

#include<bits/stdc++.h>
using namespace std;
#define rint register int
#define ll long long
#define rll register long long
#define db long double
const int N=1e4+10;
struct Edge{
	int fr,to,val;
}e[N<<1];
int Min[N],fa[N],col[N],vis[N],exis[N];
int main(){
	rint n,m,r;
	scanf("%d%d%d",&n,&m,&r);
	for(rint i=1;i<=m;i++)
		scanf("%d%d%d",&e[i].fr,&e[i].to,&e[i].val);
	rint res=0;
	for(rint i=1;i<=n;i++)
		exis[i]=1;
	while(1){
		for(rint i=1;i<=n;i++)
			col[i]=i,vis[i]=-1,Min[i]=1e9;
		for(rint i=1;i<=m;i++){
			if(e[i].fr==e[i].to)continue;
			if(e[i].val<Min[e[i].to])
				Min[e[i].to]=e[i].val,fa[e[i].to]=e[i].fr;
		}
		bool chk=1;
		for(rint i=1;i<=n;i++){
			if(i==r||!exis[i])continue;
			if(Min[i]==1e9)return puts("-1"),0;
			res+=Min[i];
			rint x;
			for(x=i;x!=r&&vis[x]==-1;x=fa[x])
				vis[x]=i;
			if(vis[x]==i){
				chk=0;
				for(rint t=fa[x];t!=x;t=fa[t])
					col[t]=x;
			}
		}
		if(chk)break;
		for(rint i=1;i<=n;i++)
			exis[i]=0;
		for(rint i=1;i<=m;i++){
			e[i].val-=Min[e[i].to],e[i].fr=col[e[i].fr],e[i].to=col[e[i].to];
			exis[e[i].fr]=exis[e[i].to]=1;
		}
		r=col[r];
	}
	printf("%d\n",res);
	return 0;
}

exgcd

#include<bits/stdc++.h>
using namespace std;
#define LL long long
int qread(){
	int w=1,c,ret=0;
	while((c=getchar())> '9'||c< '0')
	w=(c=='-'?-1:1); ret=c-'0';
	while((c=getchar())>='0'&&c<='9')
	ret=ret*10+c-'0';
	return ret*w;
}
LL exgcd(LL a,LL b,LL &x,LL &y){
	LL d=a; if(b==0) x=1,y=0; else{
		d=exgcd(b,a%b,y,x),y-=a/b*x;
	}
	return d;
}
int main(){
    int T=qread();
    while(T--){
		LL a=qread(),b=qread(),c=qread(),x,y;
		LL d=exgcd(a,b,x,y);
		if(c%d!=0) puts("-1"); else{
			x*=c/d,y*=c/d; LL p=b/d,q=a/d,k;
			if(x<0) k=ceil((1.0-x)/p),x+=p*k,y-=q*k; else	//将x提高到最小正整数 
			if(x>=0)k=(x-1)/p ,x-=p*k,y+=q*k;		//将x降低到最小正整数 
			if(y>0){	//有正整数解 
				printf("%lld ",(y-1)/q+1);	//将y减到1的方案数即为解的个数 
				printf("%lld ",x);			//当前的x即为最小正整数x
				printf("%lld ",(y-1)%q+1); 	//将y取到最小正整数 
				printf("%lld ",x+(y-1)/q*p);	//将x提升到最大 
				printf("%lld ",y);			//特解即为y最大值 
			} else{		//无整数解 
				printf("%lld " ,x);			//当前的x即为最小的正整数x 
				printf("%lld",y+q*(LL)ceil((1.0-y)/q)); //将y提高到正整数 
			}
			puts("");
		}
	}
	return 0;
}

fwt

#include<cstdio>
const int N=1e6+10;
const int p=998244353;
int a[N],b[N],t1[N],t2[N];
#define rg register
void fwtor(int *a,int n,int typ){
	for(rg int r=2,k=1;r<=n;k<<=1,r<<=1)
	for(rg int i=0;i<n;i+=r)
	for(rg int j=0;j<k;j++)
		a[i+j+k]=((a[i+j+k]+a[i+j]*typ)%p+p)%p;
}
void fwtand(int *a,int n,int typ){
	for(rg int r=2,k=1;r<=n;k<<=1,r<<=1)
	for(rg int i=0;i<n;i+=r)
	for(rg int j=0;j<k;j++)
		a[i+j]=((a[i+j]+a[i+j+k]*typ)%p+p)%p;
}
void fwtxor(int *a,int n,int typ){
	for(rg int r=2,k=1;r<=n;k<<=1,r<<=1)
	for(rg int i=0;i<n;i+=r)
	for(rg int j=0;j<k;j++){
		int x=a[i+j],y=a[i+j+k];
		a[i+j]=(x+y)%p;
		a[i+j+k]=(x-y+p)%p;
		if(typ)a[i+j]=1ll*a[i+j]*typ%p,a[i+j+k]=1ll*a[i+j+k]*typ%p;
	}
}
int main(){
	int n;scanf("%d",&n);
	n=1<<n;
	for(rg int i=0;i<n;i++)scanf("%d",&t1[i]);
	for(rg int i=0;i<n;i++)scanf("%d",&t2[i]);
	for(rg int i=0;i<n;i++)a[i]=t1[i],b[i]=t2[i];
	fwtor(a,n,1);fwtor(b,n,1);
	for(rg int i=0;i<n;i++)a[i]=1ll*a[i]*b[i]%p;
	fwtor(a,n,-1);
	for(rg int i=0;i<n;i++)printf("%d ",a[i]);
	puts("");
	for(rg int i=0;i<n;i++)a[i]=t1[i],b[i]=t2[i];
	fwtand(a,n,1);fwtand(b,n,1);
	for(rg int i=0;i<n;i++)a[i]=1ll*a[i]*b[i]%p;
	fwtand(a,n,-1);
	for(rg int i=0;i<n;i++)printf("%d ",a[i]);
	puts("");
	for(rg int i=0;i<n;i++)a[i]=t1[i],b[i]=t2[i];
	fwtxor(a,n,0);fwtxor(b,n,0);
	for(rg int i=0;i<n;i++)a[i]=1ll*a[i]*b[i]%p;
	fwtxor(a,n,p+1>>1);
	for(rg int i=0;i<n;i++)printf("%d ",a[i]);
	return 0;
}

模板

#模板

算法总览

https://suzipei.github.io/2023/02/22/template/

作者

Su_Zipei

发布于

2023年2月22日

许可协议

电工技术学习笔记上一篇

回忆·贰零贰贰下一篇

目录

数学

组合计数

数据结构

网络流

多项式

字符串

杂项

数学

高斯消元

拉格朗日插值

原根

bsgs(ex)

lucas(ex)

crt(excrt)

二次剩余

杜教筛，min25筛，洲阁筛

Miller-Rabin,Pollard-rho

类欧几里得

线性基

矩阵求逆

行列式,矩阵树定理

组合计数

莫比乌斯反演

二项式反演

Min-Max容斥

子集反演

伯努利数

斯特林数

欧拉数

如何优雅的求和

数据结构

lct维护子树信息

莫队

带修莫队

回滚莫队

树上莫队

二次离线莫队

Splay

无旋Treap

可持久化平衡树

线段树合并

线段树分裂

势能分析线段树

点分治/树

边分治/树

cdq分治

树状数组套主席树

线段树套平衡树

虚树

圆方树

长链剖分

KD-Tree

LCT

左偏树（可并堆）

猫树

支配树

李超树

划分树

笛卡尔树

分块/树分块

克鲁斯卡尔重构树

扫描线

欧拉游览树 ETT

Top Tree

Sqrt Tree

O(n)-O(1) ST表

析合树

网络流

最大流

最小割树

spfa 费用流

dinic 费用流

有源汇上下界最大流

无源汇上下界可行流

有源汇最小费用可行流

有源汇上下界可行流

有源汇最小流

最大独立集/最小点覆盖