zoukankan      html  css  js  c++  java
  • 模板大全/代码库/...

     

    读入优化 - 慢

    #define int long long
    int read(){
        int x=0;
        char ch=getchar();
        while (!isdigit(ch))
            ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+ch-48,ch=getchar();
        return x;
    }
    #undef int
    View Code

    FastIO

    namespace IO{
        const int Len=1<<21;
        char Ibuf[Len+1],*Is=Ibuf,*It=Ibuf;
        char gc(){
            if (Is==It){
                It=(Is=Ibuf)+fread(Ibuf,1,Len,stdin);
                if (Is==It)
                    return EOF;
            }
            return *Is++;
        }
        int read(){
            int x=0,f=0;
            char ch=gc();
            while (!isdigit(ch))
                f|=ch=='-',ch=gc();
            while (isdigit(ch))
                x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return f?-x:x;
        }
        char Obuf[Len+1],*Ot=Obuf;
        void pc(char ch){
            if (Ot==Obuf+Len){
                fwrite(Obuf,1,Len,stdout);
                Ot=Obuf;
            }
            *Ot++=ch;
        }
        void flush(){
            fwrite(Obuf,1,Ot-Obuf,stdout);
            Ot=Obuf;
        }
        void write(int x){
            static int buf[100],d;
            if (x==0)
                pc('0');
            else {
                if (x<0)
                    pc('-'),x=-x;
                for (d=0;x;x/=10)
                    buf[++d]=x%10+48;
                while (d)
                    pc((char)buf[d--]);
            }
        }
    }
    View Code

    UPD(2019-04-21): 一个更短的IO模板(带自动flush)

    namespace IO{
        const int S=1<<20;
        char I[S+1],*Is=I,*It=I,O[S+1],*Ot=O;
        char gc(){return Is==It?((It=(Is=I)+fread(I,1,S,stdin))==I?EOF:*Is++):*Is++;}
        void flush(){fwrite(O,1,Ot-O,stdout),Ot=O;}
        void pc(char ch){Ot==O+S?flush(),*Ot++=ch:*Ot++=ch;}
        struct flusher{ ~flusher(){flush();}}Flusher;
        #define getchar gc
        #define putchar pc
    }
    View Code

    UPD(2020-04-06): 加了个输出的函数

    namespace IO{
        const int S=1<<20;
        char I[S+1],*Is=I,*It=I,O[S+1],*Ot=O;
        char gc(){return Is==It?((It=(Is=I)+fread(I,1,S,stdin))==I?EOF:*Is++):*Is++;}
        void flush(){fwrite(O,1,Ot-O,stdout),Ot=O;}
        void pc(char ch){Ot==O+S?flush(),*Ot++=ch:*Ot++=ch;}
        struct flusher{ ~flusher(){flush();}}Flusher;
        void write(LL x){if(x<0)pc('-'),write(-x);else{if(x>9)write(x/10);pc('0'+x%10);}}
        void writeln(LL x){write(x),pc('
    ');}
        #define getchar gc
        #define putchar pc
    }
    using namespace IO;
    View Code

    手写 vector

    namespace vecint{
        typedef int T;
        const int maxlen=1e7;
        T _memp[maxlen],*memp=_memp;
        const int maxrec=1e4;
        T *p[30][maxrec];
        int pcnt[30];
        T *getmemp(int b){
            int d=0;
            while ((1<<d)<b)
                d++;
            if (pcnt[d])
                return p[d][--pcnt[d]];
            T *res=memp;
            memp+=b;
            return res;
        }
        void recycmemp(T *a,int b){
            if (!b)
                return;
            int d=0;
            while ((1<<d)<b)
                d++;
            if (pcnt[d]<maxrec)
                p[d][pcnt[d]++]=a;
        }
        struct vec{
            int n,b;
            T *a;
            vec(){
                n=0,b=0;
                a=memp;
                memp+=b;
            }
            void clear(){
                n=0;
            }
            int size(){
                return n;
            }
            int empty(){
                return n==0;
            }
            T back(){
                return a[n-1];
            }
            void push_back(T v){
                if (n==b){
                    b=max(b*2,1);
                    T *nxt=getmemp(b);
                    memcpy(nxt,a,n*sizeof(T));
                    recycmemp(a,n);
                    a=nxt;
                }
                a[n++]=v;
            }
            T *begin(){
                return a;
            }
            T *end(){
                return a+n;
            }
            T &operator [](int i){
                return a[i];
            }
        };
    }
    View Code

    Hash表

    struct hash_map{
        static const int Ti=233,mod=1<<16;
        int cnt,k[mod+1],v[mod+1],nxt[mod+1],fst[mod+1];
        int Hash(int x){
            int v=x&(mod-1);
            return v==0?mod:v;    
        }
        void clear(){
            cnt=0;
            memset(fst,0,sizeof fst);
        }
        void update(int x,int a){
            int y=Hash(x);
            for (int p=fst[y];p;p=nxt[p])
                if (k[p]==x){
                    v[p]=a;
                    return;
                }
            k[++cnt]=x,nxt[cnt]=fst[y],fst[y]=cnt,v[cnt]=a;
            return;
        }
        int find(int x){
            int y=Hash(x);
            for (int p=fst[y];p;p=nxt[p])
                if (k[p]==x)
                    return v[p];
            return 0;
        }
        int &operator [] (int x){
            int y=Hash(x);
            for (int p=fst[y];p;p=nxt[p])
                if (k[p]==x)
                    return v[p];
            k[++cnt]=x,nxt[cnt]=fst[y],fst[y]=cnt;
            return v[cnt]=0;
        }
    }Map;
    View Code

    矩阵快速幂

    namespace Matrix{
        int t;
        struct Mat{
            int v[N][N];
            Mat(){}
            Mat(int x){
                memset(v,0,sizeof v);
                for (int i=1;i<=t;i++)
                    v[i][i]=x;
            }
            void Print(){
                for (int i=1;i<=t;i++,puts(""))
                    for (int j=1;j<=t;j++)
                        printf("%3d ",v[i][j]);
                puts("");
            }
        }M(0);
        Mat operator * (Mat A,Mat B){
            Mat C(0);
            for (int i=1;i<=t;i++)
                for (int j=1;j<=t;j++)
                    for (int k=1;k<=t;k++)
                        C.v[i][j]=(1LL*A.v[i][k]*B.v[k][j]+C.v[i][j])%mod;
            return C;
        }
        Mat Pow(Mat x,LL y){
            Mat ans(1);
            for (;y;y>>=1,x=x*x)
                if (y&1LL)
                    ans=ans*x;
            return ans;
        }
    }
    View Code

    堆优化 dijkstra

    int dis[N],vis[N];
    struct Node{
        int x,d;
        Node(){}
        Node(int _x,int _d){
            x=_x,d=_d;
        }
        friend bool operator < (Node x,Node y){
            return x.d>y.d;
        }
    };
    priority_queue <Node> Q;
    void Dijkstra(){
        while (!Q.empty())
            Q.pop();
        for (int i=1;i<=n;i++)
            dis[i]=2e9+5;
        dis[1]=0;
        memset(vis,0,sizeof vis);
        Q.push(Node(1,0));
        while (!Q.empty()){
            Node now=Q.top();
            Q.pop();
            int x=now.x;
            if (vis[x])
                continue;
            vis[x]=1,dis[x]=now.d;
            for (int i=g.fst[x];i;i=g.nxt[i])
                Q.push(Node(g.y[i],dis[x]+g.z[i]));
        }
    }
    View Code

    欧拉函数

    int phi(int n){
        int res=n;
        for (int i=2;i*i<=n;i++)
            if (n%i==0){
                res=res/i*(i-1);
                while (n%i==0)
                    n/=i;
            }
        if (n>1)
            res=res/n*(n-1);
        return res;
    }
    View Code

    BSGS (需要加入 Hash 表模板)

    int BSGS(int A,int B,int P){
        int M=max((int)(0.8*sqrt(1.0*P)),1),AM=Pow(A,M,P);
        Map.clear();
        for (int b=0,pw=B;b<M;b++,pw=1LL*pw*A%P)
            Map.update(pw,b+1);
        for (int a=M,pw=AM;a-M<P;a+=M,pw=1LL*pw*AM%P){
            int v=Map.find(pw);
            if (v)
                return a-(v-1);
        }
        return -1;
    }
    View Code


    Berlekamp_Massey (BM)

    #include <bits/stdc++.h>
    #define clr(x) memset(x,0,sizeof (x))
    #define For(i,a,b) for (int i=a;i<=b;i++)
    #define Fod(i,b,a) for (int i=b;i>=a;i--)
    #define pb(x) push_back(x)
    #define mp(x,y) make_pair(x,y)
    #define fi first
    #define se second
    #define _SEED_ ('C'+'L'+'Y'+'A'+'K'+'I'+'O'+'I')
    #define outval(x) printf(#x" = %d
    ",x)
    #define outvec(x) printf("vec "#x" = ");for (auto _v : x)printf("%d ",_v);puts("")
    #define outtag(x) puts("----------"#x"----------")
    #define outarr(a,L,R) printf(#a"[%d...%d] = ",L,R);
                            For(_v2,L,R)printf("%d ",a[_v2]);puts("");
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef vector <int> vi;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f|=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=0x1233,mod=1e9+7;
    void Add(int &x,int y){
        if ((x+=y)>=mod)
            x-=mod;
    }
    void Del(int &x,int y){
        if ((x-=y)<0)
            x+=mod;
    }
    int Pow(int x,int y){
        int ans=1;
        for (;y;y>>=1,x=(LL)x*x%mod)
            if (y&1)
                ans=(LL)ans*x%mod;
        return ans;
    }
    int n,cnt;
    int a[N];
    int Fail[N],delta[N];
    vector <int> R[N];
    int main(){
        n=read();
        For(i,1,n)
            a[i]=read();
        R[0].clear();
        cnt=0;
        For(i,1,n){
            if (cnt==0){
                if (a[i]){
                    Fail[cnt++]=i;
                    delta[i]=a[i];
                    R[cnt].resize(0);
                    R[cnt].resize(i,0);
                }
                continue;
            }
            int sum=0,m=R[cnt].size();
            delta[i]=a[i];
            Fail[cnt]=i;
            For(j,0,m-1)
                Add(sum,(LL)a[i-j-1]*R[cnt][j]%mod);
            Del(delta[i],sum);
            if (!delta[i])
                continue;
            int id=cnt-1,v=i-Fail[id]+(int)R[id].size();
            For(j,0,cnt-1)
                if (i-Fail[j]+(int)R[j].size()<v)
                    id=j,v=i-Fail[j]+(int)R[j].size();
            int tmp=(LL)delta[i]*Pow(delta[Fail[id]],mod-2)%mod;
            R[cnt+1]=R[cnt];
            while (R[cnt+1].size()<v)
                R[cnt+1].pb(0);
            Add(R[cnt+1][i-Fail[id]-1],tmp);
            For(j,0,(int)R[id].size()-1)
                Del(R[cnt+1][i-Fail[id]+j],(LL)tmp*R[id][j]%mod);
            cnt++;
        }
        printf("%d
    ",(int)R[cnt].size());
        For(i,0,(int)R[cnt].size()-1)
            printf("%d ",R[cnt][i]);
        puts("");
        return 0;
    }
    View Code

    Cayley-Hamilton定理优化常系数线性递推(CH) (BZOJ4161)

    #include <bits/stdc++.h>
    #define clr(x) memset(x,0,sizeof x)
    #define For(i,a,b) for (int i=a;i<=b;i++)
    #define Fod(i,b,a) for (int i=b;i>=a;i--)
    #define fi first
    #define se second
    #define pb(x) push_back(x)
    #define mp(x,y) make_pair(x,y)
    #define outval(x) printf(#x" = %d
    ",x)
    #define outtag(x) puts("---------------"#x"---------------")
    #define outarr(a,L,R) printf(#a"[%d..%d] = ",L,R);
                            For(_x,L,R)printf("%d ",a[_x]);puts("")
    using namespace std;
    typedef long long LL;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f|=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=2005*2,mod=1e9+7;
    int n,k;
    int a[N],b[N];
    void Add(int &x,int y){
        if ((x+=y)>=mod)
            x-=mod;
    }
    void Del(int &x,int y){
        if ((x-=y)<0)
            x+=mod;
    }
    int c[N];
    void Mul(int *x,int *y){
        static int z[N];
        clr(z);
        For(i,0,k-1)
            For(j,0,k-1)
                Add(z[i+j],(LL)x[i]*y[j]%mod);
        Fod(i,2*k-2,k){
            if (!z[i])
                continue;
            For(j,1,k)
                Add(z[i-j],(LL)a[j]*z[i]%mod);
        }
        For(i,0,k-1)
            x[i]=z[i];
    }
    void GetPoly(){
        static int x[N];
        int y=n;
        clr(x),clr(c),c[0]=x[1]=1;
        for (;y;y>>=1,Mul(x,x))
            if (y&1)
                Mul(c,x);
    }
    int main(){
        n=read(),k=read();
        For(i,1,k)
            a[i]=(read()+mod)%mod;
        For(i,0,k-1)
            b[i]=(read()+mod)%mod;
        GetPoly();
        int ans=0;
        For(i,0,k-1)
            Add(ans,(LL)b[i]*c[i]%mod);
        cout<<ans<<endl;
        return 0;
    }
    View Code

    带花树(一般图最大匹配)

    #include <bits/stdc++.h>
    #define clr(x) memset(x,0,sizeof (x))
    #define For(i,a,b) for (int i=(a);i<=(b);i++)
    #define pb(x) push_back(x)
    using namespace std;
    typedef long long LL;
    typedef vector <int> vi;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=505;
    int n,m;
    vi e[N];
    queue <int> q;
    int match[N],vis[N],pre[N],tp[N];
    int fa[N];
    int getf(int x){
        return fa[x]==x?x:fa[x]=getf(fa[x]);
    }
    int getlca(int x,int y){
        static int Time=0;
        Time++;
        while (1){
            if (x){
                x=getf(x);
                if (vis[x]==Time)
                    return x;
                vis[x]=Time;
                x=pre[match[x]];
            }
            swap(x,y);
        }
    }
    void shrink(int x,int y,int lca){
        while (getf(x)!=lca){
            pre[x]=y,y=match[x];
            if (tp[y]==2)
                tp[y]=1,q.push(y);
            if (fa[x]==x)
                fa[x]=lca;
            if (fa[y]==y)
                fa[y]=lca;
            x=pre[y];
        }
    }
    int augment(int s){
        if (match[s])
            return 0;
        clr(tp),clr(pre);
        iota(fa+1,fa+n+1,1);
        while (!q.empty())
            q.pop();
        q.push(s);
        tp[s]=1;
        while (!q.empty()){
            int x=q.front();
            q.pop();
            assert(tp[x]==1);
            for (auto y : e[x]){
                if (getf(x)==getf(y)||tp[y]==2)
                    continue;
                if (!tp[y]){
                    pre[y]=x;
                    if (!match[y]){
                        do {
                            int z=match[x];
                            match[x]=y,match[y]=x;
                            y=z,x=pre[z];
                        } while (x);
                        return 1;
                    }
                    tp[y]=2,tp[match[y]]=1;
                    q.push(match[y]);
                }
                else {
                    int lca=getlca(x,y);
                    shrink(x,y,lca);
                    shrink(y,x,lca);
                }
            }
        }
        return 0;
    }
    int main(){
        n=read(),m=read();
        For(i,1,m){
            int x=read(),y=read();
            e[x].pb(y),e[y].pb(x);
        }
        int res=0;
        For(i,1,n)
            res+=augment(i);
        cout<<res<<endl;
        For(i,1,n)
            cout<<match[i]<<" ";
        cout<<endl;
        return 0;
    }
    View Code

    Dinic (Updated on 2019-03-24)

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    struct Edge{
        int x,y,nxt,cap;
        Edge(){}
        Edge(int a,int b,int c,int d){
            x=a,y=b,cap=c,nxt=d;
        }
    };
    struct Network{
        static const int N=105,M=5005*2,INF=0x7FFFFFFF;
        Edge e[M],tmp[M];
        int n,S,T,fst[N],cur[N],cnt;
        int q[N],dis[N],head,tail;
        LL MaxFlow;
        void clear(int _n){
            n=_n,cnt=1;
            memset(fst,0,sizeof fst);
        }
        void add(int a,int b,int c){
            e[++cnt]=Edge(a,b,c,fst[a]),fst[a]=cnt;
            e[++cnt]=Edge(b,a,0,fst[b]),fst[b]=cnt;
        }
        void init(){
            for (int i=1;i<=n;i++)
                cur[i]=fst[i];
        }
        void init(int _S,int _T){
            S=_S,T=_T,MaxFlow=0,init();
        }
        int bfs(){
            memset(dis,0,sizeof dis);
            head=tail=0,q[++tail]=T,dis[T]=1;
            while (head<tail)
                for (int x=q[++head],i=fst[x];i;i=e[i].nxt)
                    if (!dis[e[i].y]&&e[i^1].cap){
                        dis[q[++tail]=e[i].y]=dis[x]+1;
                        if (e[i].y==S)
                            return 1;
                    }
            return (bool)dis[S];
        }
        int dfs(int x,int Flow){
            if (x==T||!Flow)
                return Flow;
            int now=Flow;
            for (int &i=cur[x];i;i=e[i].nxt){
                int y=e[i].y;
                if (dis[x]==dis[y]+1&&e[i].cap){
                    int d=dfs(y,min(now,e[i].cap));
                    e[i].cap-=d,e[i^1].cap+=d,now-=d;
                    if (now==0)
                        break;
                }
            }
            return Flow-now;
        }
        LL Dinic(){
            while (bfs())
                init(),MaxFlow+=dfs(S,INF);
            return MaxFlow;
        }
        LL Auto(int _S,int _T){
            init(_S,_T);
            return Dinic();
        }
    }g;
    int n,m,S,T;
    int main(){
        scanf("%d%d%d%d",&n,&m,&S,&T);
        g.clear(n);
        while (m--){
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            g.add(a,b,c);
        }
        printf("%lld",g.Auto(S,T));
        return 0;
    }
    View Code

    Dinic - 上下界网络流

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    struct Edge{
        int x,y,nxt;
        LL cap;
        Edge(){}
        Edge(int a,int b,LL c,int d){
            x=a,y=b,cap=c,nxt=d;
        }
    };
    struct Network{
        static const int N=50010,M=175010*2;
        static const LL INF=1LL<<50;
        Edge e[M];
        int n,S,T,fst[N],cur[N],cnt;
        int q[N],head,tail;
        LL MaxFlow,dis[N];
        void clear(int _n){
            n=_n,cnt=1;
            memset(fst,0,sizeof fst);
        }
        void add(int a,int b,LL c){
            e[++cnt]=Edge(a,b,c,fst[a]),fst[a]=cnt;
            e[++cnt]=Edge(b,a,0,fst[b]),fst[b]=cnt;
        }
        void init(){
            for (int i=1;i<=n;i++)
                cur[i]=fst[i];
        }
        void init(int _S,int _T){
            S=_S,T=_T,MaxFlow=0,init();
        }
        int bfs(){
            memset(dis,0,sizeof dis);
            head=tail=0;
            q[++tail]=T,dis[T]=1;
            while (head<tail)
                for (int x=q[++head],i=fst[x];i;i=e[i].nxt)
                    if (!dis[e[i].y]&&e[i^1].cap){
                        if (e[i].y==T)
                            return 1;
                        dis[q[++tail]=e[i].y]=dis[x]+1;
                    }
            return (bool)dis[S];
        }
        LL dfs(int x,LL Flow){
            if (x==T||!Flow)
                return Flow;
            LL now=Flow;
            for (int &i=cur[x];i;i=e[i].nxt){
                int y=e[i].y;
                if (dis[x]==dis[y]+1&&e[i].cap){
                    LL d=dfs(y,min(now,e[i].cap));
                    e[i].cap-=d,e[i^1].cap+=d,now-=d;
                    if (now==0)
                        break;
                }
            }
            return Flow-now;
        }
        LL Dinic(){
            while (bfs())
                init(),MaxFlow+=dfs(S,INF);
            return MaxFlow;
        }
        LL Auto(int _S,int _T){
            init(_S,_T);
            return Dinic();
        }
    };
    struct LU_Network{
        static const int N=50010;
        static const LL INF=1LL<<50;
        Network g;
        int n,S,T;
        LL in[N],Sum;
        void clear(int _n){
            memset(in,0,sizeof in);
            n=_n,S=n+1,T=n+2,Sum=0,g.clear(T);
        }
        void add(int x,int y,int L,int U){
            g.add(x,y,U-L),in[x]-=L,in[y]+=L;
        }
        void build(){
            for (int i=1;i<=n;i++)
                if (in[i]>0)
                    g.add(S,i,in[i]),Sum+=in[i];
                else if (in[i]<0)
                    g.add(i,T,-in[i]);
        }
        bool CanFlow(){
            build();
            return g.Auto(S,T)>=Sum;
        }
        int cur,c1;
        bool CanFlow(int s,int t){
            build();
            c1=g.cnt;
            g.add(t,s,INF>>1);
            cur=g.cnt;
            return g.Auto(S,T)>=Sum;
        }
        LL MaxFlow(int s,int t){
            if (!CanFlow(s,t))
                return -1;
            return g.Auto(s,t);
        }
        LL MinFlow(int s,int t){
            if (!CanFlow(s,t))
                return -1;
            LL now=g.e[g.cnt].cap;
            g.e[g.cnt].cap=g.e[g.cnt^1].cap=0;
            return now-g.Auto(t,s);
        }
    }g;
    int read(){
        int x=0;
        char ch=getchar();
        while (!isdigit(ch))
            ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return x;
    }
    int n,m,S,T;
    signed main(){
        n=read(),m=read(),S=read(),T=read();
        g.clear(n);
        for (int i=1;i<=m;i++){
            int x=read(),y=read(),L=read(),U=read();
            g.add(x,y,L,U);
        }
        LL ans=g.MinFlow(S,T);
        if (!~ans)
            puts("please go home to sleep");
        else
            printf("%lld",ans);
        return 0;
    }
    View Code

    动态DP(洛谷P4719)

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f|=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=100005;
    const LL INF=1e17;
    int n,m;
    vector <int> e[N];
    int fa[N],depth[N],son[N],size[N],top[N],dn[N],p[N],ap[N],Time=0;
    int v[N];
    LL dp[N][2],g[N][2];
    void dfs1(int x,int pre,int d){
        depth[x]=d,fa[x]=pre,son[x]=0,size[x]=1;
        for (auto y : e[x])
            if (y!=pre){
                dfs1(y,x,d+1);
                size[x]+=size[y];
                if (!son[x]||size[y]>size[son[x]])
                    son[x]=y;
            }
    }
    void dfs2(int x,int Top){
        top[x]=Top;
        ap[p[x]=++Time]=x;
        if (son[x])
            dfs2(son[x],Top),dn[x]=dn[son[x]];
        else
            dn[x]=x;
        g[x][0]=0,g[x][1]=v[x];
        for (auto y : e[x])
            if (y!=fa[x]&&y!=son[x]){
                dfs2(y,y);
                g[x][0]+=max(dp[y][0],dp[y][1]);
                g[x][1]+=dp[y][0];
            }
        dp[x][0]=g[x][0],dp[x][1]=g[x][1];
        if (son[x]){
            dp[x][0]+=max(dp[son[x]][0],dp[son[x]][1]);
            dp[x][1]+=dp[son[x]][0];
        }
    }
    struct Mat{
        LL v00,v01,v10,v11;
        Mat(){}
        Mat(LL x){
            v00=v01=v10=v11=-INF;
            if (x)
                v00=v11=0;
        }
        void init(LL x00,LL x01,LL x10,LL x11){
            v00=x00,v01=x01,v10=x10,v11=x11;
        }
        void Print(){
            cout << "Matrix: {
        " 
            << v00 << " " << v01 << "
        " << v10 << " " << v11 << endl
            << "}
    ";
        }
    }_1(1);
    Mat operator * (Mat A,Mat B){
        static Mat C;
        C.v00=max(A.v00+B.v00,A.v01+B.v10);
        C.v01=max(A.v00+B.v01,A.v01+B.v11);
        C.v10=max(A.v10+B.v00,A.v11+B.v10);
        C.v11=max(A.v10+B.v01,A.v11+B.v11);
        return C;
    }
    Mat val[N<<2];
    void build(int rt,int L,int R){
        if (L==R){
            int x=ap[L];
            val[rt].init(g[x][0],g[x][0],g[x][1],-INF);
            return;
        }
        int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
        build(ls,L,mid);
        build(rs,mid+1,R);
        val[rt]=val[ls]*val[rs];
    }
    void update(int rt,int L,int R,int x,LL v0,LL v1){
        if (L==R){
            val[rt].init(v0,v0,v1,-INF);
            return;
        }
        int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
        if (x<=mid)
            update(ls,L,mid,x,v0,v1);
        else
            update(rs,mid+1,R,x,v0,v1);
        val[rt]=val[ls]*val[rs];
    }
    Mat query(int rt,int L,int R,int xL,int xR){
        if (xL<=L&&R<=xR)
            return val[rt];
        int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
        if (xR<=mid)
            return query(ls,L,mid,xL,xR);
        else if (xL>mid)
            return query(rs,mid+1,R,xL,xR);
        else
            return query(ls,L,mid,xL,xR)*query(rs,mid+1,R,xL,xR);
    }
    void update(int x,LL g0,LL g1){
        static Mat M;
        int f=top[x];
        while (x){
            update(1,1,n,p[x],g[x][0]=g0,g[x][1]=g1);
            M=query(1,1,n,p[f],p[dn[x]]);
            x=fa[f];
            if (x){
                g0=g[x][0]-max(dp[f][0],dp[f][1])+max(M.v00,M.v10);
                g1=g[x][1]-dp[f][0]+M.v00;
            }
            dp[f][0]=M.v00;
            dp[f][1]=M.v10;
            f=top[x];
        }
    }
    int main(){
        n=read(),m=read();
        for (int i=1;i<=n;i++)
            v[i]=read();
        for (int i=1;i<n;i++){
            int x=read(),y=read();
            e[x].push_back(y);
            e[y].push_back(x);
        }
        dfs1(1,0,0);
        dfs2(1,1);
        build(1,1,n);
        while (m--){
            int x=read(),y=read();
            update(x,g[x][0],g[x][1]-v[x]+y);
            v[x]=y;
            printf("%lld
    ",max(dp[1][1],dp[1][0]));
        }
        return 0;
    }
    View Code

    动态DP - 全局平衡二叉树优化(洛谷P4751)

    #pragma GCC optimize("Ofast","inline")
    #include <bits/stdc++.h>
    using namespace std;
    typedef int LL;
    namespace IO{
        const int Len=1<<21;
        char Ibuf[Len+1],*Is=Ibuf,*It=Ibuf;
        char gc(){
            if (Is==It){
                It=(Is=Ibuf)+fread(Ibuf,1,Len,stdin);
                if (Is==It)
                    return EOF;
            }
            return *Is++;
        }
        int read(){
            int x=0,f=0;
            char ch=gc();
            while (!isdigit(ch))
                f|=ch=='-',ch=gc();
            while (isdigit(ch))
                x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return f?-x:x;
        }
        char Obuf[Len+1],*Ot=Obuf;
        void pc(char ch){
            if (Ot==Obuf+Len){
                fwrite(Obuf,1,Len,stdout);
                Ot=Obuf;
            }
            *Ot++=ch;
        }
        void flush(){
            fwrite(Obuf,1,Ot-Obuf,stdout);
            Ot=Obuf;
        }
        void write(int x){
            static int buf[100],d;
            if (x==0)
                pc('0');
            else {
                if (x<0)
                    pc('-'),x=-x;
                for (d=0;x;x/=10)
                    buf[++d]=x%10+48;
                while (d)
                    pc((char)buf[d--]);
            }
        }
    }
    using namespace IO;
    const int N=1000005;
    const LL INF=2147483647>>1;
    int n,m;
    vector <int> e[N];
    int fa[N],depth[N],son[N],size[N],top[N],dn[N];
    LL v[N],dp[N][2],g[N][2];
    void dfs1(int x,int pre,int d){
        depth[x]=d,fa[x]=pre,son[x]=0,size[x]=1;
        for (auto y : e[x])
            if (y!=pre){
                dfs1(y,x,d+1);
                size[x]+=size[y];
                if (!son[x]||size[y]>size[son[x]])
                    son[x]=y;
            }
    }
    void dfs2(int x,int Top){
        top[x]=Top;
        if (son[x])
            dfs2(son[x],Top),dn[x]=dn[son[x]];
        else
            dn[x]=x;
        g[x][0]=0,g[x][1]=v[x];
        for (auto y : e[x])
            if (y!=fa[x]&&y!=son[x]){
                dfs2(y,y);
                g[x][0]+=max(dp[y][0],dp[y][1]);
                g[x][1]+=dp[y][0];
            }
        dp[x][0]=g[x][0],dp[x][1]=g[x][1];
        if (son[x]){
            dp[x][0]+=max(dp[son[x]][0],dp[son[x]][1]);
            dp[x][1]+=dp[son[x]][0];
        }
    }
    struct Mat{
        LL v00,v01,v10,v11;
        Mat(){}
        Mat(LL x){
            v00=v01=v10=v11=-INF;
            if (x)
                v00=v11=0;
        }
        Mat(LL x00,LL x01,LL x10,LL x11){
            v00=x00,v01=x01,v10=x10,v11=x11;
        }
        void init(LL x00,LL x01,LL x10,LL x11){
            v00=x00,v01=x01,v10=x10,v11=x11;
        }
        void Print(){
            cout << "Matrix: {
        " 
            << v00 << " " << v01 << "
        " 
            << v10 << " " << v11 << "
    }
    ";
        }
        inline friend Mat operator * (Mat A,Mat &B){
            return Mat(
                max(A.v00+B.v00,A.v01+B.v10),
                max(A.v00+B.v01,A.v01+B.v11),
                max(A.v10+B.v00,A.v11+B.v10),
                max(A.v10+B.v01,A.v11+B.v11));
        }
    }_1(1);
    Mat val[N],G[N];
    int Son[N][2];
    void update(int x){
        val[x]=val[Son[x][0]]*G[x]*val[Son[x][1]];
    }
    int tmp[N],sz[N];
    int build_bin(int pre,int L,int R){
        if (L>R)
            return 0;
        int x=L,mid,l=L+1,r=R,ckv=(sz[L-1]+sz[R]+1)>>1;
        while (l<=r){
            mid=(l+r)>>1;
            if (sz[mid]<=ckv)
                l=mid+1,x=mid;
            else
                r=mid-1;
        }
        mid=x,fa[x=tmp[x]]=pre;
        Son[x][0]=build_bin(x,L,mid-1);
        Son[x][1]=build_bin(x,mid+1,R);
        update(x);
        return x;
    }
    int build(int pre,int x){
        for (int i=x;i;i=son[i])
            for (auto y : e[i])
                if (y!=son[i]&&y!=fa[i])
                    build(i,y);
        int n=tmp[0]=sz[0]=0;
        for (int i=x;i;i=son[i]){
            tmp[++n]=i;
            sz[n]=sz[n-1]+size[i]-size[son[i]];
        }
        return build_bin(pre,1,n);
    }
    int main(){
        n=read(),m=read();
        for (int i=1;i<=n;i++)
            v[i]=read();
        for (int i=1;i<n;i++){
            int x=read(),y=read();
            e[x].push_back(y);
            e[y].push_back(x);
        }
        dfs1(1,0,0);
        dfs2(1,1);
        for (int i=1;i<=n;i++)
            G[i].init(g[i][0],g[i][0],g[i][1],-INF);
        val[0]=_1;
        int root=build(0,1),lastans=0;
        while (m--){
            int x=read()^lastans,y=read();
            
            g[x][1]+=y-v[x],v[x]=y;
            G[x].v10=g[x][1];
            int f;
            while (x){
                int f=fa[x];
                if (f&&Son[f][0]!=x&&Son[f][1]!=x){
                    g[f][0]-=max(val[x].v00,val[x].v10);
                    g[f][1]-=val[x].v00;
                }
                update(x);
                if (f&&Son[f][0]!=x&&Son[f][1]!=x){
                    g[f][0]+=max(val[x].v00,val[x].v10);
                    g[f][1]+=val[x].v00;
                    G[f].v00=g[f][0];
                    G[f].v01=g[f][0];
                    G[f].v10=g[f][1];
                }
                x=f;
            }
            
            write(lastans=max(val[root].v00,val[root].v10));
            pc('
    ');
        }
        flush();
        return 0;
    }
    View Code

    动态凸包(支持插入直线)(c++11)

    const LL Q_Tag=-1.02e18;
    struct Line{
        LL k,b;
        mutable function <const Line *()> nxt;
        bool operator < (const Line &x) const {
            if (x.b!=Q_Tag)
                return k!=x.k?k<x.k:b<x.b;
            const Line *s=nxt();
            if (!s)
                return 0;
            return b-(*s).b<((*s).k-k)*x.k;//long double?
        }
    };
    struct DynamicHull:public multiset <Line>{
        int useless(iterator y){
            iterator z=next(y);
            if (y==begin()){
                if (z==end())
                    return 0;
                return (*y).k==(*z).k&&(*y).b<=(*z).b;
            }
            iterator x=prev(y);
            if (z==end())
                return (*y).k==(*x).k&&(*y).b<=(*x).b;
            return ((*x).b-(*y).b)*((*z).k-(*y).k)
                 >=((*y).b-(*z).b)*((*y).k-(*x).k);//long double?
        }
        void insert_line(LL k,LL b){
            iterator x=insert({k,b});
            x->nxt=[=]{return next(x)==end()?0:&*next(x);};
            if (useless(x)){
                erase(x);
                return;
            }
            while (next(x)!=end()&&useless(next(x)))
                erase(next(x));
            while (x!=begin()&&useless(prev(x)))
                erase(prev(x));
        }
        LL calc(LL x){
            Line L=*lower_bound({x,Q_Tag});
            return L.k*x+L.b;
        }
    };
    View Code

    多项式(求逆、开根、对数、指数、快速幂、除法、取模、多点求值、快速插值)

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f|=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=1<<18,mod=998244353;
    void Add(int &x,int y){
        if ((x+=y)>=mod)
            x-=mod;
    }
    void Del(int &x,int y){
        if ((x-=y)<0)
            x+=mod;
    }
    int del(int x,int y){
        return x-y<0?x-y+mod:x-y;
    }
    int Pow(int x,int y){
        int ans=1;
        for (;y;y>>=1,x=(LL)x*x%mod)
            if (y&1)
                ans=(LL)ans*x%mod;
        return ans;
    }
    int randint(){
        return ((rand()&65535)<<15)^(rand()&65535);
    }
    namespace Rem2{
        int INIT_TAG=0;
        int t,w;
        #define fi first
        #define se second
        void init(){
            INIT_TAG=1;
            srand('C'+'L'+'Y'+'A'+'K'+'I'+'O'+'I');
        }
        pair <int,int> Mul_pii(pair <int,int> A,pair <int,int> B){
            static int a,b;
            a=((LL)A.fi*B.fi+(LL)A.se*B.se%mod*w)%mod;
            b=((LL)A.fi*B.se+(LL)A.se*B.fi)%mod;
            return make_pair(a,b);
        }
        pair <int,int> Pow_pii(pair <int,int> x,int y){
            pair <int,int> ans=make_pair(1,0);
            for (;y;y>>=1,x=Mul_pii(x,x))
                if (y&1)
                    ans=Mul_pii(ans,x);
            return ans;
        }
        int Sqrt(int x){
            if (!INIT_TAG)
                init();
            if (x==0)
                return 0;
            if (Pow(x,(mod-1)/2)!=1)
                return -1;
            do {
                t=randint()%(mod-1)+1;
                w=((LL)t*t+mod-x)%mod;
            } while (Pow(w,(mod-1)/2)==1);
            pair <int,int> res=Pow_pii(make_pair(t,1),(mod+1)/2);
            return min(res.fi,mod-res.fi);
        }
    }
    namespace Polynomial{
        namespace Fast{
            const int N=1<<18;
            int n,Log[N+1],Fac[N+1],InvFac[N+1],Inv[N+1];
            int ww[N*2],*Ew=ww,*w[N+1];
            int iww[N*2],*Ei=iww,*iw[N+1];
            int INIT_TAG=0;
            void init(int _n){
                INIT_TAG=1;
                Log[1]=0,n=_n;
                for (int i=2;i<=N;i++)
                    Log[i]=Log[i>>1]+1;
                for (int i=Fac[0]=1;i<=N;i++)
                    Fac[i]=(LL)Fac[i-1]*i%mod;
                InvFac[N]=Pow(Fac[N],mod-2);
                for (int i=N;i>=1;i--)
                    InvFac[i-1]=(LL)InvFac[i]*i%mod;
                for (int i=1;i<=N;i++)
                    Inv[i]=(LL)InvFac[i]*Fac[i-1]%mod;
                for (int d=0;d<=Log[n];d++){
                    w[d]=Ew,iw[d]=Ei;
                    int n=1<<d;
                    w[d][0]=1,w[d][1]=Pow(3,(mod-1)/n);
                    for (int i=2;i<n;i++)
                        w[d][i]=(LL)w[d][i-1]*w[d][1]%mod;
                    iw[d][0]=1,iw[d][1]=Pow(w[d][1],mod-2);
                    for (int i=2;i<n;i++)
                        iw[d][i]=(LL)iw[d][i-1]*iw[d][1]%mod;
                    Ew+=n,Ei+=n;
                }
            }
            int Rev[N+1],A[N+1],B[N+1];
            void FFT(int a[],int n,int **w){
                if (!INIT_TAG)
                    init(N);
                for (int i=0;i<n;i++)
                    if (Rev[i]<i)
                        swap(a[i],a[Rev[i]]);
                for (int t=1,d=1;d<n;t++,d<<=1)
                    for (int i=0;i<n;i+=(d<<1))
                        for (int j=0,*W=w[t];j<d;j++){
                            int tmp=(LL)(*W++)*a[i+j+d]%mod;
                            a[i+j+d]=del(a[i+j],tmp);
                            Add(a[i+j],tmp);
                        }
            }
            vector <int> Mul(vector <int> &a,vector <int> &b){
                static vector <int> res;
                res.clear();
                LL Br=(LL)a.size()*b.size();
                LL FF=(a.size()+b.size())*Log[a.size()+b.size()]*10+100;
                if (Br<=FF){
                    for (int i=0;i<a.size()+b.size();i++)
                        res.push_back(0);
                    for (int i=0;i<a.size();i++)
                        for (int j=0;j<b.size();j++)
                            res[i+j]=((LL)a[i]*b[j]+res[i+j])%mod;
                }
                else {
                    int n=1,d=0;
                    for (;n<a.size()+b.size();n<<=1,d++);
                    for (int i=0;i<n;i++)
                        Rev[i]=(Rev[i>>1]>>1)|((i&1)<<(d-1)),A[i]=B[i]=0;
                    for (int i=0;i<a.size();i++)
                        A[i]=a[i];
                    for (int i=0;i<b.size();i++)
                        B[i]=b[i];
    //                w[0]=1,w[1]=Pow(3,(mod-1)/n);
    //                for (int i=2;i<n;i++)
    //                    w[i]=(LL)w[i-1]*w[1]%mod;
                    FFT(A,n,w),FFT(B,n,w);
                    for (int i=0;i<n;i++)
                        A[i]=(LL)A[i]*B[i]%mod;
    //                w[1]=Pow(w[1],mod-2);
    //                for (int i=2;i<n;i++)
    //                    w[i]=(LL)w[i-1]*w[1]%mod;
                    FFT(A,n,iw);
                    int inv=Pow(n,mod-2);
                    for (int i=0;i<n;i++)
                        res.push_back((int)((LL)inv*A[i]%mod));
                }
                while (!res.empty()&&!res.back())
                    res.pop_back();
                return res;
            }
            vector <int> MulInv(vector <int> &a,vector <int> &b){
                static vector <int> res;
                res.clear();
                int n=1,d=0;
                for (;n<a.size()*2+b.size();n<<=1,d++);
                for (int i=0;i<n;i++)
                    Rev[i]=(Rev[i>>1]>>1)|((i&1)<<(d-1)),A[i]=B[i]=0;
                for (int i=0;i<a.size();i++)
                    A[i]=a[i];
                for (int i=0;i<b.size();i++)
                    B[i]=b[i];
    //            w[0]=1,w[1]=Pow(3,(mod-1)/n);
    //            for (int i=2;i<n;i++)
    //                w[i]=(LL)w[i-1]*w[1]%mod;
                FFT(A,n,w),FFT(B,n,w);
                for (int i=0;i<n;i++)
                    A[i]=(LL)A[i]*A[i]%mod*B[i]%mod;
    //            w[1]=Pow(w[1],mod-2);
    //            for (int i=2;i<n;i++)
    //                w[i]=(LL)w[i-1]*w[1]%mod;
                FFT(A,n,iw);
                int inv=Pow(n,mod-2);
                for (int i=0;i<n;i++)
                    res.push_back((int)((LL)inv*A[i]%mod));
                while (!res.empty()&&!res.back())
                    res.pop_back();
                return res;
            }
        }
        struct Poly{
            vector <int> v;
            Poly(){
                v.clear();
            }
            Poly(int x){
                v.clear();
                v.push_back(x);
            }
            Poly(vector <int> x){
                v=x;
            }
            int operator ()(int x){
                int ans=0,y=1;
                for (int i=0;i<v.size();i++)
                    ans=((LL)v[i]*y+ans)%mod,y=(LL)y*x%mod;
                return ans;
            }
            int size(){
                return v.size();
            }
            void print(){
                for (int i=0;i<v.size();i++)
                    printf("%d ",v[i]);
            }
            void print(int x){
                for (int i=0;i<x;i++)
                    printf("%d ",i>=v.size()?0:v[i]);
            }
            void print(string s){
                print(),cout << s;
            }
            void clear(){
                v.clear();
            }
            void push_back(int x){
                v.push_back(x);
            }
            void pop_back(){
                v.pop_back();
            }
            int empty(){
                return v.empty();
            }
            int back(){
                return v.back();
            }
            int &operator [](int x){
                return v[x];
            }
            void operator += (Poly A){
                while (v.size()<A.size())
                    v.push_back(0);
                for (int i=0;i<A.size();i++)
                    Add(v[i],A[i]);
            }
            void operator -= (Poly &A){
                while (v.size()<A.size())
                    v.push_back(0);
                for (int i=0;i<A.size();i++)
                    Del(v[i],A[i]);
            }
            void operator *= (Poly &A);
            void Derivation(){
                for (int i=0;i<v.size()-1;i++)
                    v[i]=(LL)v[i+1]*(i+1)%mod;
                v.pop_back();
            }
            void Integral(){
                v.push_back(0);
                for (int i=v.size()-2;i>=0;i--)
                    v[i+1]=(LL)v[i]*Fast :: Inv[i+1]%mod;
                v[0]=0;
            }
            void operator *= (int x){
                for (int i=0;i<v.size();i++)
                    v[i]=(LL)v[i]*x%mod;
            }
        }pp;
        //struct Poly end-------------
        Poly operator + (Poly A,Poly B){
            pp.clear();
            for (int i=0;i<max(A.size(),B.size());i++)
                pp.push_back(0);
            for (int i=0;i<A.size();i++)
                Add(pp[i],A[i]);
            for (int i=0;i<B.size();i++)
                Add(pp[i],B[i]);
            return pp;
        }
        Poly operator - (Poly A,Poly B){
            pp.clear();
            for (int i=0;i<max(A.size(),B.size());i++)
                pp.push_back(0);
            for (int i=0;i<A.size();i++)
                Add(pp[i],A[i]);
            for (int i=0;i<B.size();i++)
                Del(pp[i],B[i]);
            return pp;
        }
        Poly operator * (Poly A,Poly B){
            return Poly(Fast :: Mul(A.v,B.v));
        }
        void Poly :: operator *= (Poly &A){
            v=Fast :: Mul(v,A.v);
        }
        Poly operator * (Poly A,int x){
            pp=A;
            for (int i=0;i<A.size();i++)
                pp[i]=(LL)pp[i]*x%mod;
            return pp;
        }
        Poly Inverse(Poly a,int n);
        Poly operator / (Poly A,Poly B){//Divide
            int n=A.size(),m=B.size();
            reverse(A.v.begin(),A.v.end());
            reverse(B.v.begin(),B.v.end());
            int k=n-m+1;
            if (k<0)
                return Poly(0);
            while (A.size()>k)
                A.pop_back();
            while (B.size()>k)
                B.pop_back();
            A=A*Inverse(B,k);
            while (A.size()>k)
                A.pop_back();
            reverse(A.v.begin(),A.v.end());
            return A;
        }
        Poly operator % (Poly A,Poly B){//Modulo
            while (!A.empty()&&!A.back())
                A.pop_back();
            while (!B.empty()&&!B.back())
                B.pop_back();
            A=A-A/B*B;
            while (A.size()>=B.size())
                A.pop_back();
            while (!A.empty()&&!A.back())
                A.pop_back();
            return A;
        }
        Poly Derivation(Poly A){
            for (int i=0;i<A.size()-1;i++)
                A[i]=(LL)A[i+1]*(i+1)%mod;
            A.pop_back();
            return A;
        }
        Poly Integral(Poly A){
            A.push_back(0);
            for (int i=A.size()-2;i>=0;i--)
                A[i+1]=(LL)A[i]*Fast :: Inv[i+1]%mod;
            A[0]=0;
            return A;
        }
        Poly Inverse(Poly a,int n){
            static Poly A,B;
            while (!a.empty()&&!a.back())
                a.pop_back();
            if (a.empty())
                return a;
            A.clear(),B.clear();
            B.push_back(a[0]);
            A.push_back(Pow(B[0],mod-2));
            for (int t=1;t<n;){
                for (int i=t;i<min(a.size(),(t<<1));i++)
                    B.push_back(a[i]);
                t<<=1;
                A=A*2-Poly(Fast :: MulInv(A.v,B.v));
                while (A.size()>t)
                    A.pop_back();
            }
            while (A.size()>n)
                A.pop_back();
            return A;
        }
        Poly Sqrt(Poly a,int n){
            static Poly A,B;
            while (!a.empty()&&!a.back())
                a.pop_back();
            if (a.empty())
                return a;
            A.clear(),B.clear();
            B.push_back(a[0]);
            A.push_back(Rem2 :: Sqrt(B[0]));
            for (int t=1;t<n;){
                for (int i=t;i<min(a.size(),(t<<1));i++)
                    B.push_back(a[i]);
                t<<=1;
                A+=B*Inverse(A,t);
                while (A.size()>t)
                    A.pop_back();
                A*=499122177;
            }
            if (A[0]>mod-A[0])
                for (int i=0;i<A.size();i++)
                    A[i]=(mod-A[i])%mod;
            while (A.size()>n)
                A.pop_back();
            return A;
        }
        Poly Ln(Poly a,int n){
            while (!a.empty()&&!a.back())
                a.pop_back();
            if (a.empty()||a[0]!=1)
                return a;
            a=Integral(Derivation(a)*Inverse(a,n));
            while (a.size()>n)
                a.pop_back();
            return a;
        }
        Poly Exp(Poly a,int n){
            static Poly A,B;
            while (!a.empty()&&!a.back())
                a.pop_back();
            if (a.empty())
                return Poly(1);
            if (a[0]!=0)
                return a;
            A.clear(),B.clear();
            B.push_back(1);
            A.push_back(a[0]);
            for (int t=1;t<n;){
                for (int i=t;i<min(a.size(),(t<<1));i++)
                    A.push_back(a[i]);
                t<<=1;
                B=B*(Poly(1)+A-Ln(B,t));
                while (B.size()>t)
                    B.pop_back();
            }
            while (B.size()>n)
                B.pop_back();
            return B;
        }
        Poly PolyPow(Poly x,int y,int n){
            static Poly A,B;
            int k0=0,kc,ivkc;
            while (!x.empty()&&!x.back())
                x.pop_back();
            if (x.empty())
                return x;
            while (k0<x.size()&&x[k0]==0)
                k0++;
            kc=x[k0],ivkc=Pow(kc,mod-2);
            A.clear();
            for (int i=k0;i<x.size();i++)
                A.push_back((int)((LL)x[i]*ivkc%mod));
            A=Exp(Ln(A,n)*y,n);
            B.clear();
            if ((LL)k0*y>=n)
                return B;
            kc=Pow(kc,y),k0*=y;
            for (int i=0;i<k0;i++)
                B.push_back(0);
            for (int i=0;i<min(A.size(),n-k0);i++)
                B.push_back((int)((LL)A[i]*kc%mod));
            while (B.size()>n)
                B.pop_back();
            return B;
        }
        namespace Qiuzhi{
            Poly P[N<<2],f[N<<2],M;
            vector <int> x,y;
            int n;
            void GetP(int rt,int L,int R){
                if (L==R){
                    P[rt].clear();
                    P[rt].push_back((mod-x[L])%mod);
                    P[rt].push_back(1);
                    return;
                }
                int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
                GetP(ls,L,mid);
                GetP(rs,mid+1,R);
                P[rt]=P[ls]*P[rs];
            }
            void qiuzhi(int rt,int L,int R){
                if (f[rt].empty())
                    f[rt].push_back(0);
                if (L==R)
                    return (void)(y[L]=f[rt][0]);
                int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
                f[ls]=f[rt]%P[ls];
                f[rs]=f[rt]%P[rs];
                qiuzhi(ls,L,mid);
                qiuzhi(rs,mid+1,R);
            }
            vector <int> Get_Val(vector <int> A,Poly F){
                n=A.size();
                x.clear(),y.clear();
                for (int i=0;i<n;i++){
                    x.push_back(A[i]);
                    y.push_back(0);
                }
                GetP(1,0,n-1);
                f[1]=F;
                qiuzhi(1,0,n-1);
                return y;
            }
        }
        namespace Chazhi{
            Poly P[N<<2],M;
            vector <int> x,y;
            int n;
            void GetP(int rt,int L,int R){
                if (L==R){
                    P[rt].clear();
                    P[rt].push_back((mod-x[L])%mod);
                    P[rt].push_back(1);
                    return;
                }
                int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
                GetP(ls,L,mid);
                GetP(rs,mid+1,R);
                P[rt]=P[ls]*P[rs];
            }
            Poly chazhi(int rt,int L,int R){
                if (L==R)
                    return Poly(y[L]);
                int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
                return chazhi(ls,L,mid)*P[rs]+chazhi(rs,mid+1,R)*P[ls];
            }
            Poly Get_Poly(vector <int> A,vector <int> B){
                n=A.size();
                x=A;
                int Product=1;
                GetP(1,0,n-1);
                M=Derivation(P[1]);
                y=Qiuzhi :: Get_Val(A,M);
                for (int i=0;i<y.size();i++)
                    y[i]=(LL)B[i]*Pow(y[i],mod-2)%mod;
                return chazhi(1,0,n-1);
            }
        }
    }// be careful about init!!!!!!
    using namespace Polynomial;
    Poly A,B;
    vector <int> x,y;
    int main(){
        Fast :: init(1<<18);
        int n=read();
        x.clear(),y.clear();
        for (int i=0;i<n;i++){
            x.push_back(read());
            y.push_back(read());
        }
        A=Chazhi :: Get_Poly(x,y);
        int m=read();
        x.clear();
        for (int i=0;i<m;i++)
            x.push_back(read());
        y=Qiuzhi :: Get_Val(x,A);
        for (int i=0;i<m;i++)
            printf("%d ",y[i]);
        return 0;
    }
    View Code

    UPD2020-07-22(vector 版):

    #include <bits/stdc++.h>
    #define clr(x) memset(x,0,sizeof (x))
    #define For(i,a,b) for (int i=(a);i<=(b);i++)
    #define Fod(i,b,a) for (int i=(b);i>=(a);i--)
    #define fi first
    #define se second
    #define kill _z_kill
    #define pb(x) push_back(x)
    #define mp(x,y) make_pair(x,y)
    #define outval(x) cerr<<#x" = "<<x<<endl
    #define outv(x) cerr<<#x" = "<<x<<"  "
    #define outtag(x) cerr<<"--------------"#x"---------------"<<endl
    #define outarr(a,L,R) cerr<<#a"["<<L<<".."<<R<<"] = ";
        For(_x,L,R) cerr<<a[_x]<<" ";cerr<<endl;
    #define User_Time ((double)clock()/CLOCKS_PER_SEC)
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef unsigned uint;
    typedef long double LD;
    typedef vector <int> vi;
    typedef pair <int,int> pii;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    template <class T> void ckmax(T &x,T y){
        if (x<y)
            x=y;
    }
    template <class T> void ckmin(T &x,T y){
        if (y<x)
            x=y;
    }
    const int mod=998244353;
    int Pow(int x,int y){
        int ans=1;
        for (;y;y>>=1,x=(LL)x*x%mod)
            if (y&1)
                ans=(LL)ans*x%mod;
        return ans;
    }
    void Add(int &x,int y){
        if ((x+=y)>=mod)
            x-=mod;
    }
    void Del(int &x,int y){
        if ((x-=y)<0)
            x+=mod;
    }
    int Add(int x){
        return x>=mod?x-mod:x;
    }
    int Del(int x){
        return x<0?x+mod:x;
    }
    const int N=(1<<18)+1;
    int Fac[N],Inv[N],Iv[N];
    namespace fft{
        int R[N];
        vi w[23];
        void prework(){
            int d=log(N)/log(2)+0.5;
            For(i,1,d){
                w[i].resize(1<<i);
                w[i][0]=1,w[i][1]=Pow(3,(mod-1)>>i);
                For(j,2,(1<<i)-1)
                    w[i][j]=(LL)w[i][j-1]*w[i][1]%mod;
            }
        }
        void init(int n){
            int d=0;
            while ((1<<d)<n)
                d++;
            For(i,0,n-1)
                R[i]=(R[i>>1]>>1)|((i&1)<<(d-1));
        }
        void FFT(int *a,int n,int flag){
            For(i,0,n-1)
                if (i<R[i])
                    swap(a[i],a[R[i]]);
            for (int t=1,d=1;d<n;d<<=1,t++)
                for (int i=0;i<n;i+=d<<1){
                    int *W=&w[t][0];
                    For(j,0,d-1){
                        int tmp=(LL)a[i+j+d]*(*W++)%mod;
                        a[i+j+d]=Del(a[i+j]-tmp);
                        a[i+j]=Add(a[i+j]+tmp);
                    }
                }
            if (flag<0){
                reverse(a+1,a+n);
                int inv=Pow(n,mod-2);
                For(i,0,n-1)
                    a[i]=(LL)a[i]*inv%mod;
            }
        }
    }
    void prework(){
        int n=N-1;
        Fac[0]=1;
        For(i,1,n)
            Fac[i]=(LL)Fac[i-1]*i%mod;
        Inv[n]=Pow(Fac[n],mod-2);
        Fod(i,n,1)
            Inv[i-1]=(LL)Inv[i]*i%mod;
        For(i,1,n)
            Iv[i]=(LL)Inv[i]*Fac[i-1]%mod;
        fft::prework();
    }
    using fft::FFT;
    vi fix(vi a,int n){
        a.resize(n);
        return a;
    }
    int calcfx(vi f,int x){
        int v=1,ans=0;
        For(i,0,(int)f.size()-1){
            Add(ans,(LL)v*f[i]%mod);
            v=(LL)v*x%mod;
        }
        return ans;
    }
    vi operator + (vi a,vi b){
        a.resize(max(a.size(),b.size()));
        For(i,0,(int)b.size()-1)
            Add(a[i],b[i]);
        return a;
    }
    vi operator - (vi a,vi b){
        a.resize(max(a.size(),b.size()));
        For(i,0,(int)b.size()-1)
            Del(a[i],b[i]);
        return a;
    }
    vi operator * (vi a,int b){
        for (auto &i : a)
            i=(LL)i*b%mod;
        return a;
    }
    vi operator * (vi a,vi b){
        if (a.empty()||b.empty())
            return vi();
        int nn=a.size()+b.size()-1,n=1<<(int)(log(nn)/log(2)+1);
        a.resize(n),b.resize(n);
        fft::init(n);
        FFT(&a[0],n,1),FFT(&b[0],n,1);
        For(i,0,n-1)
            a[i]=(LL)a[i]*b[i]%mod;
        FFT(&a[0],n,-1);
        return fix(a,nn);
    }
    const int inv2=(mod+1)/2;
    vi polyinv(vi a){
        if (a.size()==1)
            return vi(1,Pow(a[0],mod-2));
        int n=a.size();
        vi b=polyinv(fix(a,(n+1)/2));
        int len=1<<(int)(log(n+(int)b.size())/log(2)+1);
        vi c=fix(b,len),d=fix(a,len);
        fft::init(len),FFT(&c[0],len,1),FFT(&d[0],len,1);
        For(i,0,len-1)
            c[i]=(LL)c[i]*c[i]%mod*d[i]%mod;
        FFT(&c[0],len,-1);
        b.resize(n);
        For(i,(n+1)/2,n-1)
            Del(b[i],c[i]);
        return b;
    }
    vi Der(vi a){
        For(i,1,(int)a.size()-1)
            a[i-1]=(LL)a[i]*i%mod;
        a.pop_back();
        return a;
    }
    vi Int(vi a){
        a.pb(0);
        Fod(i,(int)a.size()-1,1)
            a[i]=(LL)a[i-1]*Iv[i]%mod;
        a[0]=0;
        return a;
    }
    vi polyln(vi a){
        return Int(fix(Der(a)*polyinv(a),(int)a.size()-1));
    }
    vi polyexp(vi a){
        if (a.size()==1)
            return vi(1,1);
        int n=a.size();
        vi b=polyexp(fix(a,(n+1)/2));
        return fix(b+b*(a-polyln(fix(b,n))),n);
    }
    vi operator / (vi a,vi b){
        int n=a.size(),m=b.size();
        if (n<m)
            return vi();
        reverse(a.begin(),a.end());
        reverse(b.begin(),b.end());
        a=fix(fix(a,n-m+1)*polyinv(fix(b,n-m+1)),n-m+1);
        reverse(a.begin(),a.end());
        return a;
    }
    vi operator % (vi a,vi b){
        return fix(a-a/b*b,(int)b.size()-1);
    }
    vi polypow(vi a,int k){
    //    return polyexp(polyln(a)*k);
        int p=0;
        while (p<(int)a.size()&&!a[p])
            p++;
        if ((LL)p*k>=(int)a.size())
            return vi();
        vi b(a.begin()+p,a.end());
        int coef=b[0],inv=Pow(coef,mod-2),powcoef=Pow(coef,k);
        b=polyexp(polyln(b*inv)*k)*powcoef;
        vi c(p*k+(int)b.size());
        For(i,0,(int)b.size()-1)
            c[i+p*k]=b[i];
        return fix(c,a.size());
    }
    int Sqrt(int x){
        const int g=3;
        unordered_map <int,int> Map;
        int sq=(int)sqrt(mod-1)+1,pv=Pow(g,sq);
        for (int i=0,v=1;i<=sq;i++,v=(LL)v*pv%mod)
            Map[v]=i;
        for (int i=0,v=x;i<sq;i++,v=(LL)v*g%mod)
            if (Map.count(v)){
                int ind=((LL)Map[v]*sq-i+mod-1)%(mod-1);
                if (ind&1)
                    return -1;
                ind/=2;
                int res=Pow(g,ind);
                return min(res,(mod-res)%mod);
            }
        return -1;
    }
    vi polysqrt(vi a){
        if (a.size()==1)
            return vi(1,Sqrt(a[0]));
        int n=a.size();
        vi b=fix(polysqrt(fix(a,(n+1)/2)),n);
        return fix((b+a*polyinv(b))*((mod+1)/2),n);
    }
    namespace eval_inter{
        int n;
        vi f,x,y,prod[N*4];
        void getprod(int rt,int L,int R){
            if (L==R){
                prod[rt]={Del(-x[L]),1};
                return;
            }
            int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
            getprod(ls,L,mid);
            getprod(rs,mid+1,R);
            prod[rt]=prod[ls]*prod[rs];
        }
        void gety(int rt,int L,int R,vi f){
            f=fix(f%prod[rt],(int)prod[rt].size()-1);
            if (L==R)
                return (void)(y[L]=f[0]);
            int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
            gety(ls,L,mid,f);
            gety(rs,mid+1,R,f);
        }
        vi eval(vi _f,vi _x){
            n=_x.size();
            if (!n)
                return vi();
            f=_f,x=_x,y.resize(n);
            getprod(1,0,n-1);
            gety(1,0,n-1,f);
            return y;
        }
        vi getf(int rt,int L,int R){
            if (L==R)
                return vi(1,y[L]);
            int mid=(L+R)>>1,ls=rt<<1,rs=ls|1;
            return getf(ls,L,mid)*prod[rs]+getf(rs,mid+1,R)*prod[ls];
        }
        vi inter(vi _x,vi _y){
            n=_x.size();
            if (!n)
                return vi();
            x=_x,y.resize(n);
            getprod(1,0,n-1);
            vi M=Der(prod[1]);
            gety(1,0,n-1,M);
            For(i,0,n-1)
                y[i]=(LL)_y[i]*Pow(y[i],mod-2)%mod;
            return getf(1,0,n-1);
        }
    }
    int main(){
        prework();
    
    //LOJ #150
        int n=read(),k=read();
        vi f(n+1);
        For(i,0,n)
            f[i]=read();
        vi g=fix(Der(polypow(vi(1,1)+polyln(vi(1,Del(2-f[0]))+f-polyexp(Int(polyinv(polysqrt(f))))),k)),n);
        For(i,0,n-1)
            printf("%d ",g[i]);
        puts("");
    
    /*
    //NFLSOJ #53
        int n=read();
        vi x(n),y(n);
        For(i,0,n-1)
            x[i]=read(),y[i]=read();
        vi f=eval_inter::inter(x,y);
    //    For(i,0,n-1)
    //        printf("%d ",f[i]);
    //    puts("");
        int m=read();
        vi z(m);
        For(i,0,m-1)
            z[i]=read();
        z=eval_inter::eval(f,z);
        For(i,0,m-1)
            printf("%d ",z[i]);
        puts("");
    */
        return 0;
    }
    View Code

    exBSGS (需要加入 Hash 表模板)

    int ExBSGS(int A,int B,int P){
        A%=P,B%=P;
        int k=0,v=1;
        while (1){
            int g=gcd(A,P);
            if (g==1)
                break;
            if (B%g)
                return -1;
            k++,B/=g,P/=g,v=1LL*v*(A/g)%P;
            if (v==B)
                return k;
        }
        if (P==1)
            return k;
        int M=max((int)sqrt(1.0*P),1),AM=Pow(A,M,P);
        Map.clear();
        for (int b=0,pw=B;b<M;b+=1,pw=1LL*pw*A%P)
            Map.update(pw,b+1);
        for (int a=M,pw=1LL*v*AM%P;a-M<P;a+=M,pw=1LL*pw*AM%P){
            int v=Map.find(pw);
            if (v)
                return a-(v-1)+k;
        }
        return -1;
    }
    View Code

    exgcd

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

    exCRT

    LL ex_gcd(LL a,LL b,LL &x,LL &y){
        if (!b){
            x=1,y=0;
            return a;
        }
        LL res=ex_gcd(b,a%b,y,x);
        y-=x*(a/b);
        return res;
    }
    LL gcd(LL a,LL b){
        return b?gcd(b,a%b):a;
    }
    LL inv(LL v,LL p){
        LL x,y,g=ex_gcd(v,p,x,y);
        if (g>1)
            return -1;
        return (x+p)%p;
    }
    LL Mul(LL a,LL b,LL p){
        a=(a%p+p)%p;
        b=(b%p+p)%p;
        LL ans=0;
        for (;a;a>>=1,b=(b<<1)%p)
            if (a&1LL)
                ans=(ans+b)%p;
        return ans;
    }
    bool CRT(LL w1,LL p1,LL w2,LL p2,LL &w,LL &p){
        LL x,y,z=w2-w1,g=ex_gcd(p1,p2,x,y);
        if (z%g)
            return 0;
        LL t=z/g;
        x=Mul(x,t,p2/g);
        p=p1/g*p2;
        w=((w1+Mul(x,p1,p))%p+p)%p;
        return 1;
    }
    View Code

    exkmp

    #include <bits/stdc++.h>
    using namespace std;
    void exKMP(char s[],char t[],int g[],int f[],int n,int m){
        int Max=0;
        f[0]=m;
        for (int i=1;i<m;i++){
            f[i]=max(0,min(f[i-Max],Max+f[Max]-i));
            while (i+f[i]<m&&t[f[i]]==t[i+f[i]])
                f[i]++;
            if (!Max||i+f[i]>Max+f[Max])
                Max=i;
        }
        Max=0;
        for (int i=0;i<n;i++){
            g[i]=max(0,min(f[i-Max],Max+g[Max]-i));
            while (i+g[i]<n&&t[g[i]]==s[i+g[i]])
                g[i]++;
            if (!Max||i+g[i]>Max+g[Max])
                Max=i;
        }
    }
    int main(){
        
        return 0;
    }
    View Code

    exLucas

    LL Pow(LL x,LL y,LL mod){
        if (y==0)
            return 1LL;
        LL xx=Pow(x,y/2,mod);
        xx=xx*xx%mod;
        if (y&1LL)
            xx=xx*x%mod;
        return xx;
    }
    void ex_gcd(LL a,LL b,LL &x,LL &y){
        if (!b)
            x=1,y=0;
        else
            ex_gcd(b,a%b,y,x),y-=a/b*x;
    }
    LL Inv(LL X,LL mod){
        if (!X)
            return 0;
        LL a=X,b=mod,x,y;
        ex_gcd(a,b,x,y);
        x=(x%b+b)%b;
        return x;
    }
    LL ex_lucas(LL n,LL pi,LL pk){
        if (!n)
            return 1LL;
        LL ans=1;
        for (LL i=2;i<=pk;i++)
            if (i%pi)
                ans=ans*i%pk;
        ans=Pow(ans,n/pk,pk);
        for (LL i=2;i<=n%pk;i++)
            if (i%pi)
                ans=ans*i%pk;
        return ans*ex_lucas(n/pi,pi,pk)%pk;
    }
    LL C(LL n,LL m,LL pi,LL pk){
        if (m>n)
            return 0;
        LL a=ex_lucas(n,pi,pk),b=ex_lucas(m,pi,pk),c=ex_lucas(n-m,pi,pk);
        LL k=0,ans;
        for (LL i=n;i;i/=pi,k+=i);
        for (LL i=m;i;i/=pi,k-=i);
        for (LL i=n-m;i;i/=pi,k-=i);
        ans=a*Inv(b,pk)%pk*Inv(c,pk)%pk*Pow(pi,k,pk)%pk;
        return ans*(P/pk)%P*Inv(P/pk,pk)%P;
    }
    LL C(LL n,LL m){
        LL ans=0;
        for (int i=1;i<=cnt;i++)
            ans=(ans+C(n,m,px[i],Pow(px[i],py[i],P+1)))%P;
        return ans;
    }
    View Code

    二次剩余

    int Pow(int x,int y){
        int ans=1;
        for (;y;y>>=1,x=(LL)x*x%mod)
            if (y&1)
                ans=(LL)ans*x%mod;
        return ans;
    }
    int randint(){
        return ((rand()&65535)<<15)^(rand()&65535);
    }
    namespace Rem2{
        int INIT_TAG=0;
        int t,w;
        #define fi first
        #define se second
        void init(){
            INIT_TAG=1;
            srand('C'+'L'+'Y'+'A'+'K'+'I'+'O'+'I');
        }
        pair <int,int> Mul_pii(pair <int,int> A,pair <int,int> B){
            static int a,b;
            a=((LL)A.fi*B.fi+(LL)A.se*B.se%mod*w)%mod;
            b=((LL)A.fi*B.se+(LL)A.se*B.fi)%mod;
            return make_pair(a,b);
        }
        pair <int,int> Pow_pii(pair <int,int> x,int y){
            pair <int,int> ans=make_pair(1,0);
            for (;y;y>>=1,x=Mul_pii(x,x))
                if (y&1)
                    ans=Mul_pii(ans,x);
            return ans;
        }
        int Sqrt(int x){
            if (!INIT_TAG)
                init();
            if (x==0)
                return 0;
            if (Pow(x,(mod-1)/2)!=1)
                return -1;
            do {
                t=randint()%(mod-1)+1;
                w=((LL)t*t+mod-x)%mod;
            } while (Pow(w,(mod-1)/2)==1);
            pair <int,int> res=Pow_pii(make_pair(t,1),(mod+1)/2);
            return min(res.fi,mod-res.fi);
        }
    }
    View Code

    FFT

    const int N=1<<18;
    struct C{
        double r,i;
        C(){}
        C(double _r,double _i){r=_r,i=_i;}
        friend C operator + (C a,C b){return C(a.r+b.r,a.i+b.i);}
        void operator += (C a){r+=a.r,i+=a.i;}
        friend C operator - (C a,C b){return C(a.r-b.r,a.i-b.i);}
        friend C operator * (C a,double b){return C(a.r*b,a.i*b);}
        friend C operator * (C a,C b){return C(a.r*b.r-a.i*b.i,a.r*b.i+a.i*b.r);}
        C conj(){return C(r,-i);}
    }a[N],b[N];
    vector <C> w[19];
    const double pi=acos(-1.0);
    void prework(){
        int d=18;
        For(i,0,d){
            int n=1<<i;
            w[i].resize(n);
            For(j,0,n-1)
                w[i][j]=C(cos(pi*2*j/n),sin(pi*2*j/n));
        }
    }
    int R[N];
    void init(int n){
        int d=0;
        while ((1<<d)<n)
            d++;
        For(i,0,n-1)
            R[i]=(R[i>>1]>>1)|((i&1)<<(d-1));
    }
    void dft(C *a,int n,int flag){
        init(n);
        For(i,0,n-1)
            if (R[i]<i)
                swap(a[i],a[R[i]]);
        for (int t=1,d=1;d<n;d<<=1,t++)
            for (int i=0;i<n;i+=d<<1){
                C *W=w[t].data();
                for (int j=0;j<d;j++){
                    C tmp=a[i+j+d]*(*W++);
                    a[i+j+d]=a[i+j]-tmp;
                    a[i+j]+=tmp;
                }
            }
        if (flag<0){
            reverse(a+1,a+n);
            For(i,0,n-1)
                a[i].r/=n,a[i].i/=n;
        }
    }
    View Code

    FFT - 共轭常数优化(MTT?)(UOJ#34)

    #include <bits/stdc++.h>
    #define clr(x) memset(x,0,sizeof (x))
    #define For(i,a,b) for (int i=(a);i<=(b);i++)
    using namespace std;
    typedef long long LL;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=1<<18;
    struct C{
        double r,i;
        C(){}
        C(double _r,double _i){r=_r,i=_i;}
        friend C operator + (C a,C b){return C(a.r+b.r,a.i+b.i);}
        void operator += (C a){r+=a.r,i+=a.i;}
        friend C operator - (C a,C b){return C(a.r-b.r,a.i-b.i);}
        friend C operator * (C a,double b){return C(a.r*b,a.i*b);}
        friend C operator * (C a,C b){return C(a.r*b.r-a.i*b.i,a.r*b.i+a.i*b.r);}
        C conj(){return C(r,-i);}
    }a[N],b[N];
    vector <C> w[19];
    const double pi=acos(-1.0);
    void prework(){
        int d=18;
        For(i,0,d){
            int n=1<<i;
            w[i].resize(n);
            For(j,0,n-1)
                w[i][j]=C(cos(pi*2*j/n),sin(pi*2*j/n));
        }
    }
    int R[N];
    void init(int n){
        int d=0;
        while ((1<<d)<n)
            d++;
        For(i,0,n-1)
            R[i]=(R[i>>1]>>1)|((i&1)<<(d-1));
    }
    void dft(C *a,int n,int flag){
        init(n);
        For(i,0,n-1)
            if (R[i]<i)
                swap(a[i],a[R[i]]);
        for (int t=1,d=1;d<n;d<<=1,t++)
            for (int i=0;i<n;i+=d<<1){
                C *W=w[t].data();
                for (int j=0;j<d;j++){
                    C tmp=a[i+j+d]*(*W++);
                    a[i+j+d]=a[i+j]-tmp;
                    a[i+j]+=tmp;
                }
            }
        if (flag<0){
            reverse(a+1,a+n);
            For(i,0,n-1)
                a[i].r/=n,a[i].i/=n;
        }
    }
    void mtt(C *a,C *b,int n){
        static C t[N];
        For(i,0,n-1)
            t[i].r=a[i].r,t[i].i=b[i].r;
        dft(t,n,1);
        For(i,0,n-1){
            C p=t[i],q=t[(n-i)&(n-1)];
            a[i]=(p*p-(q*q).conj())*C(0,-0.25);
        }
        dft(a,n,-1);
    }
    int main(){
        prework();
        int n=read()+1,m=read()+1;
        For(i,0,n-1)
            a[i].r=read();
        For(i,0,m-1)
            b[i].r=read();
        int len=1;
        while (len<n+m-1)
            len*=2;
        mtt(a,b,len);
        For(i,0,n+m-2)
            printf("%d ",(int)(a[i].r+0.5));
        puts("");
        return 0;
    }
    View Code

    拆系数FFT

    namespace poly{
        const double pi=acos(-1.0);
        struct Comp{
            double r,i;
            Comp(){}
            Comp(double _r,double _i){
                r=_r,i=_i;
            }
            friend Comp operator + (Comp a,Comp b){
                return Comp(a.r+b.r,a.i+b.i);
            }
            friend Comp operator - (Comp a,Comp b){
                return Comp(a.r-b.r,a.i-b.i);
            }
            friend Comp operator * (Comp a,Comp b){
                return Comp(a.r*b.r-a.i*b.i,a.i*b.r+a.r*b.i);
            }
        }a0[N],a1[N],b0[N],b1[N],A[N],B[N],C[N],w[N];
        int R[N];
        void FFT(Comp a[],int n){
            for (int i=0;i<n;i++)
                if (i<R[i])
                    swap(a[i],a[R[i]]);
            for (int t=n>>1,d=1;d<n;d<<=1,t>>=1)
                for (int i=0;i<n;i+=d<<1)
                    for (int j=0;j<d;j++){
                        Comp tmp=w[t*j]*a[i+j+d];
                        a[i+j+d]=a[i+j]-tmp;
                        a[i+j]=a[i+j]+tmp;
                    }
        }
        vector <int> Mul(vector <int> a,vector <int> b){
            static vector <int> ans;
            ans.clear();
            int n,d;
            for (n=1,d=0;n<a.size()+b.size();n<<=1,d++);
            for (int i=0;i<n;i++){
                R[i]=(R[i>>1]>>1)|((i&1)<<(d-1));
                w[i]=Comp(cos(pi*2/n*i),sin(pi*2/n*i));
            }
            for (int i=0;i<n;i++)
                a0[i]=a1[i]=b0[i]=b1[i]=Comp(0,0);
            for (int i=0;i<a.size();i++)
                a0[i].r=a[i]&32767,a1[i].r=a[i]>>15;
            for (int i=0;i<b.size();i++)
                b0[i].r=b[i]&32767,b1[i].r=b[i]>>15;
            FFT(a0,n),FFT(a1,n),FFT(b0,n),FFT(b1,n);
            for (int i=0;i<n;i++){
                A[i]=a0[i]*b0[i];
                B[i]=a1[i]*b1[i];
                C[i]=a0[i]*b1[i]+a1[i]*b0[i];
                w[i].i=-w[i].i;
            }
            FFT(A,n),FFT(B,n),FFT(C,n);
            for (int i=0;i<n;i++){
                int v=0;
                Add(v,(LL)(A[i].r/n+0.5)%mod);
                Add(v,(((LL)(B[i].r/n+0.5)%mod)<<30)%mod);
                Add(v,(((LL)(C[i].r/n+0.5)%mod)<<15)%mod);
                ans.pb(v);
            }
            while (!ans.empty()&&!ans.back())
                ans.pop_back();
            return ans;
        }
    }
    View Code

    求p的原根G(getg::calc(p))

    namespace getg{
        int power(int x,int y,int p){
            int ans=1;
            for (;y;y>>=1,x=(LL)x*x%p)
                if (y&1)
                    ans=(LL)ans*x%p;
            return ans;
        }
        int p;
        vector <int> v;
        int check(int g){
            if (__gcd(g,p)!=1)
                return 0;
            for (auto i : v)
                if (power(g,(p-1)/i,p)==1)
                    return 0;
            return 1;
        }
        int calc(int _p){
            p=_p-1;
            v.clear();
            for (int i=2;i*i<=p;i++)
                if (p%i==0){
                    v.push_back(i);
                    while (p%i==0)
                        p/=i;
                }
            if (p>1)
                v.push_back(p);
            p=_p;
            for (int g=1;g<p;g++)
                if (check(g))
                    return g;
            return -1;
        }
    }
    View Code

    $$n!mod P \ sum_{i=1}^{n} frac 1 i mod P\ P = 998244353$$

    #include <bits/stdc++.h>
    #define clr(x) memset(x,0,sizeof (x))
    #define For(i,a,b) for (int i=(a);i<=(b);i++)
    #define Fod(i,b,a) for (int i=(b);i>=(a);i--)
    #define fi first
    #define se second
    #define pb(x) push_back(x)
    #define mp(x,y) make_pair(x,y)
    #define outval(x) cerr<<#x" = "<<x<<endl
    #define outv(x) cerr<<#x" = "<<x<<"  "
    #define outtag(x) cerr<<"--------------"#x"---------------"<<endl
    #define outarr(a,L,R) cerr<<#a"["<<L<<".."<<R<<"] = ";
        For(_x,L,R) cerr<<a[_x]<<" ";cerr<<endl;
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef unsigned uint;
    typedef long double LD;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=1<<19,mod=998244353;
    int Pow(int x,int y){
        int ans=1;
        for (;y;y>>=1,x=(LL)x*x%mod)
            if (y&1)
                ans=(LL)ans*x%mod;
        return ans;
    }
    void Add(int &x,int y){
        if ((x+=y)>=mod)
            x-=mod;
    }
    void Del(int &x,int y){
        if ((x-=y)<0)
            x+=mod;
    }
    int Add(int x){
        return x>=mod?x-mod:x;
    }
    int Del(int x){
        return x<0?x+mod:x;
    }
    namespace fft{
        int w[N],R[N];
        void init(int n){
            int d=0;
            while ((1<<d)<n)
                d++;
            For(i,0,n-1)
                R[i]=(R[i>>1]>>1)|((i&1)<<(d-1));
            w[0]=1,w[1]=Pow(3,(mod-1)/n);
            For(i,2,n-1)
                w[i]=(LL)w[i-1]*w[1]%mod;
        }
        void FFT(int *a,int n,int flag){
            For(i,0,n-1)
                if (i<R[i])
                    swap(a[i],a[R[i]]);
            for (int t=n>>1,d=1;d<n;d<<=1,t>>=1)
                for (int i=0;i<n;i+=d<<1)
                    for (int j=0;j<d;j++){
                        int tmp=(LL)w[t*j]*a[i+j+d]%mod;
                        a[i+j+d]=Del(a[i+j]-tmp);
                        Add(a[i+j],tmp);
                    }
            if (flag<0){
                reverse(a+1,a+n);
                int inv=Pow(n,mod-2);
                For(i,0,n-1)
                    a[i]=(LL)a[i]*inv%mod;
            }
        }
        vector <int> Mul(vector <int> a,vector <int> b){
            int n=1,t=a.size()+b.size()-1;
            while (n<t)
                n<<=1;
            a.resize(n),b.resize(n);
            init(n);
            FFT(&a[0],n,1),FFT(&b[0],n,1);
            For(i,0,n-1)
                a[i]=(LL)a[i]*b[i]%mod;
            FFT(&a[0],n,-1);
            a.resize(t);
            return a;
        }
    }
    vector <int> calinv(vector <int> a){
        static int pre[N];
        int n=a.size()-1;
        pre[0]=1;
        For(i,1,n)
            pre[i]=(LL)pre[i-1]*a[i-1]%mod;
        int v=Pow(pre[n],mod-2);
        vector <int> b(n);
        Fod(i,n,1){
            b[i-1]=(LL)v*pre[i-1]%mod;
            v=(LL)v*a[i-1]%mod;
        }
        return b;
    }
    int cal(vector <int> f,int x){
        int v=1,ans=0;
        for (auto i : f)
            Add(ans,(LL)v*i%mod),v=(LL)v*x%mod;
        return ans;
    }
    int Fac[N],Inv[N],Iv[N];
    void prework(){
        int n=N-1;
        Fac[0]=1;
        For(i,1,n)
            Fac[i]=(LL)Fac[i-1]*i%mod;
        Inv[n]=Pow(Fac[n],mod-2);
        Fod(i,n,1)
            Inv[i-1]=(LL)Inv[i]*i%mod;
        For(i,1,n)
            Iv[i]=(LL)Inv[i]*Fac[i-1]%mod;
    }
    int bs;
    vector <int> cal0(vector <int> g){
        int d=(int)g.size()-1;
        if (!d)
            return g;
        vector <int> a,b;
        a.resize(d+1);
        For(i,0,d){
            a[i]=(LL)g[i]*Inv[i]%mod*Inv[d-i]%mod;
            if ((d-i)&1)
                a[i]=Del(-a[i]);
        }
        b.resize(d*2+1);
        b[0]=0;
        For(i,1,d*2)
            b[i]=Iv[i];
        a=fft::Mul(a,b);
        a.resize(d*2+1);
        For(i,0,d)
            a[i]=g[i];
        For(i,d+1,d*2)
            a[i]=(LL)a[i]*Fac[i]%mod*Inv[i-d-1]%mod;
        return a;
    }
    vector <int> cal1(vector <int> g){
        int d=(int)g.size()-1;
        if (!d)
            return g;
        vector <int> a,b;
        int tmp=Pow(bs,mod-(d+1));
        a.resize(d+1);
        For(i,0,d){
            a[i]=(LL)g[i]*tmp%mod*Inv[i]%mod*Inv[d-i]%mod;
            if ((d-i)&1)
                a[i]=Del(-a[i]);
        }
        b.resize(d*3+1);
        For(i,0,d)
            b[i]=Pow(Add((LL)Del(i-d)*bs%mod+d),mod-2);
        For(i,1,d*2)
            b[i+d]=Pow(Add((LL)i*bs%mod+d),mod-2);
        a=fft::Mul(a,b);
        int v=1;
        For(i,0,d)
            v=(LL)v*Add(d+(LL)Del(-i)*bs%mod)%mod;
        For(i,0,d*2){
            a[i+d]=(LL)a[i+d]*v%mod;
            v=(LL)v*Add(d+(LL)(i+1)*bs%mod)%mod
                *Pow(Add(d+(LL)Del(i-d)*bs%mod),mod-2)%mod;
        }
        For(i,0,d*2)
            a[i]=a[i+d];
        a.resize(d*2+1);
        return a;
    }
    vector <int> Fac_super_double(vector <int> g){
        int d=(int)g.size()-1;
        vector <int> a=cal0(g),b=cal1(g);
        For(i,0,d*2)
            a[i]=(LL)a[i]*b[i]%mod;
        return a;
    }
    int Fac_n(int n){
        bs=max((int)sqrt(n),1)+1;
        vector <int> g(1,1);
        Fod(t,20,0){
            g=Fac_super_double(g);
            if (bs>>t&1){
                int d=(int)g.size();
                For(i,0,d-1)
                    g[i]=(LL)g[i]*(i*bs+d)%mod;
                int v=1;
                For(i,d*bs+1,d*bs+d)
                    v=(LL)v*i%mod;
                g.pb(v);
            }
        }
        int t=0,ans=1;
        For(i,0,(int)g.size()-1)
            if (t+bs<=n)
                ans=(LL)ans*g[i]%mod,t+=bs;
            else
                break;
        For(i,t+1,n)
            ans=(LL)ans*i%mod;
        return ans;
    }
    void Har_super_double(vector <int> &g,vector <int> &f){
        assert(g.size()==f.size());
        int d=(int)g.size()-1;
        vector <int> g0=cal0(g),g1=cal1(g);
        vector <int> f0=cal0(f),f1=cal1(f);
        g.resize(d*2+1),f.resize(d*2+1);
        For(i,0,d*2){
            g[i]=(LL)g0[i]*g1[i]%mod;
            f[i]=((LL)g0[i]*f1[i]+(LL)g1[i]*f0[i])%mod;
        }
    }
    int Har_n(int n){
        bs=max((int)sqrt(n),1)+1;
        vector <int> g(1,1),f(1,0);
        Fod(t,20,0){
            Har_super_double(g,f);
            if (bs>>t&1){
                int d=(int)g.size();
                For(i,0,d-1)
                    f[i]=((LL)f[i]*(i*bs+d)+g[i])%mod;
                For(i,0,d-1)
                    g[i]=(LL)g[i]*(i*bs+d)%mod;
                int v=1;
                For(i,d*bs+1,d*bs+d)
                    v=(LL)v*i%mod;
                g.pb(v);
                v=0;
                For(i,d*bs+1,d*bs+d)
                    Add(v,Pow(i,mod-2));
                f.pb((LL)v*g[d]%mod);
            }
        }
        int t=0,ans=0;
        For(i,0,(int)g.size()-1)
            if (t+bs<=n)
                Add(ans,(LL)f[i]*Pow(g[i],mod-2)%mod),t+=bs;
            else
                break;
        For(i,t+1,n)
            Add(ans,Pow(i,mod-2));
        return ans;
    }
    //namespace Test{
    //    int bFac_n(int n){
    //        int ans=1;
    //        For(i,1,n)
    //            ans=(LL)ans*i%mod;
    //        return ans;
    //    }
    //    int bHar_n(int n){
    //        int ans=0;
    //        For(i,1,n)
    //            Add(ans,Pow(i,mod-2));
    //        return ans;
    //    }
    //    void main(){
    //        while (1){
    //            int n=read();
    //            int v=Har_n(n);
    //            cout<<v<<endl;
    //            assert(v==bHar_n(n));
    //        }
    //    }
    //}
    int main(){
        prework();
    //    Test::main();
        return 0;
    }
    View Code

    KM (UOJ#80)

    #include <bits/stdc++.h>
    #define clr(x) memset(x,0,sizeof (x))
    using namespace std;
    typedef long long LL;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f|=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=405,INF=2e9;
    int n,m;
    int g[N][N];
    void Getg(){
        int c=read();
        while (c--){
            int x=read(),y=read(),z=read();
            g[x][y]=z;
        }
    }
    int match[N],visx[N],visy[N],ex[N],ey[N],Min[N];
    int dfs(int x){
        visx[x]=1;
        for (int y=1;y<=m;y++)
            if (!visy[y]){
                int d=ex[x]+ey[y]-g[x][y];
                if (!d){
                    visy[y]=1;
                    if (!match[y]||dfs(match[y]))
                        return match[y]=x,1;
                }
                else
                    Min[y]=min(Min[y],d);
            }
        return 0;
    }
    LL KM(){
        clr(match),clr(ex),clr(ey);
        for (int i=1;i<=n;i++)
            for (int j=1;j<=m;j++)
                ex[i]=max(ex[i],g[i][j]);
        for (int i=1;i<=n;i++){
            for (int j=1;j<=m;j++)
                Min[j]=INF;
            clr(visx),clr(visy);
            if (!dfs(i)){
                while (1){
                    int d=INF,y=0;
                    for (int j=1;j<=m;j++)
                        if (!visy[j])
                            d=min(d,Min[j]);
                    for (int j=1;j<=n;j++)
                        if (visx[j])
                            ex[j]-=d;
                    for (int j=1;j<=m;j++)
                        if (visy[j])
                            ey[j]+=d;
                        else
                            if (!(Min[j]-=d))
                                y=j;
                    if (!match[y])
                        break;
                    int x=match[y];
                    visx[x]=visy[y]=1;
                    for (int j=1;j<=m;j++)
                        Min[j]=min(Min[j],ex[x]+ey[j]-g[x][j]);
                }
                clr(visx),clr(visy);
                assert(dfs(i));
            }
        }
        LL ans=0;
        for (int i=1;i<=n;i++)
            ans+=ex[i];
        for (int i=1;i<=m;i++)
            ans+=ey[i];
        return ans;
    }
    int id[N];
    int main(){
        n=read(),m=max(n,(int)read());
        Getg();
        cout<<KM()<<endl;
        for (int i=1;i<=m;i++)
            id[match[i]]=i;
        for (int i=1;i<=n;i++)
            if (g[i][id[i]])
                cout<<id[i]<<" ";
            else
                cout<<0<<" ";
        return 0;
    }
    View Code

    KMP

    int Fail[N];
    char S1[N],S2[N];
    int KMP(string &s1,string &s2){
        int n=s1.size(),m=s2.size();
        for (int i=1;i<=n;i++)
            S1[i]=s1[i-1];
        for (int i=1;i<=m;i++)
            S2[i]=s2[i-1];
        Fail[0]=Fail[1]=0;
        for (int i=2;i<=m;i++){
            int k=Fail[i-1];
            while (k&&S2[i]!=S2[k+1])
                k=Fail[k];
            if (S2[i]==S2[k+1])
                k++;
            Fail[i]=k;
        }
        int ans=0,k=0;
        for (int i=1;i<=n;i++){
            while (k&&S1[i]!=S2[k+1])
                k=Fail[k];
            if (S1[i]==S2[k+1])
                k++;
            if (k==m){
                ans++;
                k=Fail[k];
            }
        }
        return ans;
    }
    View Code

    Lucas

    LL Lucas(LL n,LL m){
        if (n<mod&&m<mod)
            return C(n,m);
        return C(n%mod,m%mod)*Lucas(n/mod,m/mod);
    }
    View Code

    Lyndon分解

    vector <int> lyndon(char *s,int n){
        //s[1..n] -> s[1..p1][p1+1..p2]...
        vector <int> res;
        for (int i=1;i<=n;){
            int j=i,k=i+1;
            while (k<=n&&s[j]<=s[k]){
                if (s[j]==s[k])
                    j++,k++;
                else if (s[j]<s[k])
                    j=i,k++;
            }
            while (i<=j){
                i+=k-j;
                res.push_back(i-1);
            }
        }
        return res;
    }
    View Code

    NTT

    namespace fft{
        int w[N],rev[N];
        void init(int n){
            int d=0;
            while ((1<<d)<n)
                d++;
            w[0]=1,w[1]=Pow(3,(mod-1)/n);
            For(i,2,n-1)
                w[i]=(LL)w[i-1]*w[1]%mod;
            For(i,0,n-1)
                rev[i]=(rev[i>>1]>>1)|((i&1)<<(d-1));
        }
        void dft(int *a,int n,int flag){
            For(i,0,n-1)
                if (i<rev[i])
                    swap(a[i],a[rev[i]]);
            for (int t=n>>1,d=1;d<n;d<<=1,t>>=1)
                for (int i=0;i<n;i+=d<<1)
                    For(j,0,d-1){
                        int tmp=(LL)a[i+j+d]*w[t*j]%mod;
                        a[i+j+d]=Del(a[i+j]-tmp);
                        a[i+j]=Add(a[i+j]+tmp);
                    }
            if (flag<0){
                reverse(a+1,a+n);
                int inv=Pow(n,mod-2);
                For(i,0,n-1)
                    a[i]=(LL)a[i]*inv%mod;
            }
        }
    }
    using fft::dft;
    View Code

    NTT + 基础寻址优化

    namespace fft{
        int R[N];
        vi w[23];
        void prework(){
            int d=log(N)/log(2)+0.5;
            For(i,1,d){
                w[i].resize(1<<i);
                w[i][0]=1,w[i][1]=Pow(3,(mod-1)>>i);
                For(j,2,(1<<i)-1)
                    w[i][j]=(LL)w[i][j-1]*w[i][1]%mod;
            }
        }
        void init(int n){
            int d=0;
            while ((1<<d)<n)
                d++;
            For(i,0,n-1)
                R[i]=(R[i>>1]>>1)|((i&1)<<(d-1));
        }
        void FFT(int *a,int n,int flag){
            For(i,0,n-1)
                if (i<R[i])
                    swap(a[i],a[R[i]]);
            for (int t=1,d=1;d<n;d<<=1,t++)
                for (int i=0;i<n;i+=d<<1){
                    int *W=&w[t][0];
                    For(j,0,d-1){
                        int tmp=(LL)a[i+j+d]*(*W++)%mod;
                        a[i+j+d]=Del(a[i+j]-tmp);
                        a[i+j]=Add(a[i+j]+tmp);
                    }
                }
            if (flag<0){
                reverse(a+1,a+n);
                int inv=Pow(n,mod-2);
                For(i,0,n-1)
                    a[i]=(LL)a[i]*inv%mod;
            }
        }
    }
    View Code

    MinRep 最小表示法

    int MinRep(char *s,int n){
        int i=1,j=2,k=0,t;
        while (i<=n&&j<=n&&k<n)
            (!(t=s[i+k>n?i+k-n:i+k]-s[j+k>n?j+k-n:j+k]))?(k++):((t>0?i:j)+=k+1,j+=(i==j),k=0);
        return min(i,j);
    }
    只供抄取,不供学习
    int MinRep(char *s,int n){
        #define _(x) ((x)>n?(x)-n:(x))
        int i=1,j=2,k=0,t;
        while (i<=n&&j<=n&&k<n){
            int t=t=s[_(i+k)]-s[_(j+k)];
            if (t)
                (t>0?i:j)+=k+1,j+=(i==j),k=-1;
            k++;
        }
        #undef _
        return min(i,j);
    }
    正常的

    (Updated 2019-03-25)

    void Get_Min_Rep(char *s,int n){
        static char tmp[N];
        #define S(x) s[(x)>n?(x)-n:(x)]
        int i=1,j=2,k=0;
        while (i<=n&&j<=n&&k<n){
            if (S(i+k)==S(j+k))
                k++;
            else {
                (S(i+k)<S(j+k)?j:i)+=k+1;
                k=0;
                if (i==j)
                    j++;
            }
        }
        int p=min(i,j);
        For(i,1,n)
            tmp[i]=S(p+i-1);
        For(i,1,n)
            s[i]=tmp[i];
        #undef S
    }
    再来一个直接把字符串转成最小表示的

    Manachar (Updated on 2019-03-25)

    int n;
    char s[N],ss[N];
    int len[N];
    void Manachar(char *s,int n){
        For(i,0,n+1)
            len[i]=0;
        int mx=1;
        For(i,1,n){
            len[i]=max(1,min(mx+len[mx]-i,len[mx*2-i]));
            while (s[i-len[i]]==s[i+len[i]])
                len[i]++;
            if (i+len[i]>mx+len[mx])
                mx=i;
        }
    }
    int solve(char *s,int n){
        ss[0]='#',ss[1]='*';
        For(i,1,n){
            ss[i<<1]=s[i];
            ss[i<<1|1]='*';
        }
        ss[n*2+2]='$';
        Manachar(ss,n*2+1);
        int ans=0;
        For(i,1,n*2+1)
            ans=max(ans,len[i]-1);
        return ans;
    }
    View Code

    PAM

    namespace PAM{
        int len[N],Fail[N],Next[N][26];
        int size[N];
        int cnt;
        void init(){
            cnt=2;
            len[1]=-1,Fail[1]=1;
            len[2]=0,Fail[2]=1;
            clr(Next),clr(size);
        }
        void build(char *s,int n){
            init();
            s[0]='*';
            int x=1;
            for (int i=1;i<=n;i++){
                while (s[i-len[x]-1]!=s[i])
                    x=Fail[x];
                int c=s[i]-'a';
                if (Next[x][c])
                    x=Next[x][c];
                else {
                    int y=Next[x][c]=++cnt;
                    len[y]=len[x]+2;
                    if (len[y]==1)
                        Fail[y]=2;
                    else {
                        x=Fail[x];
                        while (s[i-len[x]-1]!=s[i])
                            x=Fail[x];
                        Fail[y]=Next[x][c];
                    }
                    x=y;
                }
                size[x]++;
            }
            for (int i=cnt;i>=1;i--)
                size[Fail[i]]+=size[i];
        }
    }
    View Code

    Pollard_Rho 

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef long double LD;
    namespace Pollard_Rho{
        int prime[9]={2,3,5,7,11,13,17,19,23};
        ULL RR;
        int Pcnt;
        LL p[70];
        vector <LL> res;
        LL R(LL mod){
            return (RR+=4179340454199820289LL)%mod;
        }
        LL Mul(LL x,LL y,LL mod){
            LL d=(LL)floor((LD)x*y/mod+0.5);
            LL res=x*y-d*mod;
            if (res<0)
                res+=mod;
            return res;
        }
        LL Pow(LL x,LL y,LL mod){
            LL ans=1%mod;
            for (;y;y>>=1,x=Mul(x,x,mod))
                if (y&1)
                    ans=Mul(ans,x,mod);
            return ans;
        }
        bool Miller_Rabin(LL n){
            if (n<=1)
                return 0;
            for (int i=0;i<9;i++)
                if (n==prime[i])
                    return 1;
            LL d=n-1;
            int tmp=0;
            while (!(d&1))
                d>>=1,tmp++;
            for (int i=0;i<9;i++){
                LL x=Pow(prime[i],d,n),p=x;
                for (int j=1;j<=tmp;j++){
                    x=Mul(x,x,n);
                    if (x==1&&p!=1&&p!=n-1)
                        return 0;
                    p=x;
                }
                if (x!=1)
                    return 0;
            }
            return 1;
        }
        LL f(LL x,LL c,LL mod){
            return (Mul(x,x,mod)+c)%mod;
        }
        LL gcd(LL x,LL y){
            return y?gcd(y,x%y):x;
        }
        LL Get_Factor(LL c,LL n){
            LL x=R(n),y=f(x,c,n),p=n;
            while (x!=y&&(p==n||p==1)){
                p=gcd(n,max(x-y,y-x));
                x=f(x,c,n);
                y=f(f(y,c,n),c,n);
            }
            return p;
        }
        void Pollard_Rho(LL n){
            if (n<=1)
                return;
            if (Miller_Rabin(n)){
                res.push_back(n);
                return;
            }
            while (1){
                LL v=Get_Factor(R(n-1)+1,n);
                if (v!=n&&v!=1){
                    Pollard_Rho(v);
                    Pollard_Rho(n/v);
                    return;
                }
            }
        }
        void work(LL n){
            res.clear();
            Pollard_Rho(n);
        }
    }
    int main(){
        LL n;
        scanf("%lld",&n);
        Pollard_Rho :: work(n);
        vector <LL> ans=Pollard_Rho :: res;
        sort(ans.begin(),ans.end());
        printf("%d
    ",(int)ans.size());
        for (int i=0;i<ans.size();i++)
            printf("%lld ",ans[i]);
        puts("");
        return 0;
    }
    View Code

    SA (附带 ST 表查询区间 LCP )

    int SA[N],rank[N],tmp[N],height[N],tax[N];
    int ST[N][20];
    void Sort(int n,int m){
        for (int i=0;i<=m;i++)
            tax[i]=0;
        for (int i=1;i<=n;i++)
            tax[rank[i]]++;
        for (int i=1;i<=m;i++)
            tax[i]+=tax[i-1];
        for (int i=n;i>=1;i--)
            SA[tax[rank[tmp[i]]]--]=tmp[i];
    }
    bool cmp(int rk[],int x,int y,int w){
        return rk[x]==rk[y]&&rk[x+w]==rk[y+w];
    }
    void Suffix_Array(int s[],int n){
        memset(SA,0,sizeof SA);
        memset(tmp,0,sizeof tmp);
        memset(rank,0,sizeof rank);
        memset(height,0,sizeof height);
        for (int i=1;i<=n;i++)
            rank[i]=s[i],tmp[i]=i;
        int m=234;
        Sort(n,m);
        for (int w=1,p=0;p<n;w<<=1,m=p){
            p=0;
            for (int i=n-w+1;i<=n;i++)
                tmp[++p]=i;
            for (int i=1;i<=n;i++)
                if (SA[i]>w)
                    tmp[++p]=SA[i]-w;
            Sort(n,m);
            swap(rank,tmp);
            rank[SA[1]]=p=1;
            for (int i=2;i<=n;i++)
                rank[SA[i]]=cmp(tmp,SA[i],SA[i-1],w)?p:++p;
        }
        for (int i=1,j,k=0;i<=n;height[rank[i++]]=k)
            for (k=max(k-1,0),j=SA[rank[i]-1];s[i+k]==s[j+k];k++);
        height[1]=0;
    }
    void Get_ST(int n){
        memset(ST,0,sizeof ST);
        for (int i=1;i<=n;i++){
            ST[i][0]=height[i];
            for (int j=1;j<20;j++){
                ST[i][j]=ST[i][j-1];
                if (i-(1<<(j-1))>0)
                    ST[i][j]=min(ST[i][j],ST[i-(1<<(j-1))][j-1]);
            }
        }
    }
    int Query(int L,int R){
        int val=floor(log(R-L+1)/log(2));
        return min(ST[L+(1<<val)-1][val],ST[R][val]);
    }
    int LCP(int x,int y){
        x=rank[x],y=rank[y];
        return Query(min(x,y)+1,max(x,y));
    }
    View Code

    SAM

    struct Node{
        int Next[26],fa,Max;
    }t[N<<1];
    int size,last;
    void init(){
        size=last=1;
    }
    void extend(int c){
        int p=last,np=++size,q,nq;
        t[np].Max=t[p].Max+1;
        for (;p&&!t[p].Next[c];p=t[p].fa)
            t[p].Next[c]=np;
        if (!p)
            t[np].fa=1;
        else {
            q=t[p].Next[c];
            if (t[q].Max==t[p].Max+1)
                t[np].fa=q;
            else {
                nq=++size;
                t[nq]=t[q],t[nq].Max=t[p].Max+1;
                t[q].fa=t[np].fa=nq;
                for (;p&&t[p].Next[c]==q;p=t[p].fa)
                    t[p].Next[c]=nq;
            }
        }
        last=np;
    }
    View Code

    SAM(广义)

    struct SAM{
        int Next[26],fa,Max;
    }t[N<<1];
    int size;
    void init(){
        memset(t,0,sizeof t);
        size=1,t[0].Max=-1;
        for (int i=0;i<26;i++)
            t[0].Next[i]=1;
    }
    int extend(int p,int c){
        if (t[p].Next[c]&&t[p].Max+1==t[t[p].Next[c]].Max)
            return t[p].Next[c];
        int np=++size,q,nq;
        t[np].Max=t[p].Max+1;
        for (;!t[p].Next[c];p=t[p].fa)
            t[p].Next[c]=np;
        q=t[p].Next[c];
        if (t[p].Max+1==t[q].Max)
            t[np].fa=q;
        else {
            nq=++size;
            t[nq]=t[q],t[nq].Max=t[p].Max+1;
            t[q].fa=t[np].fa=nq;
            for (;t[p].Next[c]==q;p=t[p].fa)
                t[p].Next[c]=nq;
        }
        return np;
    }
    View Code

    求第一类 Stirling 数

    namespace str{
        vector <int> rpow[22];
        //attention: rpow[i].size() = pow(2, n) + 1 
        //using: Fac[i] = i!, Inv[i] = 1/Fac[i]; mxd = pow(2, n) + 1 
        vector <int> Get_Add(vector <int> f,int n,int v){
            vector <int> g,h;
            g.clear();
            for (int i=-n;i<=0;i++)
                g.pb((LL)Pow(v,-i)*Inv[-i]%mod);
            for (int i=0;i<=n;i++)
                f[i]=(LL)f[i]*Fac[i]%mod;
            h=poly::Mul(f,g);
            while (h.size()<n*2+1)
                h.pb(0);
            g.clear();
            for (int i=0;i<=n;i++)
                g.pb((LL)h[i+n]*Inv[i]%mod);
            return g;
        }
        vector <int> Get_rpow(int x){
            if (!rpow[x].empty())
                return rpow[x];
            if (x==0){
                rpow[x].pb(0),rpow[x].pb(1);
                return rpow[x];
            }
            int n=1<<(x-1);
            vector <int> f=Get_rpow(x-1);
            rpow[x]=poly::Mul(f,Get_Add(f,n,n));
            while (rpow[x].size()>n*2+1)
                rpow[x].pop_back();
            while (rpow[x].size()<n*2+1)
                rpow[x].pb(0);
            return rpow[x];
        }
        void Get_s1(int n,int res[]){
            static vector <int> ans;
            if (n==0)
                return (void)(res[0]=1);
            ans.clear();
            ans.pb(1);
            for (int i=0;i<20;i++)
                if (n>>i&1)
                    ans=poly::Mul(Get_rpow(i),Get_Add(ans,(int)ans.size()-1,1<<i));
            for (int i=0;i<=n;i++)
                res[i]=ans[i];
        }
    }
    View Code

    ISAP (Updated in 2018-09-11)

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    struct Edge{
        int x,y,cap,nxt;
        Edge(){}
        Edge(int a,int b,int c,int d){
            x=a,y=b,cap=c,nxt=d;
        }
    };
    struct gragh{
        static const int N=105,M=5005*2,INF=0x7fffffff;
        int n,S,T,fst[N],cnt;
        int dist[N],num[N],cur[N],p[N];
        LL MaxFlow;
        Edge e[M];
        void clear(int _n){
            cnt=1,n=_n;
            memset(fst,0,sizeof fst);
        }
        void add(int a,int b,int c){
            e[++cnt]=Edge(a,b,c,fst[a]),fst[a]=cnt;
            e[++cnt]=Edge(b,a,0,fst[b]),fst[b]=cnt;
        }
        void init(){
            memset(dist,0,sizeof dist);
            memset(num,0,sizeof num);
            for (int i=1;i<=n;i++)
                num[dist[i]]++,cur[i]=fst[i];
        }
        void init(int _S,int _T){
            S=_S,T=_T;
            MaxFlow=0;
            init();
        }
        int Augment(int &x){
            int Flow=INF;
            for (int i=T;i!=S;i=e[p[i]].x)
                if (e[p[i]].cap<=Flow)
                    Flow=e[p[i]].cap,x=e[p[i]].x;
            for (int i=T;i!=S;i=e[p[i]].x)
                e[p[i]].cap-=Flow,e[p[i]^1].cap+=Flow;
            return Flow;
        }
        LL ISAP(){
            int x=S,y;
            while (dist[S]<n){
                if (x==T){
                    MaxFlow+=Augment(x);
                    continue;
                }
                bool found=0;
                for (int i=cur[x];i;i=e[i].nxt)
                    if (dist[y=e[i].y]+1==dist[x]&&e[i].cap){
                        cur[x]=p[y]=i,x=y,found=1;
                        break;
                    }
                if (!found){
                    int d=n+1;
                    for (int i=fst[x];i;i=e[i].nxt)
                        if (e[i].cap)
                            d=min(d,dist[e[i].y]+1);
                    if (!--num[dist[x]])
                        return MaxFlow;
                    num[dist[x]=d]++,cur[x]=fst[x],x=x==S?x:e[p[x]].x;
                }
            }
            return MaxFlow;
        }
        LL Auto(int _S,int _T){
            init(_S,_T);
            return ISAP();
        }
    }g;
    View Code

     

    Tarjan(有向图强联通分量)

    int dfn[N],low[N],bh[N],st[N],inst[N],vis[N],Time,top,tot;
    void Tarjan_Prepare(){
        Time=top=tot=0;
        memset(bh,0,sizeof bh);
        memset(st,0,sizeof st);
        memset(dfn,0,sizeof dfn);
        memset(low,0,sizeof low);
        memset(vis,0,sizeof vis);
        memset(inst,0,sizeof inst);
    }
    void Tarjan(int x){
        dfn[x]=low[x]=++Time;
        inst[x]=vis[x]=1;
        st[++top]=x;
        for (int i=g.fst[x];i;i=g.nxt[i])
            if (!vis[g.y[i]]){
                Tarjan(g.y[i]);
                low[x]=min(low[x],low[g.y[i]]);
            }
            else if (inst[g.y[i]])
                low[x]=min(low[x],low[g.y[i]]);
        if (dfn[x]==low[x]){
            tot++;
            bh[st[top]]=tot;
            inst[st[top]]=0;
            while (st[top--]!=x){
                bh[st[top]]=tot;
                inst[st[top]]=0;
            }
        }
    }
    View Code

    SPFA 费用流 (Updated on 2018-09-12)

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    struct Edge{
        int x,y,c,nxt,cap;
        Edge(){}
        Edge(int a,int b,int _c,int d,int e){
            x=a,y=b,c=_c,cap=d,nxt=e;
        }
    };
    struct Network{
        static const int N=405,M=15005*2,INF=0x7FFFFFFF;
        Edge e[M];
        int n,S,T,fst[N],cur[N],cnt;
        int q[N],vis[N],head,tail;
        int MaxFlow,MinCost,dis[N];
        void clear(int _n){
            n=_n,cnt=1;
            memset(fst,0,sizeof fst);
        }
        void add(int a,int b,int c,int d){
            e[++cnt]=Edge(a,b,d,c,fst[a]),fst[a]=cnt;
            e[++cnt]=Edge(b,a,-d,0,fst[b]),fst[b]=cnt;
        }
        void init(){
            for (int i=1;i<=n;i++)
                cur[i]=fst[i];
        }
        void init(int _S,int _T){
            S=_S,T=_T,MaxFlow=MinCost=0,init();
        }
        int SPFA(){
            for (int i=1;i<=n;i++)
                dis[i]=INF;
            memset(vis,0,sizeof vis);
            head=tail=0;
            dis[q[++tail]=T]=0;
            while (head!=tail){
                if ((++head)>=n)
                    head-=n;
                int x=q[head];
                vis[x]=0;
                for (int i=fst[x];i;i=e[i].nxt){
                    int y=e[i].y;
                    if (e[i^1].cap&&dis[x]-e[i].c<dis[y]){
                        dis[y]=dis[x]-e[i].c;
                        if (!vis[y]){
                            if ((++tail)>=n)
                                tail-=n;
                            vis[q[tail]=y]=1;
                        }
                    }
                }
            }
            memset(vis,0,sizeof vis);
            return dis[S]<INF;
        }
        int dfs(int x,int Flow){
            if (x==T||!Flow)
                return Flow;
            vis[x]=1;
            int now=Flow;
            for (int &i=cur[x];i;i=e[i].nxt){
                int y=e[i].y;
                if (!vis[y]&&e[i].cap&&dis[x]-e[i].c==dis[y]){
                    int d=dfs(y,min(now,e[i].cap));
                    e[i].cap-=d,e[i^1].cap+=d;
                    if (!(now-=d))
                        break;
                }
            }
            vis[x]=0;
            return Flow-now;
        }
        void Dinic(){
            while (SPFA()){
                init();
                int now=dfs(S,INF);
                MaxFlow+=now,MinCost+=now*dis[S];
            }
        }
        void MCMF(int &_MinCost,int &_MaxFlow){
            Dinic(),_MinCost=MinCost,_MaxFlow=MaxFlow;
        }
        void Auto(int _S,int _T,int &_MinCost,int &_MaxFlow){
            init(_S,_T),MCMF(_MinCost,_MaxFlow);
        }
    }g;
    int read(){
        int x=0;
        char ch=getchar();
        while (!isdigit(ch))
            ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return x;
    }
    int n,m,S,T;
    int main(){
        n=read(),m=read(),S=1,T=n;
        g.clear(n);
        while (m--){
            int a=read(),b=read(),c=read(),cap=read();
            g.add(a,b,c,cap);
        }
        int MinCost,MaxFlow;
        g.Auto(S,T,MinCost,MaxFlow);
        printf("%d %d
    ",MaxFlow,MinCost);
        return 0;
    }
    View Code

     

    主席树 - 二维数点

    int root[N],sum[S],ls[S],rs[S],tot=0;
    void build(int &rt,int L,int R){
        sum[rt=++tot]=0;
        if (L==R)
            return;
        int mid=(L+R)>>1;
        build(ls[rt],L,mid);
        build(rs[rt],mid+1,R);
    }
    void update(int prt,int &rt,int L,int R,int x){
        if (!rt||rt==prt)
            sum[rt=++tot]=sum[prt];
        sum[rt]++;
        if (L==R)
            return;
        if (!ls[rt])
            ls[rt]=ls[prt];
        if (!rs[rt])
            rs[rt]=rs[prt];
        int mid=(L+R)>>1;
        if (x<=mid)
            update(ls[prt],ls[rt],L,mid,x);
        else
            update(rs[prt],rs[rt],mid+1,R,x);
    }
    int query(int rt,int L,int R,int xL,int xR){
        if (!rt||R<xL||L>xR)
            return 0;
        if (xL<=L&&R<=xR)
            return sum[rt];
        int mid=(L+R)>>1;
        return query(ls[rt],L,mid,xL,xR)+query(rs[rt],mid+1,R,xL,xR);
    }
    int Query(int x1,int x2,int y1,int y2){
        return query(root[x2],1,yt,y1,y2)-query(root[x1-1],1,yt,y1,y2);
    }
    View Code

    最小圆覆盖

    #pragma GCC optimize("Ofast","inline")
    #include <bits/stdc++.h>
    #define clr(x) memset(x,0,sizeof (x))
    #define outval(x) printf(#x" = %d
    ",x)
    #define For(x,a,b) for (int x=a;x<=b;x++)
    #define Fod(x,b,a) for (int x=b;x>=a;x--)
    #define pb push_back
    using namespace std;
    typedef long long LL;
    LL read(){
        LL x=0,f=0;
        char ch=getchar();
        while (!isdigit(ch))
            f|=ch=='-',ch=getchar();
        while (isdigit(ch))
            x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
        return f?-x:x;
    }
    const int N=500005;
    const double Eps=1e-9,pi=acos(-1.0);
    struct Point{
        double x,y;
        Point(){}
        Point(double _x,double _y){    
            x=_x,y=_y;
        }
    }p[N];
    int Dcmp(double x){
        if (fabs(x)<Eps) return 0;return x<0?-1:1;
    }
    int Dcmp(double x,double y){
        return Dcmp(x-y);
    }
    Point operator + (Point A,Point B){
        return Point(A.x+B.x,A.y+B.y);
    }
    Point operator - (Point A,Point B){
        return Point(A.x-B.x,A.y-B.y);
    }
    Point operator * (Point A,double x){
        return Point(A.x*x,A.y*x);
    }
    Point operator / (Point A,double x){
        return Point(A.x/x,A.y/x);
    }
    Point Rotate(Point A,double B){
        return Point(A.x*cos(B)-A.y*sin(B),A.x*sin(B)+A.y*cos(B));
    }
    double cross(Point A,Point B){
        return A.x*B.y-A.y*B.x;
    }
    double cross(Point A,Point B,Point C){
        return cross(B-A,C-A);
    }
    double Dot(Point A,Point B){
        return A.x*B.x+A.y*B.y;
    }
    double Dis(Point A,Point B){
        return sqrt(Dot(A-B,A-B));
    }
    Point Center(Point A,Point B,Point C){
        Point a=(A+B)/2,b=(A+C)/2;
        Point u=Rotate(B-A,pi/2),v=Rotate(C-A,pi/2);
        if (Dcmp(cross(u,v))==0){
            if (Dcmp(Dis(A,C),Dis(A,B)+Dis(B,C))==0)
                return (A+C)/2;
            if (Dcmp(Dis(A,B),Dis(A,C)+Dis(B,C))==0)
                return (A+B)/2;
            if (Dcmp(Dis(B,C),Dis(A,B)+Dis(A,C))==0)
                return (B+C)/2;
        }
        return a+u*cross(v,a-b)/cross(u,v);
    }
    int n;
    int main(){
        srand(233);
        n=read();
        for (int i=1;i<=n;i++)
            p[i].x=read(),p[i].y=read();
        Point c=p[1];
        double r=0;
        random_shuffle(p+1,p+n+1);
        for (int i=2;i<=n;i++){
            if (Dcmp(Dis(p[i],c),r)<=0)
                continue;
            c=p[i],r=0;
            for (int j=1;j<i;j++){
                if (Dcmp(Dis(p[j],c),r)<=0)
                    continue;
                c=(p[i]+p[j])/2,r=Dis(p[i],c);
                for (int k=1;k<j;k++){
                    if (Dcmp(Dis(p[k],c),r)<=0)
                        continue;
                    c=Center(p[i],p[j],p[k]);
                    r=Dis(p[i],c);
                }
            }
        }
        printf("%.6lf %.6lf %.6lf
    ",r,c.x,c.y);
        return 0;
    }
    View Code

    Splay - BZOJ3224

    #include <bits/stdc++.h>
    using namespace std;
    const int N=100005;
    int n,root=1,size=1,val[N],cnt[N],son[N][2],fa[N],tot[N];
    int wson(int x){
        return son[fa[x]][1]==x;
    }
    void pushup(int x){
        tot[x]=cnt[x]+tot[son[x][0]]+tot[son[x][1]];
    }
    void rotate(int x){
        if (!x)
            return;
        int y=fa[x],z=fa[y],L=wson(x),R=L^1;
        if (z)
            son[z][wson(y)]=x;
        fa[x]=z,fa[y]=x,fa[son[x][R]]=y;
        son[y][L]=son[x][R],son[x][R]=y;
        pushup(y),pushup(x);
    }
    void splay(int x,int k){
        if (!x)
            return;
        if (!k)
            root=x;
        for (int y=fa[x];fa[x]!=k;rotate(x),y=fa[x])
            if (fa[y]!=k)
                rotate(wson(x)==wson(y)?y:x);
    }
    int find(int x,int v){
        return val[x]==v?x:find(son[x][v>val[x]],v);
    }
    int findkth(int x,int k){
        if (k<=tot[son[x][0]])
            return findkth(son[x][0],k);
        k-=tot[son[x][0]];
        if (k<=cnt[x])
            return x;
        k-=cnt[x];
        return findkth(son[x][1],k);
    }
    int findnxt(int x,int v){
        if (!x)
            return 0;
        if (val[x]<=v)
            return findnxt(son[x][1],v);
        else {
            int res=findnxt(son[x][0],v);
            return res?res:x;
        }
    }
    int findpre(int x,int v){
        if (!x)
            return 0;
        if (val[x]>=v)
            return findpre(son[x][0],v);
        else {
            int res=findpre(son[x][1],v);
            return res?res:x;
        }
    }
    void insert(int &x,int pre,int v){
        if (!x){
            x=++size;
            val[x]=v,cnt[x]=tot[x]=1,fa[x]=pre;
            splay(x,0);
            return;
        }
        tot[x]++;
        if (val[x]==v){
            cnt[x]++;
            return;
        }
        insert(son[x][v>val[x]],x,v);
    }
    void Insert(int v){insert(root,0,v);}
    void Delete(int v){
        int x;
        splay(x=find(root,v),0);
        if (--cnt[x])
            return;
        splay(findnxt(root,v),root);
        root=son[x][1];
        son[root][0]=son[x][0];
        fa[son[x][0]]=root;
        fa[root]=son[x][0]=son[x][1]=0;
        pushup(root);
    }
    int Rank(int v){
        splay(find(root,v),0);
        return tot[son[root][0]]+1;
    }
    int main(){
        val[1]=2147483647;
        cnt[1]=tot[1]=1;
        scanf("%d",&n);
        while (n--){
            int opt,x;
            scanf("%d%d",&opt,&x);
            if (opt==1) Insert(x);
            if (opt==2) Delete(x);
            if (opt==3) printf("%d
    ",Rank(x));
            if (opt==4) printf("%d
    ",val[findkth(root,x)]);
            if (opt==5) printf("%d
    ",val[findpre(root,x)]);
            if (opt==6) printf("%d
    ",val[findnxt(root,x)]);
        }
        return 0;
    }
    View Code

    匈牙利算法

    int g[N][N],vis[N],match[N];
    bool Match(int x){
        for (int i=1;i<=n;i++)
            if (!vis[i]&&g[x][i]){
                vis[i]=1;
                if (!match[i]||Match(match[i])){
                    match[i]=x;
                    return 1;
                }
            }
        return 0;
    }
    int hungary(){
        int res=0;
        memset(match,0,sizeof match);
        for (int i=1;i<=n;i++){
            memset(vis,0,sizeof vis);
            if (Match(i))
                res++;
        }
        return res;
    }
    View Code

    高精度整数运算 (由于懒和菜,没有更新带 FFT 的乘法和快速的除法)(UPD2020-08-10:这个模板可能有锅?(尽管它能过很多高精度模板题))

      支持负数

    #include <cstring>
    #include <cstdio>
    #include <cstdlib>
    #include <algorithm>
    #include <cmath>
    #define max(a,b) ((a)>(b)?(a):(b))
    #define min(a,b) ((a)<(b)?(a):(b))
    using namespace std;
    typedef long long LL;
    /* 
    大数模板8 
    1.  支持大数之间的+、-、*、/、%运算,以及一些基础的计算 
    2.  支持部分大数与整型之间的运算 (仅限于大数在前整型在后) +、-、*、/、% 
    3.  支持负数的运算,不会出现减法的时候由于被减数小于减数所造成的报错。 
    4.  快速幂,代码优化 
    5.  可以进行正常的比较。 
    6.  新增 abs(), read()两个方便的函数 
                    read() 具体用法 
                    1. read('c',Var_Name<BigInt>)  读入一个char数组类型转化而来的大数 
                    2. read('i',Var_Name<BigInt>)  读入一个int类型转化而来的大数 
                    3. read('L',Var_Name<BigInt>)  读入一个long long类型转化而来的大数 
    7.  修复原先模板中关于0的bug 
    8.  新增构造函数 
    9.  修复若干bug,新增一些功能  
    10. progress: 9位压位,大大提高效率 
    *///版权所有--周镇东 
    const int MaxLen=1200;
    const LL mod=1e9;
    const LL Pow10[9]={1e0,1e1,1e2,1e3,1e4,1e5,1e6,1e7,1e8};
    struct BigInt{
        LL d,v[MaxLen+5];
        bool f;//保存正负性 ,0当作正数看待 
        BigInt (){}
        BigInt (LL x){(*this)=x;}
        BigInt (int x){(*this)=x;}
        BigInt (char x[]){(*this)=x;}
        BigInt (const BigInt &x){(*this)=x;}
        void Print(){
            if (f)
                putchar('-');
            if (d==0){
                putchar('0');
                return;
            }
            printf("%lld",v[d]);
            for (int i=d-1;i>=1;i--)
                printf("%09lld",v[i]);
        }
        void Print(char c){//输出数字 
            (*this).Print();
            printf("%c",c);
        }
        void ya(){
            LL v_[MaxLen];
            memset(v_,0,sizeof v_);
            for (int i=1;i<=d;i++)
                v_[(i-1)/9+1]+=Pow10[(i-1)%9]*v[i];
            d=(d-1)/9+1;
            memset(v,0,sizeof v);
            for (int i=1;i<=d;i++)
                v[i]=v_[i];
            while (d>0&&v[d]==0)
                d--;
        }
        void operator =(char x[]){
            f=x[0]=='-',d=strlen(x)-f,memset(v,0,sizeof v);
            for (int i=f;i<d+f;i++)
                v[i-f+1]+=(x[d+f-(i-f)-1]-48);
            while (d>0&&v[d]==0)
                d--;
            (*this).ya();
        }
        void operator =(int x){
            (*this)=(LL)x;
        }
        void operator =(LL x){
            d=0,f=x<0,x=abs(x);
            memset(v,0,sizeof v);
            while (x)
                v[++d]=x%mod,x/=mod;
        }
        bool equ(BigInt &x){//cmp abs
            if (d!=x.d)
                return 0;
            for (int i=1;i<=d;i++)
                if (v[i]!=x.v[i])
                    return 0;
            return 1;
        }
        bool operator ==(BigInt &x){//cmp abs
            if (f!=x.f) 
                return 0;
            return (*this).equ(x);
        }
        bool nequ(BigInt &x){
            return !(*this).equ(x);
        }
        bool operator !=(BigInt &x){
            return !(*this==x);
        }
        bool smaller(BigInt &x){
            if (d!=x.d)
                return d<x.d;
            for (int i=d;i>=1;i--)
                if (v[i]!=x.v[i])
                    return v[i]<x.v[i];
            return 0;
        }
        bool bigger(BigInt &x){
            if (d!=x.d)    
                return d>x.d;
            for (int i=d;i>=1;i--)
                if (v[i]!=x.v[i])
                    return v[i]>x.v[i];
            return 0;
        }
        bool operator <(BigInt &x){//cmp abs
            if (f!=x.f) 
                return f;
            if (f&&x.f) 
                return (*this).bigger(x);
            return (*this).smaller(x);
        }
        bool operator >(BigInt &x){
            if (f!=x.f) 
                return x.f;
            if (f&&x.f) 
                return (*this).smaller(x);
            return (*this).bigger(x);
        }
        bool smqu(BigInt &x){
            return !(*this).bigger(x);
        }
        bool bgqu(BigInt &x){
            return !(*this).smaller(x);
        }
        bool operator <=(BigInt &x){
            return !(*this>x);
        }
        bool operator >=(BigInt &x){
            return !(*this<x);
        }
        BigInt operator +(BigInt x){//加法运算 
            BigInt Ans=*this;
            if (f!=x.f){
                Ans.f=x.f=0;
                if (f)
                    return x-Ans;
                else
                    return Ans-x;
            }
            memset(Ans.v,0,sizeof Ans.v);
            Ans.f=f,Ans.d=max(d,x.d);
            for (int i=1;i<=Ans.d;i++)
                Ans.v[i]=v[i]+x.v[i];
            for (int i=1;i<=Ans.d;i++)
                Ans.v[i+1]+=Ans.v[i]/mod,Ans.v[i]%=mod;
            if (Ans.v[Ans.d+1])    
                Ans.d++;
            if (Ans.d==0)
                Ans.f=0;
            return Ans;
        }
        BigInt operator +(const LL x){
            BigInt X(x);
            return X+(*this);
        }
        BigInt operator -(BigInt y){//减法运算 
            BigInt Ans=*this;
            if (f!=y.f){
                y.f=Ans.f,Ans=Ans+y;
                return Ans;
            }
            if (Ans.equ(y)){
                Ans=0;
                return Ans;
            }
            if (Ans.smaller(y)){
                Ans=y-Ans,Ans.f=!f;
                return Ans;
            }
            for (int i=1;i<=max(Ans.d,y.d);i++)
                if (Ans.v[i]-y.v[i]<0)
                    Ans.v[i]+=mod-y.v[i],Ans.v[i+1]--;
                else
                    Ans.v[i]-=y.v[i];
            while (Ans.d>0&&Ans.v[Ans.d]==0)
                Ans.d--;
            if (Ans.d==0) 
                Ans.f=0;
            return Ans;
        }
        BigInt operator -(const LL x){
            BigInt Ans(x);
            return (*this)-Ans;
        }
        BigInt operator *(const BigInt &y){//乘法运算 
            BigInt x=*this,Ans(0);
            Ans=0,Ans.f=f^y.f;
            for (int i=1;i<=x.d;i++)
                for (int j=1;j<=y.d;j++){
                    LL now=Ans.v[i+j-1]+x.v[i]*y.v[j];
                    Ans.v[i+j-1]=now%mod;
                    Ans.v[i+j]+=now/mod;
                }
            Ans.d=x.d+y.d-1;
            for (int i=1;i<=Ans.d;i++)
                Ans.v[i+1]+=Ans.v[i]/mod,Ans.v[i]%=mod;
            if (Ans.v[Ans.d+1])    
                Ans.d++;
            if (Ans.d==0)
                Ans.f=0;
            return Ans;
        }
        BigInt operator *(LL y){
            BigInt Ans=*this;
            if (y<0) 
                Ans.f^=1;
            y=abs(y);
            for (int i=1;i<=d;i++)
                Ans.v[i]*=y;
            for (int i=1;i<=d||Ans.v[i]>0;i++)
                Ans.v[i+1]+=Ans.v[i]/mod,Ans.v[i]%=mod,Ans.d=max(d,i);
            if (Ans.d==0) 
                Ans.f=0;
            return Ans;
        }
        BigInt operator /(BigInt y){//除法运算 
            BigInt Ans(0),x=*this,minus;
            bool Ansf=f^y.f;
            x.f=y.f=0,minus=y;
            while ((minus*10).smqu(x))
                minus=minus*10;
            while (minus.bgqu(y)){
                Ans=Ans*10;
                while (minus.smqu(x))
                    x=x-minus,Ans=Ans+1;
                minus=minus/10;
            }
            Ans.f=Ansf;
            if (Ans.d==0) 
                Ans.f=0;
            return Ans;
        }
        BigInt operator /(LL x){
            BigInt Ans(0);
            LL prev=0;
            Ans.f=f^(x<0),Ans.d=0,x=abs(x);
            for (int i=d;i>0;i--){
                prev=prev*mod+v[i];
                if (prev>=x)
                    Ans.v[i]=prev/x,prev%=x,Ans.d=max(Ans.d,i);
            }
            if (Ans.d==0) 
                Ans.f=0;
            return Ans;
        }
        BigInt operator %(BigInt y){//取模运算 
            BigInt x=*this,minus;
            bool xfz=f^y.f;
            x.f=y.f=0,minus=y;
            if (x<y){
                x.f=xfz;
                return x;
            }
            while ((minus*10).smqu(x))
                minus=minus*10;
            while (minus.bgqu(y)){
                while (minus.smqu(x))
                    x=x-minus;
                minus=minus/10;
            }
            x.f=xfz;
            if (x.d==0) 
                x.f=0;
            return x;
        }
        LL operator %(LL x){
            LL prev=0;
            bool flag=f^(x<0);
            x=abs(x);
            for (int i=d;i>0;i--)
                prev=prev*mod+v[i],prev%=x;
            if (flag) 
                prev=-prev;
            return prev;
        }
        BigInt operator ^(int x){
            BigInt Ans;
            Ans=1;
            if (x==0)
                return Ans;
            Ans=*this^(x/2);
            Ans=Ans*Ans;
            if (x&1)
                Ans=Ans**this;
            return Ans;
        }
    }zero(0),one(1);
    BigInt GcdY(BigInt x,BigInt y){
        return y!=zero?GcdY(y,x%y):x;
    }
    BigInt Gcd(BigInt x,BigInt y){
        x.f=y.f=0;
        if (x==zero)
            return y;
        if (y==zero)
            return x;
        return GcdY(x,y);
    }
    BigInt LcmY(BigInt x,BigInt y){
        return x/GcdY(x,y)*y;
    }
    BigInt Lcm(BigInt x,BigInt y){
        x.f=y.f=0;
        if (x==zero)
            return y;
        if (y==zero)
            return x;
        return LcmY(x,y);
    }
    BigInt abs(BigInt x){
        x.f=0;
        return x;
    }
    void read(char ch,BigInt &x){
        if (ch=='c'){
            char str[MaxLen];
            scanf("%s",str),x=str;
        }
        if (ch=='i'){
            int y;
            scanf("%d",&y),x=y;
        }
        if (ch=='L'){
            LL y;
            scanf("%lld",&y),x=y;
        }
    }
    void readint(BigInt &x,int &ret){
        scanf("%d",&ret),x=ret;
    }
    void readLL(BigInt &x,LL &ret){
        scanf("%lld",&ret),x=ret;
    }
    int main(){
        BigInt A,B;
        LL a,b;
        readLL(A,a),readLL(B,b);                    A.Print(' ');    B.Print(' ');
    //    read('c',A),read('c',B);                    A.Print(' ');    B.Print(' ');
        printf("Gcd(A,B)=");Gcd(A,B).Print('
    ');    A.Print(' ');    B.Print(' ');
        printf("Lcm(A,B)=");Lcm(A,B).Print('
    ');    A.Print(' ');    B.Print(' ');
        printf("A*B=");(A*B).Print('
    ');            A.Print(' ');    B.Print(' ');
        printf("A-B=");(A-B).Print('
    ');            A.Print(' ');    B.Print(' ');
        printf("A+B=");(A+B).Print('
    ');            A.Print(' ');    B.Print(' ');
        printf("A/B=");(A/B).Print('
    ');            A.Print(' ');    B.Print(' ');
    //    printf("A/b=");(A/b).Print('
    ');            A.Print(' ');    B.Print(' ');
    //    printf("A^b=");(A^b).Print('
    ');            A.Print(' ');    B.Print(' ');
    //    printf("A*b=");(A*b).Print('
    ');            A.Print(' ');    B.Print(' ');
    //    printf("A+b=");(A+b).Print('
    ');            A.Print(' ');    B.Print(' ');
    //    printf("A-b=");(A-b).Print('
    ');            A.Print(' ');    B.Print(' ');
    //    printf("A%b=");printf("%d
    ",A%b);            A.Print(' ');    B.Print(' ');
        printf("A%B=");(A%B).Print('
    ');            A.Print(' ');    B.Print(' ');
        printf("A<B?->");printf("%d
    ",A<B);        A.Print(' ');    B.Print(' ');
        printf("A>B?->");printf("%d
    ",A>B);        A.Print(' ');    B.Print(' ');
        printf("A<=B?->");printf("%d
    ",A<=B);        A.Print(' ');    B.Print(' ');
        printf("A>=B?->");printf("%d
    ",A>=B);        A.Print(' ');    B.Print(' ');
        printf("A==B?->");printf("%d
    ",A==B);        A.Print(' ');    B.Print(' ');
        printf("A!=B?->");printf("%d
    ",A!=B);        A.Print(' ');    B.Print(' ');
        printf("abs(A)=");abs(A).Print('
    ');        A.Print(' ');    B.Print(' ');
        printf("abs(B)=");abs(B).Print('
    ');        A.Print(' ');    B.Print(' ');
        printf("
    优先级判断:
    ");                    A.Print(' ');    B.Print(' ');
        printf("A*B^2=");(A*B^2).Print('
    ');        A.Print(' ');    B.Print(' ');//ans:A*B^2=(A*B)^2
        printf("A/A*B=");(A/A*B).Print('
    ');        A.Print(' ');    B.Print(' ');//ans:A/A*B=B
        return 0;
    }
    View Code

    高精度分数运算(基于上一份高精度整数运算)

    struct BigDouble{
        BigInt a,b;
        void Print(){
            (*this).Smaller();
            if (a==zero){
                printf("0");
                return;
            }
            if (b==one)
                a.Print();
            else
                a.Print('/'),b.Print();
        }
        void Print(char ch){
            (*this).Smaller();
            if (a==zero){
                printf("0%c",ch);
                return;
            }
            if (b==one)
                a.Print(ch);
            else
                a.Print('/'),b.Print(ch);
        }
        void Smaller(){
            if (a==zero||b==zero)
                return;
    //        a.Print('
    ');b.Print('
    ');
            BigInt gcd=Gcd(a,b);
    //        puts("Small achieve");
            a=a/gcd,b=b/gcd;
    //        puts("div adchive");
            if (b.f)
                a.f^=1,b.f^=1;
        }
        void operator = (BigInt x){
            a=x,b=one;
        }
        BigDouble operator * (BigDouble x){
            BigDouble ans;
            ans.a=a*x.a,ans.b=b*x.b;
            ans.Smaller();
            return ans;
        }
        BigDouble operator * (BigInt x){
            BigDouble ans;
            ans.a=x*a,ans.b=b;
            ans.Smaller();
            return ans;
        }
        BigDouble operator - (BigDouble x){
            BigDouble ans;
            BigInt lcm=Lcm(b,x.b),tt=lcm/b,tx=lcm/x.b;
    //        BigInt lcm=one,tt=one,tx=one;
            ans.b=lcm,ans.a=a*tt-x.a*tx;
            ans.Smaller();
            return ans;
        }
        BigDouble operator / (BigInt x){
            BigDouble ans;
    //        b.Print('
    ');x.Print('
    ');
            ans.a=a,ans.b=b*x;
    //        puts("still alive");
            ans.Smaller();
            return ans;
        }
        BigDouble operator / (BigDouble x){
            BigDouble ans;
            ans.a=a*x.b,ans.b=b*x.a;
            ans.Smaller();
            return ans;
        }
    }x[N];
    View Code

    Topcoder 模板

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    class $CLASSNAME${
    public:
    $RC$ $METHODNAME$($METHODPARMS$){
        
    }
    
    
    $TESTCODE$
    };
    
    
    $BEGINCUT$
    
    int main(){
        $CLASSNAME$ ___test;
        ___test.run_test(-1);
        return 0;
    }
    
    $ENDCUT$
    
    TopCoder Template
    View Code

    Adblock-Plus 配置 

    blog.csdn.net##.recommend-item-box.recommend-box-ident.type_blog.clearfix
    blog.csdn.net##.right_box.footer_box.csdn-tracking-statistics
    blog.csdn.net##.recommend-item-box.recommend-box-ident.recommend-download-box.clearfix
    blog.csdn.net##.recommend-item-box.recommend-ad-box
    blog.csdn.net##.recommend-item-box.blog-expert-recommend-box
    blog.csdn.net##.recommend-item-box.type_hot_word
    blog.csdn.net##.pulllog-box
    blog.csdn.net##.meau-list
    blog.csdn.net###reportContent
    blog.csdn.net##.btn-meau
    blog.csdn.net##.showBig.animated.bounceInRight
    blog.csdn.net##.btn-close-ad
    blog.csdn.net##.recommend-item-box.baiduSearch
    blog.csdn.net##.mediav_ad
    blog.csdn.net###csdn-toolbar
    blog.csdn.net##.csdn-toolbar.tb_disnone
    blog.csdn.net##.aside-title
    blog.csdn.net##.aside-content.clearfix
    blog.csdn.net##.aside-content
    ||pos.baidu.com/fcpm?conwid=300&conhei=250&rdid=3032528&dc=3&exps=110011&psi=3804dbd7f0cda1b989b5fada13af8a2f&di=u3032528&dri=0&dis=0&dai=3&ps=328x101&enu=encoding&dcb=___adblockplus&dtm=HTML_POST&dvi=0.0&dci=-1&dpt=none&tsr=0&tpr=1550628280860&ti=bzoj1001%5B%E7%8B%BC%E7%88%AA%E5%85%94%E5%AD%90%5D%E6%9C%80%E5%B0%8F%E5%89%B2%E8%BD%AC%E6%9C%80%E7%9F%AD%E8%B7%AF%20-%20nikelong%E7%9A%84%E5%8D%9A%E5%AE%A2%20-%20CSDN%E5%8D%9A%E5%AE%A2&ari=2&dbv=2&drs=1&pcs=1520x730&pss=1520x1381&cfv=0&cpl=2&chi=1&cce=true&cec=UTF-8&tlm=1550628281&prot=2&rw=730&ltu=https%3A%2F%2Fblog.csdn.net%2Fnikelong0%2Farticle%2Fdetails%2F50727840&ltr=https%3A%2F%2Fwww.baidu.com%2Flink%3Furl%3DtXeCtlFHKSI-g80lHRJJU5wQQTLkRXwzB9I_1Yv55g1AELHBDINXkK7We0Q8oyGYd8I_1w9_NJEIVWJPc4VB7c77R35Xvq6ZKjliEwJH_dK%26wd%3D%26eqid%3Da9c9876500035954000000045c6cb5b6&ecd=1&uc=1535x876&pis=-1x-1&sr=1600x900&tcn=1550628281&qn=c1321737e08fd10c&tt=1550628280838.463.464.472
    ||ubmcmm.baidustatic.com/media/v1/0f000ji9GQBTQif4KlS1n6.jpg
    ||ubmcmm.baidustatic.com/media/v1/0f000Dd29O15Im4ZGHXmF0.jpg
    ||pos.baidu.com/fcpm?conwid=300&conhei=250&rdid=3163270&dc=3&exps=110011&psi=3804dbd7f0cda1b989b5fada13af8a2f&di=u3163270&dri=0&dis=0&dai=4&ps=2124x101&enu=encoding&dcb=___adblockplus&dtm=HTML_POST&dvi=0.0&dci=-1&dpt=none&tsr=0&tpr=1550628280860&ti=bzoj1001%5B%E7%8B%BC%E7%88%AA%E5%85%94%E5%AD%90%5D%E6%9C%80%E5%B0%8F%E5%89%B2%E8%BD%AC%E6%9C%80%E7%9F%AD%E8%B7%AF%20-%20nikelong%E7%9A%84%E5%8D%9A%E5%AE%A2%20-%20CSDN%E5%8D%9A%E5%AE%A2&ari=2&dbv=2&drs=1&pcs=1520x730&pss=1520x2124&cfv=0&cpl=2&chi=1&cce=true&cec=UTF-8&tlm=1550628281&prot=2&rw=730&ltu=https%3A%2F%2Fblog.csdn.net%2Fnikelong0%2Farticle%2Fdetails%2F50727840&ltr=https%3A%2F%2Fwww.baidu.com%2Flink%3Furl%3DtXeCtlFHKSI-g80lHRJJU5wQQTLkRXwzB9I_1Yv55g1AELHBDINXkK7We0Q8oyGYd8I_1w9_NJEIVWJPc4VB7c77R35Xvq6ZKjliEwJH_dK%26wd%3D%26eqid%3Da9c9876500035954000000045c6cb5b6&ecd=1&uc=1535x876&pis=-1x-1&sr=1600x900&tcn=1550628281&qn=4c64b7f9f4152e7f&tt=1550628280838.552.552.554
    View Code

    Ublock 配置

    ! 2/20/2019 https://blog.csdn.net
    blog.csdn.net##.box-shadow.mb8.csdn-tracking-statistics
    blog.csdn.net###asideHotArticle > .aside-content
    blog.csdn.net###asideCategory > .text-center
    blog.csdn.net###asideArchive > .text-center
    ! 2018/10/28 下午12:34:10 https://blog.csdn.net/forever_shi/article/details/83109712
    blog.csdn.net###asideArchive > .text-center
    
    ! 2018/10/28 下午12:35:08 https://blog.csdn.net/forever_shi/article/details/83109712
    blog.csdn.net##.d-flex.badge-box
    
    ! 2018/10/28 下午12:35:12 https://blog.csdn.net/forever_shi/article/details/83109712
    blog.csdn.net###asideCategory > .text-center
    
    ! 2018/10/28 下午12:35:17 https://blog.csdn.net/forever_shi/article/details/83109712
    blog.csdn.net##.meau-list > li:nth-of-type(2)
    
    ! 2018/10/28 下午12:35:21 https://blog.csdn.net/forever_shi/article/details/83109712
    blog.csdn.net##.meau-list > li:nth-of-type(3)
    
    ! 2018/10/28 下午12:35:31 https://blog.csdn.net/forever_shi/article/details/83109712
    blog.csdn.net###btnAttent
    
    ! 2018/10/28 下午3:16:50 https://www.baidu.com/
    www.baidu.com##.qrcodeCon
    
    ! 2018/10/28 下午3:16:54 https://www.baidu.com/
    www.baidu.com##.ftCon-Wrapper
    
    ! 2018/10/28 下午3:16:59 https://www.baidu.com/
    www.baidu.com###ftCon
    
    ! 2018/10/28 下午3:17:04 https://www.baidu.com/
    www.baidu.com###u1
    
    ! 2018/10/28 下午3:17:34 https://www.baidu.com/s?ie=utf-8&f=8&rsv_bp=0&rsv_idx=1&tn=baidu&wd=bzoj1117&rsv_pq=f8dd55a00004d7ff&rsv_t=193aOBVWE2eZF6R2NN3h0DU7J8Gd8a3U%2FRiSF1tHowxIyesTo4%2B4In05Au0&rqlang=cn&rsv_enter=1&rsv_sug3=8&rsv_sug1=4&rsv_sug7=100&rsv_sug2=0&inputT=3603&rsv_sug4=3603
    www.baidu.com###content_right
    
    ! 2018/10/28 下午4:33:48 https://blog.csdn.net/ocgcn2010/article/details/50939261?fps=1&locationNum=9
    blog.csdn.net##.clearfix.t0
    
    ! 2018/10/28 下午6:24:05 https://blog.csdn.net/clover_hxy/article/details/53834702
    blog.csdn.net##div.recommend-item-box:nth-of-type(20)
    
    ! 2018/10/28 下午6:26:00 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    ||simg.sinajs.cn/blog7newtpl/image/30/30_4/images/sg_newsp.png$image
    
    ! 2018/10/28 下午6:26:07 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    blog.sina.com.cn##.turnBoxzz > .SG_turn.SG_aBtn_ico.SG_aBtn > cite
    
    ! 2018/10/28 下午6:27:10 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    blog.sina.com.cn##.formBtn
    
    ! 2018/10/28 下午6:27:15 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    blog.sina.com.cn##.writeComm
    
    ! 2018/10/28 下午6:27:21 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    blog.sina.com.cn##.SG_floatR > .SG_txtb
    
    ! 2018/10/28 下午6:27:52 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    ||simg.sinajs.cn/blog7style/images/common/btn_share.png$image
    
    ! 2018/10/28 下午6:27:56 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    blog.sina.com.cn##.SG_txtb.share > span
    
    ! 2018/10/28 下午6:28:00 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    ||simg.sinajs.cn/blog7style/images/common/multishow_bg.png$image
    
    ! 2018/10/28 下午6:28:05 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    blog.sina.com.cn##.picSlideBtn > [href="javascript:void(0);"]
    
    ! 2018/10/28 下午6:28:14 http://blog.sina.com.cn/s/blog_8f06da99010125ol.html
    ||sjs.sinajs.cn/blog7swf/lookViewSpring.swf?2$object
    
    ! 2018/10/28 下午6:29:28 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com###activity-tg
    
    ! 2018/10/28 下午6:29:45 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com###activity-tg
    
    ! 2018/10/28 下午6:30:09 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.reader-backToTop
    
    ! 2018/10/28 下午6:30:16 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com###activity-tg
    
    ! 2018/10/28 下午6:30:30 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.notvip-book.side-a.novip.vip-wrap
    
    ! 2018/10/28 下午6:30:35 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.reader-download-app
    
    ! 2018/10/28 下午6:30:39 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.help-feedback.log-xsend.reader-feedback
    
    ! 2018/10/28 下午6:32:26 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.ui-nav
    
    ! 2018/10/28 下午6:32:32 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.wk-other-new-cntent
    
    ! 2018/10/28 下午6:32:51 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.bg-opacity
    
    ! 2018/10/28 下午6:32:57 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.fix-searchbar-wrap
    
    ! 2018/10/28 下午6:33:03 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.ui-btn-btc.ui-btn-p-16
    
    ! 2018/10/28 下午6:33:19 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.view_change_test.top-down-load-container
    
    ! 2018/10/28 下午6:33:23 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.open-in-phone
    
    ! 2018/10/28 下午6:33:27 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.addUnderLine.log-xsend.banner-text
    
    ! 2018/10/28 下午6:33:31 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    ||wkstatic.bdimg.com/static/wkview/widget/fix_searchbar/images/add-money-icon.svg$image
    
    ! 2018/10/28 下午6:33:37 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    ||wkstatic.bdimg.com/static/wkview/widget/fix_searchbar/images/bg_fix_search_5ea1707.png$image
    
    ! 2018/10/28 下午6:33:43 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.s_ipt_wr-with535.s_ipt_wr
    
    ! 2018/10/28 下午6:33:48 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.clearfix.sb-con
    
    ! 2018/10/28 下午6:33:53 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.fix-searchbar-wrap
    
    ! 2018/10/28 下午6:34:07 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.doc-tag-ticket.doc-tag
    
    ! 2018/10/28 下午6:34:12 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.fix-searchbar-wrap
    
    ! 2018/10/28 下午6:34:30 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.clearfix.reader-tools-bar-center
    
    ! 2018/10/28 下午6:35:34 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.tools-bar-small.reader-tools-bar-wrap
    
    ! 2018/10/28 下午6:35:57 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.crubms-wrap
    
    ! 2018/10/28 下午6:36:11 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.doc-bottom-text
    
    ! 2018/10/28 下午6:36:18 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.fix-searchbar-wrap
    
    ! 2018/10/28 下午6:36:34 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.doc-banner-rights-wrap
    
    ! 2018/10/28 下午6:36:38 https://wenku.baidu.com/view/379e8baaa58da0116d174924.html
    wenku.baidu.com##.doc-banner-text
    
    ! 2018/10/28 下午6:37:22 https://www.cnblogs.com/cjyyb/p/9495131.html
    www.cnblogs.com###green_channel
    
    ! 2018/10/28 下午6:38:05 https://www.cnblogs.com/cjyyb/p/9495131.html
    www.cnblogs.com###under_post_news > .c_ad_block.itnews
    
    ! 2018/10/28 下午6:38:10 https://www.cnblogs.com/cjyyb/p/9495131.html
    www.cnblogs.com###kb_block
    
    ! 2018/10/28 下午6:38:16 https://www.cnblogs.com/cjyyb/p/9495131.html
    www.cnblogs.com###HistoryToday
    
    ! 2018/11/2 下午6:32:46 https://blog.csdn.net/qq_31759205/article/details/75008659
    blog.csdn.net##.recommend-end-box
    
    ! 2018/11/10 下午6:21:30 https://fanyi.qq.com/
    ||fanyi.qq.com/images/footer-bg.ba7ba223.png$image
    
    ! 2018/11/10 下午6:21:36 https://fanyi.qq.com/
    fanyi.qq.com##.foot-container
    
    ! 2018/11/23 下午10:42:31 https://blog.csdn.net/qq_34531807/article/details/80627568
    blog.csdn.net###asideColumn > .text-center
    
    ! 2018/11/27 下午10:14:44 https://baike.baidu.com/item/%E9%98%BF%E6%88%BF%E5%AE%AB%E8%B5%8B/2196073?fr=aladdin
    ||gss1.bdstatic.com/-vo3dSag_xI4khGkpoWK1HF6hhy/baike/whfpf%3D268%2C152%2C50/sign=1061dd6d03f41bd5da06bbb437e7b3fd/d4628535e5dde711fd9079b0aaefce1b9d166131.jpg$image
    
    ! 2018/12/23 下午6:09:36 https://wenku.baidu.com/view/75906f160b4e767f5acfcedb
    wenku.baidu.com##.doc-tag-vip-free.doc-tag > span
    
    ! 2018/12/23 下午6:09:42 https://wenku.baidu.com/view/75906f160b4e767f5acfcedb
    wenku.baidu.com##.doc-tag-vip-free.doc-tag
    
    ! 2018/12/23 下午6:10:18 https://wenku.baidu.com/view/75906f160b4e767f5acfcedb
    ||wkstatic.bdimg.com/static/wkcommon/widget/user_bar/images/spr_bg_tip_c0ad547.png$image
    
    ! 2018/12/23 下午6:10:23 https://wenku.baidu.com/view/75906f160b4e767f5acfcedb
    ||wkstatic.bdimg.com/static/wkcommon/widget/user_bar/images/spr_bg_tip_c0ad547.png$image
    
    ! 2018/12/23 下午6:10:27 https://wenku.baidu.com/view/75906f160b4e767f5acfcedb
    ||wkstatic.bdimg.com/static/wkcommon/widget/user_bar/images/spr_bg_tip_c0ad547.png$image
    
    ! 2019/1/24 上午9:34:04 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideNewArticle > .aside-content
    
    ! 2019/1/24 上午9:34:10 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net##.clearfix.aside-content
    
    ! 2019/1/24 上午9:34:15 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideCustom67105890 > .aside-title
    
    ! 2019/1/24 上午9:34:18 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideNewArticle > .aside-title
    
    ! 2019/1/24 上午9:34:24 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideCategory > .aside-content
    
    ! 2019/1/24 上午9:34:29 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideCategory > .aside-title
    
    ! 2019/1/24 上午9:34:32 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideArchive > .aside-title
    
    ! 2019/1/24 上午9:34:38 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideArchive > .aside-content
    
    ! 2019/1/24 上午9:34:43 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideHotArticle > .aside-content
    
    ! 2019/1/24 上午9:34:47 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideHotArticle > .aside-title
    
    ! 2019/1/24 上午9:34:52 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideNewComments > .aside-title
    
    ! 2019/1/24 上午9:34:56 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net###asideNewComments > .aside-content
    
    ! 2019/1/24 上午9:35:14 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net##.item-tiling.d-flex.data-info
    
    ! 2019/1/24 上午9:35:18 https://blog.csdn.net/qq_39972971/article/details/80609097
    blog.csdn.net##.clearfix.grade-box
    
    ! 2019/1/24 下午1:40:46 https://blog.csdn.net/ez_yww/article/details/79320152
    blog.csdn.net###asideCustom59895271 > .aside-title
    
    ! 2019/1/24 下午1:53:20 https://blog.csdn.net/ez_yww/article/details/79320152
    blog.csdn.net###csdn-toolbar
    View Code

     

  • 相关阅读:
    蓝桥杯历届试题 打印十字图 文字图形
    Cuckoo Hashing
    2006 飞行员配对(二分图最大匹配)
    Bad Hair Day(求数组中元素和它后面离它最近元素之间的元素个数)
    2019CCPC江西省赛
    字典树系统学习
    ac自动机学习
    项目管理(把与某点相邻边分为两类 是复杂度降为(n^(3/2))
    Ultra-QuickSort(离散化)
    Chika and Friendly Pairs(莫队+树状数组+离散化+预处理上下界)
  • 原文地址:https://www.cnblogs.com/zhouzhendong/p/Templates.html
Copyright © 2011-2022 走看看