• https://vjudge.net/contest/321565#problem/C 超时代码


    #include <iostream>
    #include <cstdio>
    #include <queue>
    #include <algorithm>
    #include <cmath>
    #include <cstring>
    #define inf 2147483647
    #define N 10100
    #define p(a) putchar(a)
    #define For(i,a,b) for(register long long i=a;i<=b;++i)
    //by war
    //2019.8.22
    using namespace std;
    long long T,n,x,y,cnt,tot;
    long long prime[N],mu[N],ans[55];
    bool vis[N];
    
    struct dian{
        long long l,r,t;
    }a[N];
    
    struct node{
        long long n;
        node *next;
    }*e[N];
    
    inline void in(long long &x){
        long long y=1;char c=getchar();x=0;
        while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
        while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
        x*=y;
    }
    inline void o(long long x){
        if(x<0){p('-');x=-x;}
        if(x>9)o(x/10);
        p(x%10+'0');
    }
    
    inline void push(long long x,long long y){
        node *p;
        p=new node();
        p->n=y;
        if(e[x]==0)
            e[x]=p;
        else{
            p->next=e[x]->next;
            e[x]->next=p;
        }
    }
    
    void Euler(){
        mu[1]=1;
        For(i,2,50){
            if(!vis[i]) prime[++cnt]=i,mu[i]=-1;
            for(register long long j=1;j<=cnt&&i*prime[j]<=50;j++){
                vis[i*prime[j]]=1;
                if(i%prime[j]==0){
                    mu[i*prime[j]]=0;
                    break;
                }
                mu[i*prime[j]]=-mu[i];
            }
        }
    }
    
    long long dfs(long long x,long long fa,long long w){
        long long res=0;
        for(node *i=e[x];i;i=i->next)
            if(i->n!=fa)
                res+=dfs(i->n,x,w*a[i->n].t);
        return res+w;
    }
    
    inline long long F(register long long d){
        long long res=0,kk=0;
        For(i,1,n){
            a[i].t=a[i].r/d-a[i].l/d;
            if(a[i].l%d==0)
                a[i].t++;
        }
        For(i,1,n){
            res+=dfs(i,i,a[i].t);
            res-=a[i].t;
            kk+=a[i].t;
        }
        return res/2+kk;
    }
    
    inline void clear(){
        For(i,1,50)
            e[i]=0;
        memset(ans,0,sizeof(ans));
    }
    
    signed main(){
        in(T);
        Euler();
        while(T--){
            clear();
            in(n);
            For(i,1,n-1){
                in(x);in(y);
                push(x,y);
                push(y,x);
            }
            For(i,1,n) in(a[i].l);
            For(i,1,n) in(a[i].r);
            For(i,1,50)
                for(register long long d=i;d<=50;d+=i)
                    ans[i]+=mu[d/i]*F(d);
            printf("Case %lld:
    ",++tot);
            For(i,1,50){
                o(i);p(':');p(' ');
                o(ans[i]);p('
    ');
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/war1111/p/11397482.html
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