P2257 YY的GCD

第一次做的时候没有优化到底,T了

我竟然蠢蠢地去枚举质数了

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#define inf 2147483647
#define N 10000010
#define mod 1000000007
#define p(a) putchar(a)
#define For(i,a,b) for(long long i=a;i<=b;++i)
//by war
//2020.8.6
using namespace std;
long long T,n,m,cnt,ans;
long long prime[N],mu[N];
bool vis[N];
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;
}
void o(long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

void Euler(){
    mu[1]=1;
    For(i,2,1e7){
        if(!vis[i]) prime[++cnt]=i,mu[i]=-1;
        for(long long j=1;j<=cnt&&i*prime[j]<=1e7;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    For(i,1,1e7) mu[i]+=mu[i-1];
}

int solve(int n,int m){
    int r=0;
    for(int i=1,j;i<=min(n,m);i=j+1){
        j=min(n/(n/i),m/(m/i));
        r+=(mu[j]-mu[i-1])*(n/i)*(m/i);
    }   
    return r;
}

signed main(){
    Euler();
    in(T);
    while(T--){
        in(n);in(m);
        ans=0;
        For(i,1,cnt){
            if(prime[i]>n || prime[i]>m) break;
            ans+=solve(n/prime[i],m/prime[i]);
        }
        o(ans);p('\n');
    }
    return 0;
}

 

 

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstring>
#define inf 2147483647
#define N 10000010
#define mod 1000000007
#define p(a) putchar(a)
#define For(i,a,b) for(register int i=a;i<=b;++i)
//by war
//2020.8.6
using namespace std;
long long T,n,m,cnt,ans;
int prime[N],mu[N],g[N];
bool vis[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 Euler(){
    mu[1]=1;
    For(i,2,1e7){
        if(!vis[i]) prime[++cnt]=i,mu[i]=-1;
        for(register int j=1;j<=cnt&&i*prime[j]<=1e7;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;
                break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    For(j,1,cnt)
        for(register int i=1;i*prime[j]<=1e7;i++){
            g[i*prime[j]]+=mu[i];
        }
    For(i,1,1e7) g[i]+=g[i-1];
}

inline long long solve(long long n,long long m){
    long long r=0;
    for(register int i=1,j;i<=n;i=j+1){
        j=min(n/(n/i),m/(m/i));
        r+=(n/i)*(m/i)*(g[j]-g[i-1]);
    }   
    return r;
}

signed main(){
    Euler();
    in(T);
    while(T--){
        in(n);in(m);
        if(n>m) swap(n,m);
        o(solve(n,m));p('\n');
    }
    return 0;
}

 

posted @ 2020-08-06 22:03  WeiAR  阅读(82)  评论(0编辑  收藏  举报