[BZOJ2820]YY的GCD|莫比乌斯反演

Description

神犇YY虐完数论后给傻×kAc出了一题

给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对

kAc这种傻×必然不会了，于是向你来请教……

多组输入

 

　　题面依然很亲切，和之前不同的在于是所有质数

　　然后把式子化一化，对莫比乌斯函数求一个奇怪的前缀和

　　这里的时间复杂度比较神奇..

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
const int maxn = 10000010,maxp = 664580;
int vis[maxn],prime[maxp],mu[maxn];
long long sum[maxn];
int min(int a,int b){if (a<b) return a;else return b;}
void build(){
    memset(vis,1,sizeof(vis));
    prime[0]=0;mu[1]=1;
    for (int i=2;i<=maxn;i++){
        if (vis[i]){
            prime[++prime[0]] = i;
            mu[i] = -1;
        }
        for (int j=1;j<=prime[0]&&i*prime[j]<=maxn;j++){
            vis[i*prime[j]] = 0;
            if (i%prime[j]==0){
                mu[i*prime[j]] = 0;
                break;
            }
            mu[i*prime[j]] = -mu[i];
        }
    }
    memset(sum,0,sizeof(sum));
    for (int i=1;i<=prime[i];i++)
        for (int j=1;j*prime[i]<=maxn;j++) sum[j*prime[i]]+=mu[j];
    for (int i=2;i<=maxn;i++) sum[i]+=sum[i-1];
}

long long solve(int n,int m){
    if (n>m){int tem=n;n=m;m=tem;}
    int las;long long ans=0;
    for (int i=1;i<=n;i=las+1){
        las = min(n/(n/i),m/(m/i));
        ans+=(sum[las]-sum[i-1])*(long long)(n/i)*(m/i);
    }
    return ans;
}
int main(){
    build();
    int test,n,m;
    scanf("%d",&test);
    while (test--){
        scanf("%d%d",&n,&m);
        printf("%lld\n",solve(n,m));
    }
}