洛谷

https://www.luogu.org/problemnew/show/P2257

求 (n,m) 中 (gcd(i,j)==p) 的数对的个数

求 $sumlimits_p sumlimits_{i=1}^{n}sumlimits_{j=1}^{m}[gcd(i,j)==p] $

由套路:
(=sumlimits_p sumlimits_{k=1}^{N}mu(k) lfloorfrac{n}{kp} floor lfloorfrac{m}{kp} floor)

再套路:
(=sumlimits_p sumlimits_{T=kp}^{N}mu(frac{T}{p}) lfloorfrac{n}{T} floor lfloorfrac{m}{T} floor)

交换求和:
(=sumlimits_{T=1}^{N} sumlimits_{p|T} mu(frac{T}{p}) lfloorfrac{n}{T} floor lfloorfrac{m}{T} floor)

提T:
(=sumlimits_{T=1}^{N} lfloorfrac{n}{T} floor lfloorfrac{m}{T} floor sumlimits_{p|T} mu(frac{T}{p}))

后面的式子可以预处理,方法是在筛出质数表和莫比乌斯函数表之后,枚举每个质数p,再枚举倍数k,给kp加上 (mu(k)) .

前面的式子可以整除分块 (r=min(n/(n/l),m/(m/l))) .

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define MAXN 10000000+5

/* 莫比乌斯函数筛 begin */

int mu[MAXN];
int pri[MAXN],pritop;
bool notpri[MAXN];
//pritop从1开始计数

int sumdmu[MAXN],prefixsumdmu[MAXN];

void sieve3(int n) {
    notpri[1]=mu[1]=1;
    for(int i=2; i<=n; i++) {
        if(!notpri[i])
            pri[++pritop]=i,mu[i]=-1;
        for(int j=1; j<=pritop&&i*pri[j]<=n; j++) {
            notpri[i*pri[j]]=1;
            //略有不同
            if(i%pri[j])
                mu[i*pri[j]]=-mu[i];
            else {
                mu[i*pri[j]]=0;
                break;
            }
        }
    }

    for(int j=1; j<=pritop; j++) {
        for(int i=1;i*pri[j]<=n;i++){
            sumdmu[i*pri[j]]+=mu[i];
        }
    }
    for(int i=1;i<=n;i++)
        prefixsumdmu[i]=prefixsumdmu[i-1]+sumdmu[i];
}

/* 莫比乌斯函数筛 end */


//整除分块,n,m版
ll aliquot_patition(int n,int m) {
    ll ans=0;
    int N=min(n,m);
    for(int l=1,r; l<=N; l=r+1) {
        r=min(n/(n/l),m/(m/l));
        ans+=1ll*(n/l)*(m/l)*(prefixsumdmu[r]-prefixsumdmu[l-1]);
    }
    return ans;
}

int main() {
    sieve3(10000000);
    int T;
    scanf("%d",&T);
    while(T--){
        int n,m;
        scanf("%d%d",&n,&m);
        printf("%lld
",aliquot_patition(n,m));
    }
}