[BZOJ 2820]YY的GCD 莫比乌斯反演

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define pos(i,a,b) for(int i=(a);i<=(b);i++)
#define N 10000100
#define LL long long
int t,n,m;
int mu[N],notprime[N],prime[N];
LL sum[N];
int cun[N];
void getmu(){
    mu[1]=1;notprime[1]=1;
    pos(i,2,N-10){
        if(!notprime[i]){
            mu[i]=-1;
            prime[++prime[0]]=i;
        }
        for(int j=1;j<=prime[0]&&prime[j]*i<=N-10;j++){
            notprime[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    pos(i,1,prime[0]){
        for(int j=1;prime[i]*j<=N-10;j++){
            cun[prime[i]*j]+=mu[j];
        }
    }
    pos(i,1,N-10) sum[i]=sum[i-1]+cun[i];
}
LL getans(){
    if(n>m) swap(n,m);
    int last(0);LL ans(0);
    for(int i=1;i<=n;i=last+1){
        last=min(n/(n/i),m/(m/i));
        ans+=(n/i)*1ll*(m/i)*1ll*(sum[last]-sum[i-1]);
    }
    return ans;
}
int main(){
    scanf("%d",&t);
    getmu();
    while(t--){
        scanf("%d%d",&n,&m);
        printf("%lld\n",getans());
    }
    return 0;
}