BZOJ 2820 YY的GCD

#include<iostream>
#include<cstdio>
using namespace std;
const int N=10000005;
int L=10000000,T,n,m,tot,pri[N],miu[N],f[N];
bool mrk[N];
long long ans;
void init()
{miu[1]=1;for(int i=2;i<=L;i++){if(!mrk[i])pri[++tot]=i,miu[i]=-1;for(int j=1;j<=tot&&i*pri[j]<=L;j++){mrk[i*pri[j]]=1;if(i%pri[j]==0){miu[i*pri[j]]=0;break;}miu[i*pri[j]]=-miu[i];}}for(int i=1;i<=tot;i++)for(int j=1;j*pri[i]<=L;j++)f[j*pri[i]]+=miu[j];for(int i=1;i<=L;i++)f[i]+=f[i-1];
}
int main()
{init();scanf("%d",&T);while(T--){ans=0;scanf("%d%d",&n,&m);for(int i=1,nxt;i<=min(n,m);i=nxt+1){nxt=min(n/(n/i),m/(m/i));ans+=(long long)(n/i)*(m/i)*(f[nxt]-f[i-1]);}printf("%lld\n",ans);}return 0;
}