SPOJ PGCD(莫比乌斯反演)

#pragma comment(linker,"/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <limits.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define LL long long
#define mod 100000000
#define inf 0x3f3f3f3f
#define eps 1e-6
#define N 10000000
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define PII pair<int,int>
using namespace std;
inline int read()
{
    char ch=getchar();int x=0,f=1;
    while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch<=‘9‘&&ch>=‘0‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
bool vis[N+5];
int mu[N+5],prime[N+5],sum[N+5],num[N+5];
void Mobius()
{
    memset(vis,false,sizeof(vis));
    mu[1]=1;
    int tot=0;
    for(int i=2;i<=N;i++)
    {
        if(!vis[i])
        {
            prime[tot++]=i;
            mu[i]=-1;
        }
        for(int j=0;j<tot;j++)
        {
            if(i*prime[j]>N)break;
            vis[i*prime[j]]=true;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                break;
            }
            else
            {
                mu[i*prime[j]]=-mu[i];
            }
        }
    }
    for(int i=0;i<tot;i++)
        for(int j=prime[i];j<=N;j+=prime[i])
        num[j]+=mu[j/prime[i]];//预处理出对于所有质数p，sigma(f(p))对应的F(i)的系数，用num[i]表示
    for(int i=1;i<=N;i++)sum[i]=sum[i-1]+num[i];
}
LL solve(int n,int m)
{
    LL res=0;
    if(n>m)swap(n,m);
    for(int i=1,last=0;i<=n;i=last+1)
    {
        last=min(n/(n/i),m/(m/i));
        res+=(LL)(sum[last]-sum[i-1])*(n/i)*(m/i);
    }
    return res;
}

int main()
{
    int T,n,m;
    Mobius();
    T=read();
    while(T--)
    {
        n=read();m=read();
        LL ans=solve(n,m);
        printf("%lld\n",ans);
    }
}