https://cn.vjudge.net/contest/158362#problem/O

考慮每個質數的貢獻
目測不用線性篩也能過

#include<stdio.h>
#include<algorithm>
using namespace std;
#define lim 10000000
#define Int inline int
#define mod 1000000007
typedef long long ll;

bool vis[lim+5];
int pri[664580],ed,T,n,m;

Int pow(int a,int b)
{
    int rt=1;
    for
 
 (;b;a=(ll)a*a%mod,b>>=1) if (b&1) rt=(ll)rt*a%mod;
    return rt;
}

Int get(int p)
{
    int rt=0;
    for (ll i=p;i<=min(n,m);i*=p) rt=(rt+(ll)(n/i)*(m/i))%(mod-1);
    return rt;
}

int main()
{
    for (int i=2;i<=lim;i++)
    {
        if (!vis[i]) pri[++ed]=i;
        for (int j=1
 
;j<=ed && i*pri[j]<=lim;j++)
        {
            vis[i*pri[j]]=1;
            if (i%pri[j]==0) break;
        }
    }
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&n,&m);
        int rt=1;
        for (int i=1;i<=ed && pri[i]<=min(n,m);i++) rt=(ll)rt*pow
 
(pri[i],get(pri[i]))%mod;
        printf("%d\n",rt);
    }
}