Product it again

题意：求解 $$\prod_{1 \leq i \leq n} \prod_{1 \leq j \leq m} {(i,j)}$$

解法：

满脑子的反演

考虑对于第一个质数 $p$ 的贡献为 $p^{[\frac{n}{p}][\frac{m}{p}] + [\frac{n}{p^2}][\frac{m}{p^2}] ... }$

这样1~n的质数大概有$O(\frac{n}{logn})$，对于每一个质数$O(logn)$，总效率大概为 $O(n)$

#include <iostream>
#include <cstdio>
#include <cstring>

#define LL long long
#define N 10000010
#define P 1000000007LL

using namespace std;

int n, m, tot, prime[N];
bool v[N];

LL qpow(LL x, LL n)
{
    LL ans = ;
    for(; n; n >>= , x = x * x % P)
        if(n & )
            ans = ans * x % P;
    return ans;
}

int main()
{
    int T;
    for(int i = ; i < N; i++)
        if(!v[i])
        {
            prime[++tot] = i;
            for(int j = i+i; j < N; j += i)
                v[j] = ;
        }
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d %d", &n, &m);
        if(n > m) swap(n, m);
        LL ans = ;
        for(int i = ; i <= tot; i++)
            if(prime[i] <= n)
            {
                LL tmp = prime[i];
                LL cnt = ;
                while(tmp <= n)
                {
                    cnt += (n/tmp) * (m/tmp);
                    tmp *= prime[i];
                }
                ans = ans * qpow(prime[i], cnt) % P;
            }
        printf("%lld\n", ans);
    }
}