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 $O(n)$
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#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 = 1; for(; n; n >>= 1, x = x * x % P) if(n & 1) ans = ans * x % P; return ans; } int main() { int T; for(int i = 2; i < N; i++) if(!v[i]) { prime[++tot] = i; for(int j = i+i; j < N; j += i) v[j] = 1; } scanf("%d", &T); while(T--) { scanf("%d %d", &n, &m); if(n > m) swap(n, m); LL ans = 1; for(int i = 1; i <= tot; i++) if(prime[i] <= n) { LL tmp = prime[i]; LL cnt = 0; while(tmp <= n) { cnt += (n/tmp) * (m/tmp); tmp *= prime[i]; } ans = ans * qpow(prime[i], cnt) % P; } printf("%lld ", ans); } }