題意 "題目鏈接" Sol 一道咕咕咕了好長時間的題 題解可以看 "這裡" cpp include define LL long long using namespace std; const int MAXN = 1e7 + 5e6 + 10, mod = 1e9 + 7, mod2 = 1e9 ...
題意
Sol
一道咕咕咕了好長時間的題
題解可以看這裡
#include<bits/stdc++.h>
#define LL long long
using namespace std;
const int MAXN = 1e7 + 5e6 + 10, mod = 1e9 + 7, mod2 = 1e9 + 6;
int N, M, vis[MAXN], prime[MAXN], mu[MAXN], f[MAXN], tot;
int add(int x, int y) {
if(x + y < 0) return x + y + mod;
return x + y >= mod ? x + y - mod : x + y;
}
int mul(int x, int y) {
return 1ll * x * y % mod;
}
int fp(int a, int p) {
int base = 1;
while(p) {
if(p & 1) base = mul(base, a);
a = mul(a, a); p >>= 1;
}
return base;
}
void sieve(int N) {
vis[1] = 1; mu[1] = 1;
for(int i = 2; i <= N; i++) {
if(!vis[i]) prime[++tot] = i, mu[i] = -1;
for(int j = 1; j <= tot && i * prime[j] <= N; j++) {
vis[i * prime[j]] = 1;
if(!(i % prime[j])) {mu[i * prime[j]] = 0; break;}
else mu[i * prime[j]] = -mu[i];
}
}
for(int i = 1; i <= tot; i++)
for(LL j = prime[i]; j <= N; j *= prime[i])
f[j] =prime[i];
for(int i = 1; i <= N; i++) if(!f[i]) f[i] = 1;
}
signed main() {
cin >> N >> M;
sieve(1e7 + 5e6);
//for(int i = 1; i <= 100; i++) printf("%d %d\n", i, f[i]);
int ans = 1;
for(int i = 1; i <= N; i++) {
if(f[i] == 1) continue;
ans = mul(ans, fp(f[i], 1ll * (N / i) * (M / i) % mod2));
}
cout << ans;
return 0;
}
/*
100000 50000 200 300
100 2 1 1
*/