a^b%c = a^(phi(c)+b%phi(c))%c

1e9看似吓人，其实中间模数为1时可以直接返回0，算是个强剪枝

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL pow(LL x, LL y, LL mod) {
    LL ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod;
        y >>= 1;
    }
    return ans;
}
LL phi(LL x) {
    LL ans = x;
    for (LL i = 2; i*i <= x; i++) {
        if (x % i == 0) {
            ans = ans/i*(i-1);
            while (x % i == 0) x /= i;
        }
    }
    if (x > 1) ans = ans / x * (x-1);
    return ans;
}
LL F(LL x, LL y) {
    if (y == 1) return 0;
    else if (x == 1) return 1 % y;
    else if (x == 2) return 2 % y;
    else if (x == 3) return 9 % y;
    else if (x == 4) return 262144 % y;
    LL t = phi(y);
    return pow(x, t+F(x-1, t), y);
}
int main() {
    LL n, m;
    scanf("%lld%lld", &n, &m);
    printf("%lld\n", F(n, m));
    return 0;
}