快速幂

注意：long long ，同余

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 10, INF = 0x3f3f3f3f, MOD = 1E9 + 7;
#define int long long

int fpow(int a, int b, int p) {
    if (b == 0) return 1;
    int t = fpow(a, b / 2, p);
    t = t * t % p;
    if (b & 1) t = t * a %p;
    return t%p;
}

// int  = [signed] int
signed main() {
    int a, b, p;
    scanf("%lld%lld%lld", &a, &b, &p);
    printf("%lld^%lld mod %lld=%lld\n", a, b, p, fpow(a, b, p));
    return 0;
}