快速模幂(c++)

求m^n % k

#include<iostream>using namespace std;long long quickpow(long long m, long long n, long long k){long long ans = 1;while (n){if (n & 1)ans = (ans * m) % k;n = n >> 1;m = (m*m) % k;}return ans;}int main(){long long m, n, k;cin >> m >> n >> k;cout<<quickpow(m,n,k)<<endl;}