快速幂

原题链接 https://www.acwing.com/problem/content/877/
模版代码

#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

//求a ^ k % p
int qmi(int a, int k, int p) {
    int res = 1;
    while(k) {
        if(k & 1) res = (LL)res * a % p;
        k >>= 1;
        a = (LL)a * a % p;
    }
    
    return res;
}


int main() {
    int n;
    scanf("%d", &n);
    while(n -- ) {
        int a, k, p;
        scanf("%d%d%d", &a, &k, &p);
        printf("%d\n", qmi(a, k, p));
    }
    
    return 0;
}

快速幂

原文：https://www.cnblogs.com/huhu555/p/14613580.html