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最新推荐文章于 2025-03-30 19:45:32 发布

原创最新推荐文章于 2025-03-30 19:45:32 发布 · 967 阅读

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#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int pow_mod(int a, int n, int m) { //时间复杂度：O(logn)
    if(n == 1) { return a%m; }
    int x = pow_mod(a, n/2, m);
    long long ans = (long long)x * x % m;
    if(n%2 == 1) {
        ans = ans * a %m;
    }
    return (int)ans;
}

int mod(int a, int n, int m) {
    int ans = 1, i;
    for(i = 1; i <= n; i++) {
        ans = ans*a%m;
    }
    return ans;
}

long long quick_mod(long long a, long long b, long long m) { //时间复杂度：O(logn)

    long long ans = 1;
    while(b) {
        if(b&1) {
            ans = (ans*a)%m;
            b--; //我认为可以省略。
        }
        b /= 2;
        a = a*a%m;
    }
    return ans;
}

int main()
{
    int a, n, m;
    while(scanf("%d%d%d", &a, &n,&m) == 3) {
        int res1 = pow_mod(a, n, m);
        int res2 = mod(a, n, m);
        int res3 = quick_mod((long long)a, (long long)n, (long long)m);
        cout << "res1 = " << res1 << endl;
        cout << "res2 = " << res2 << endl;
        cout << "res3 = " << res3 << endl;
    }
    return 0;
}