分治之快速幂模版

/*
快速幂模版
求x^t%mod
时间复杂度：O(logn)
注：递归可能爆栈
原理：x^t = (x^(t/2))^2 + x^(n%2);
*/
#include <iostream>
using namespace std;
long long quick_pow(long long x, long long t, long long mod)
{
    if(t == 0) return 1;
    long long temp = quick_pow(x, t / 2, mod);
    temp = temp * temp % mod;
    if(t % 2 == 1) temp = temp * x % mod;
    return temp;
}
long long quick_pow2(long long x, long long t, long long mod)
{
    long long ret = 1;
    while(t)
    {
        if(t % 2)ret = ret * x % mod;
        x = x * x % mod;
        t >>= 1;
    }
    return ret;
}
int main()
{
    long long x, t, mod;
    cin >> x >> t >> mod;
    cout << quick_pow(x, t, mod) << endl;
    cout << quick_pow2(x, t, mod) << endl;
    return 0;
}