（二分法）求X的N次方

#include<iostream>using namespace std;int Pow(int x, int n){if (n == 1)return x;else if (n > 1){int s;int m = n / 2;s = Pow (x, m);if (n % 2 == 0)//偶次return (s * s);elsereturn (s * s * x);//奇次}}int main(){int x, n;while (cin >> x >> n){cout << Pow(x, n) << endl;}return 0;}