【快速幂】O(logn)时间复杂度

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

int qpow(int base, int n){
    int ans = 1;
    while(n){
        if(n&1) ans*=base;
        base = base * base;
        n/=2;
    }
    return ans;
}

int main(){
    cout << qpow(2,1) << endl;
    cout << qpow(2,2) << endl;
    cout << qpow(2,10) << endl;
    return 0;
}