快速冪

思路:將k看做二進制數,每次與1,若獲得結果爲1,就相乘.

#include<iostream>
using namespace std;
int main()
{
    int n = 2, k = 8;
    int res = 1;
    cout<<n<<"的"<<k<<"次方等於:";
    while(k)
    {
        if(k & 1)
        {
            res *= n;
        }
        n *= n;
        k >>= 1;
    }
    cout<<res<<endl;
    return 0;
}