Power of Cryptography

cryptography（密码学）

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    double x,y;//calculate y^(1/x)
    int k;
    while(scanf("%lf%lf",&x,&y)!=EOF)//double型输入用 %lf 
    {
        //pow函数的参数原型都是double型的

        //实数a^b四舍五入取整的方法:
        //double a,b; int num=(int)(pow(a,b)+0.5); 
        k=(int)(pow(y,1.0/x)+0.5);//四舍五入的功效! 
        printf("%d
",k);
    }
    return 0;
}

Reference:

https://cn.vjudge.net/contest/319360#problem/A