POJ 2109 Power of Cryptography

1<=k<=10^{9， 1<=n<= 200，所以输入的p不会有那么大的范围，所以直接上。}

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

int main()
{
    double k,p,n;
    while(~scanf("%lf%lf",&n,&p))
    {
        k=pow(p,1/n);
        printf("%.0lf
",k);
    }
    return 0;
}