POJ2109 Power of Cryptography

http://poj.org/problem?id=2109

题意：输入n、p，求满足k^n=p的k。

这题武大怎么归成贪心了。。。。。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<vector>#include<algorithm>#include<map>#include<cmath>using namespace std;int main(){    freopen("a","r",stdin);    double n,p;    while (cin>>n>>p)    {        double ans=pow(p,1/n);        cout<<(int)(ans+0.5)<<endl;    }    return 0;}