poj 2109 Power of Cryptography

这个题用double就能过；

View Code

#include<iostream>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<queue>

#include<set>

#include<map>

#include<vector>

using namespace std;



int main(  )

{

    double k,p;

    while( scanf( "%lf %lf",&k,&p )==2 )

    {

        printf( "%.0f\n",pow( p,1.0/k ) );    

    }

    //system( "pause" );

    return 0;

}