#UVa：113－Power of Cryptography

善用pow()函式來解題。

C++(0.012)

/*******************************************************/
/* UVa 113 Power of Cryptography                       */
/* Author: Maplewing [at] knightzone.studio            */
/* Version: 2012/03/14                                 */
/*******************************************************/
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;

int main(){
  double n, p, k;
  while( scanf( "%lf%lf", &n, &p ) != EOF )
    printf( "%.0lf\n", pow( p, 1.0/n ) );

  return 0;
}