POJ 2109 Power of Cryptography 开根

题意：给定n, p, 求pow(p, 1/n)

 

10955211

NY_lv10

2109

Accepted

196K

0MS

C++

192B

2012-10-25 18:56:50

 

#include <iostream>
#include <math.h>
using namespace std;

int main()
{
    double n, p;
    while (scanf("%lf %lf", &n, &p) !=EOF)
    {
        printf("%0.0lf\n", pow(p, 1/n));
    }
    return 0;
}