POJ 2109 Power of Cryptography

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be only of theoretical interest. 
This problem involves the efficient computation of integer roots of numbers. 
Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the n^th. power, for an integer k (this integer is what your program must find).

The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10¹⁰¹ and there exists an integer k, 1<=k<=10⁹ such that kⁿ = p.

For each integer pair n and p the value k should be printed, i.e., the number k such that k n =p.

2 16
3 27
7 4357186184021382204544

4
3
1234

题意：过滤掉废话，就是要你求n的多少次幂为m

AC代码：

#include<stdio.h>
#include<math.h>
int main()
{
    double n,m;
    while(scanf("%lf %lf",&n,&m)!=EOF)
    {
        printf("%.0lf\n",pow(m,1.0/n));
    }
    return 0;
}