D - Power of Cryptography

D - Power of Cryptography

题目描述：

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).

Input：

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<10 101 and there exists an integer k, 1<=k<=10 9 such that k n = p.

Output：

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

Sample Input：

Sample Output：

题目大意：

世
界上许多密码门都有一种输入模式，你输入2与16，2需要多少次幂变成16，答案是4，所以输出4.

思路分析：

这道题可以辛苦的用数组一个一个存，但是这道题可以直接用暴力解法，因为double类型的负值取值范围是可以满足题意的。

代码：

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