poj2109 Power of Cryptography【坑~泪目】【水过~】【刷题计划】

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 26249		Accepted: 13121

Description

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<10¹⁰¹ and there exists an integer k, 1<=k<=10⁹ such that kⁿ = 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

2 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234

Source

题意：输入n,p找到满足k^n = p的k。

数据范围给的这么大，吓死我了，以前没有写过这种类型的题，一点思路都没有，没忍住找了一下题解，发现double都能过哎。

然后，老师说，我这么水过的不算，自己出数据然后自己写标程，啊！！！！

#include<stdio.h>
#include<math.h>

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