Power of Cryptography poj

Power of Cryptography

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 25320 Accepted: 12686

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

2 163 277 4357186184021382204544

Sample Output

431234

题意：

给两个数n，p，就是求一下n的多少次方等于p

。。。。。。。我写了个很水的代码就不解释了 直接看吧

ac代码

#include<stdio.h>#include<math.h>#include<iostream>using namespace std;int main(){double n,p;while(scanf("%lf%lf",&n,&p)!=EOF){cout<<(double)pow(p,1/n)<<endl;}return 0;}