POJ 2109-Power of Cryptography(double乘方)-蒲公英云

POJ 2109-Power of Cryptography(double乘方)

亦凉 2022-08-21 10:58 117阅读 0赞

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 22464		Accepted: 11327

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<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

2 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234

Source

México and Central America 2004

题目意思：

就是给出n和p，求一个k，使得 k n = p。

解题思路：

若k^n=p，则p^(1/n)=k。

本来以为要用高精度+二分优化，结果试了一下double一下子就水过去了……

/*
* Copyright (c) 2016, 烟台大学计算机与控制工程学院
* All rights reserved.
* 文件名称：np.cpp
* 作    者：单昕昕
* 完成日期：2016年4月28日
* 版 本 号：v1.0
*/
#include <iostream>
#include <malloc.h>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
   double n,p;
   while(cin>>n>>p)
   {
       cout<<pow(p,1.0/n)<<endl;
   }
    return 0;
}