POJ2109,Power of Cryptography,神题，火钳刘明

最新推荐文章于 2017-08-20 08:21:40 发布

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本文介绍了一种通过编程实现的高效计算整数根的方法。针对给定的大整数p和正整数n，找到使得k^n = p成立的整数k。文章提供了具体的输入输出示例，并附带了一个简洁的C++代码示例，该代码利用标准库函数pow完成计算。

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Power of Cryptography

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 nth. 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

double能过！double能过！！double能过！！!

10^101是什么东东。。。精度是什么东东。。。

不要在意这些细节。。。

code:

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
    double n,p;
    while(scanf("%lf %lf",&n,&p)==2)
        printf("%.0lf\n",pow(p,1/n));
    return 0;
}