Poj 2109 / OpenJudge 2109 Power of Cryptography

1.Link：

http://poj.org/problem?id=2109

http://bailian.openjudge.cn/practice/2109/

2.Content：

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 18872		Accepted: 9520

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

México and Central America 2004

3.Method:

推导过程：

K^N = P

logk P = N

log(P) / log(k) = N (换底公式）

log(K) = 1/N * log(P)

K = e ^ ( 1/N * log(P)) = ( e ^ log(P)) ^ (1/N) = P ^ (1/N) = pow(P,1/N)

因此采用double读入，会损失部分精度，但是这题貌似精度没有那么高，所以可行

4.Code:

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 
 5 using namespace std;
 6 
 7 int main()
 8 {
 9     //freopen("D://input.txt","r",stdin);
10     
11     double p,n;
12     
13     while(cin >> n >> p)
14     {
15         cout << pow(p,1.0/n) << endl;
16     }
17     
18     return 0;
19 }

5.Reference:

Poj 2109 / OpenJudge 2109 Power of Cryptography

原文：http://www.cnblogs.com/mobileliker/p/3937836.html