ACM POJ 2109 Power of Cryptography

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

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10904		Accepted: 5626

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

 

 

题目分析：求一个数的n次幂等于p,n、p均给定

所用算法：注意double的精度约为1^-300-1^300,注意神奇的pow函数

 

 

#include<stdio.h>
#include<cmath>
int main()
{
    double n,p;
    while(scanf("%lf%lf",&n,&p)!=EOF)
    {
        printf("%.0f\n",pow(p,1/n));
    } 
    return 0;   
}