POJ-2109-Power of Cryptography

最新推荐文章于 2021-07-18 10:45:37 发布

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最新推荐文章于 2021-07-18 10:45:37 发布

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本文介绍了一种通过编程实现高效计算整数根的方法，利用C语言和数学库函数解决特定形式的大整数开根问题，并给出了解题示例。

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 22947		Accepted: 11578

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

当存入double型变量的值超出了它的精度表示范围时，将低位的数据截断。

题目中,测试数据为：7 4357186184021382204544

实际上所处理数据是：7 4357186184021382000000

4357186184021382000000开7次方的结果自然就是1234。

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