poj 2109 Power of Cryptography

最新推荐文章于 2021-07-18 10:45:37 发布

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本文介绍了一种高效计算整数根的方法，该方法在密码学领域中用于处理涉及大质数和数字幂的问题。文章提供了一个具体的编程实现案例，使用C++语言通过pow函数计算给定数值p的n次方根。

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 25311		Accepted: 12679

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

没想到double的范围这么大：

　类型                        比特数       有效数字                数值范围
　　float 类型                 32            6-7                -3.4*10(-38)～3.4*10(38)
　　double类型              64            15-16            -1.7*10(-308)～1.7*10(308)
　　long double类型     128           18-19           -1.2*10(-4932)～1.2*10(4932)

代码如下：（注意G++只认%f）

#include<stdio.h>
#include<math.h>
int main()
{
	double a,n;
	double sum;
	while(~scanf("%lf%lf",&n,&a))
	{
		sum=pow(a,1.0/n);
		printf("%.0f\n",sum);
	}
}