Power of Cryptography

最新推荐文章于 2020-01-31 13:30:25 发布

jerans

最新推荐文章于 2020-01-31 13:30:25 发布

阅读量498

点赞数

分类专栏：思维

本文链接：https://blog.youkuaiyun.com/jerans/article/details/52781287

版权

思维专栏收录该内容

55 篇文章

订阅专栏

本文介绍了一种通过双精度浮点数计算特定形式整数根的有效方法，利用数学库中的幂函数并结合四舍五入技巧，实现了在限定条件下的整数n次根求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

C - Power of Cryptography

Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%lld & %llu

Submit Status Practice POJ 2109

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

double最大可到1e308，注：double占8个字节，它的范围大，但精度小，但因为是指数，double精度小不会影响，故可以用double

四舍五入方法：（int）（x+0.5）或（round）x；

#include<stdio.h>
#include<math.h>
int main()
{
	double n,k,x;
	while(scanf("%lf %lf",&n,&k)!=EOF)
	{
		x=pow(k,1/n);
		x+=0.5; 
		printf("%d\n",(int )x);
	}
	return 0;
}