您的位置:首页 > 其它

poj 2109 Power of Cryptography

2013-08-19 14:28 417 查看
点击打开链接

Power of Cryptography

Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 16388 Accepted: 8285
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 nth. 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


给你两个数n和p,那么让你求p的根号n的结果,即k^n = p,让我们求出K的值,我是看了网上的做法才知道原来还可以这样
#include<stdio.h>
#include<math.h>
int main()
{
double a, b;
while(scanf("%lf %lf", &a, &b) != EOF)
{
printf("%g\n", pow(b, 1.0/a));
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: