POJ2109——Power of Cryptography

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Power of Cryptography

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

题目大意：输入k,p，输出n，使kⁿ==p

结题思路：

　　　　看Discuss说用double能水过 于是。。。。

　　　　PS：据说要用到二分法+高精度 等以后有空再研究一下吧

Code：

1 #include<cstdio>
2 #include<cmath>
3 int main()
4 {
5     double n,m;
6     while(scanf("%lf%lf",&n,&m)!=EOF)
7         printf("%.0lf\n",pow(m,1/n));
8     return 0;
9 }

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POJ2109——Power of Cryptography

标签：des   c   style   class   blog   code   

原文地址：http://www.cnblogs.com/Enumz/p/3764805.html