UVA 113 Power of Cryptography

第一道ACM的数学题。。。做了好久的字符串排序搜索高精度，终于到了数学了 ，

起初以为要用高精度来做。。后来看看发现double就够了 double的范围-1.7*10^(-308) ~ 1.7*10^308 没超过p的范围 输出时注意用%.0lf消去小数尾数0

然后开n次方就是root的1/n次方 UVA目前瘫痪ing POJ AC  题目及AC代码如下：

Background

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers modulo functions of 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 of only theoretical interest.

This problem involves the efficient computation of integer roots of numbers.

The Problem

Given an integer and an integer you are to write a program that determines , the positive root of p. In this problem, given such integers n and p, p will always be of the form for an integer k (this integer is what your program must find).

The 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 , and there exists an integer k, such that .

The Output

For each integer pair n and p the value should be printed, i.e., the number k such that .

Sample Input

2
16
3
27
7
4357186184021382204544

Sample Output

4
3
1234

 1 #include<math.h>
 2 #include<cstdio>
 3 #include<string>
 4 using namespace std;
 5 int main(void)
 6 {
 7     double root, power;
 8     while (scanf("%lf%lf", &power, &root) == 2)
 9     {
10         printf("%.0lf\n", pow(root, 1 / power));                //难得这么精简的代码 用double就可以了 开n次根就相当于根的 1/n次方 输出注意用.0lf去除小数位
11     }
12 }

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UVA 113 Power of Cryptography

原文：http://www.cnblogs.com/VOID-133/p/3573045.html