UVA 113 Power of Cryptography (数学)

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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 integerk (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

double能表示的范围是-1.7e308 ~ 1.7e308，精度至少为15位，而输出结果在int的范围内，即10位，所以可以输出可以用double

至于计算过程中为什么能用double而不会影响到结果，我也暂时没搞懂，因为double的精度只有15位，最多也才有16位，但是p的范围是10^101，输入过程中用的也不是科学计数法，15位后的值肯定被抹掉了，结果却是对的。

正确的分析应该在这：http://blog.csdn.net/synapse7/article/details/11672691，用到了误差分析，得出的结果是在这一题里失去的精度不会影响答案，等我数学补上来之后来研究(吐槽一下网上的好多人，风轻云淡的就发上来了，真的懂了么？自己不扎实不要紧，关键在于误导了新手)

 1 #include<stdio.h>
 2 #include<math.h>
 3 
 4 int    main(void)
 5 {
 6     double    n,p;
 7 
 8     while(scanf("%lf%lf",&n,&p) != EOF)
 9         printf("%.lf\n",pow(p,1 / n));
10 
11     return    0;
12 }

UVA 113 Power of Cryptography (数学)

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原文地址：http://www.cnblogs.com/xz816111/p/4076893.html