POJ：2109-Power of Cryptography（关于double的误差）

Power of Cryptography

Time Limit: 1000MS Memory Limit: 30000K

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

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• 刚看到的时候还是想用大数硬怼，但是明显不可能啊。看了一下网上别人的代码，发现代码很简单。但是关于double的误差问题都没怎么说清楚，想关注double详情的可以看看大神的分析：http://blog.csdn.net/synapse7/article/details/11672691

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#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std;
int main()
{
    double a,b;
    while(cin>>a>>b)
    {
        cout<<pow(b,1.0/a)<<endl;
    }
    return 0;
}