Power of Cryptography poj 2109 精度问题

Power of Cryptography

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 16480 Accepted: 8342

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 n^th. 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

2 163 277 4357186184021382204544

Sample Output

431234

Source

México and Central America 2004

 

 

我在poj说这个题，说它是个贪心题，我就做了，被赤裸裸的坑了，唉，不带这样，这个题目的难度在于对数的精度的理解，我现在才知道原来double 是如此的强大啊!!!

 

 

类型          长度 (bit)           有效数字          绝对值范围
float             32                      6~7                  10^(-37) ~ 10^38
double          64                     15~16               10^(-307) ~10^308
long double   128                   18~19                10^(-4931) ~ 10 ^ 4932

 

 

第一次听说还有 long double 这种类型啊！！！

 

 

#include<iostream>#include<cstdio>#include<cmath>using namespace std;int main(){    double n,p;    while(cin>>n>>p)    {        //cout<<pow(p,1/n)<<endl;        printf("%.f\n",pow(p,1/n));    }    return 0;}