poj 2109 Power of Cryptography(用double避开高精度)

Power of Cryptography

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16238		Accepted: 8195

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 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234

Source

México and Central America 2004

 

题意：给一对数 n(1<=n<=200)和p(1<=p<=10^101)，求k使k^n=p。

分析：1、很自然的，因为觉得数据很大，会去想高精度。然后加二分猜数。

                 然后不会高精度啊。。

           2、于是想到转换数学运算：指对互化。用double存，但是double 精确位只有6—7。而没有logx Y，只有先转化为以e为底的对数。用lognP=logn/logP。用两次函数，

                  精确度不能满足要求。

           3、换思路：k^n=p，则p^(1/n)=k。且函数可以直接用pow(x,y)去求x^y。 

收获：巩固了一下基础。启发了一下思维。

          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 

代码：

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;

int main()
{
    double n,p;
    while(scanf("%lf%lf",&n,&p)!=EOF)
    {
        printf("%.0lf\n",pow(p,1/n));
    }
    return 0;
}

 

11922275

fukan

2109

Accepted

192K

0MS

C++

221B

2013-08-05 11:12:30