（转）poj2109-Power of Cryptography

Power of Cryptography

 

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

题目大意：输入两个数，n，p。求一个整数k，使得k满足kⁿ=p。

大牛代码，膜拜！

 

float：2^23 = 8388608，一共七位，这意味着最多能有7位有效数字，但绝对能保证的为6位，也即float的精度为6~7位有效数字；

double：2^52 = 4503599627370496，一共16位，同理，double的精度为15~16位。

首先，题目中的数据强度并不弱，这一点确实如题目中所说：“For all such pairs 1<=n<= 200, 1<=p<10101，所以，double型是不能精确地表示出所给数据，但是却能表示出一个近似值。

当向double型变量中存入

4357186184021382204544

然后再输出，得到的是

4357186184021382000000

后六位的值变为了0，这一点和int型变量是有很大区别的。也就是说当存入double型变量的值超出了它的精度表示范围时，将低位的数据截断。

在本题中，如果测试数据为：

7     4357186184021382204544

实际上所处理数据是：

7     4357186184021382000000

拿4357186184021382000000开7次方的结果自然就是1234。

为什么不是1233或者1235呢？

12337=4332529576639313702577

12347=4357186184021382204544

12357=4381962969567270546875

可以看出在double型所能表示的精度范围内，它们三个值已经不同了。

#include<cstdio>
#include<cmath>

int main()
{
    int n;
    double p;
    while(scanf("%d%lf",&n,&p) != EOF)
    {
        printf("%.0lf
",pow(p,1.0/n));
    }
    return 0;
}