Power of Cryptography（double可以表示到10的-307次方到10的308次方）

╰+哭是因爲堅強的太久メ 2022-08-02 05:59 115阅读 0赞

Power of Cryptography

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;30000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;20415</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;10314</td> 
  </tr> 
 </tbody> 
</table>

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<10 101 and there exists an integer k, 1<=k<=10 9 such that k n = 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][M_xico and Central America 2004]

#include<cstdio>
    #include<cmath>
    #include<iostream>
    using namespace std;
    int main()
    {
        double a,b;
        while(scanf("%lf%lf",&a,&b)!=EOF)
        {
            printf("%.0lf\n",pow(b,1.0/a));
           // cout<<pow(b,1.0/a)<<endl;
        }
    }

G++WA C++AC怎么解释？

[M_xico and Central America 2004]: http://poj.org/searchproblem?field=source&key=M%C3%A9xico+and+Central+America+2004