1012

题目编号：1012

简单题意：

Problem 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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    AC代码：#include<iostream>
#include<cmath>
#include<cstdio>
#include<fstream>
using namespace std;
int main()
{
    //freopen("test.txt","r",stdin);
      long begin=0,end=1000000000;
      double n,p;
      while(cin>>n>>p)
     {
         begin=0;
         end=1000000000;
         while(begin<end)
          {
                        long mid=(long)((begin+end)/2);
                       if(pow(mid,n)<p)begin=mid;
                        else if(pow(mid,n)>p)end=mid;
                        else if(pow(mid,n)==p)
                        {
                                cout<<mid<<endl;
                             break;
                   }
                }
        }
        return 0;
}
   
   
   
   
   
   
   
   

    
    
    
    解题思路：已知n和p，求k使得
    
    
    
    k^n=p,上午上课的时候学过的二分枚举这里正好用到。