課程練習一Problem M

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

 

題意： 輸入n，p; 求k 並且滿足k的n次方等於p

思路： 直接定義double型用pow()函數，就ok了

個人感覺這題是想考 大數乘法或者快速乘法的，只是這道題可以用函數 水過。

代碼：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
double n,p;
while(cin>>n>>p)
{ 
   double m=1.0/n; 
        cout<<pow(p,m)<<endl;  
    }  
system("pause");
return 0;
}