acm_problem_1012

题目描述：

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

2 163 277 4357186184021382204544

 

Sample Output

431234

大意：就是给你两个数让你求前一个数的多少次方是后一个数。。

想法：直接用pow这个函数求就可以。。

代码：

#include <iostream>
#include <math.h>
using namespace std;
int main()
{
    double n,p;
    while(cin>>n>>p)
    {
        int k;
        k=pow(p,1.0/n)+0.5;
        cout<<k<<endl;
    }
}

感想：一开始看到那个好多位的数用整型发现求不出来，后来无意中发现double 可以求。。无语了。。