ACM: 简单数学推导+挑战平台的精度…

                                                                 Power of Cryptography

Description

Current work in cryptography involves (among other things) largeprime numbers and computing powers of numbers among these primes.Work in this area has resulted in the practical use of results fromnumber theory and other branches of mathematics once considered tobe only of theoretical interest.
This problem involves the efficient computation of integer roots ofnumbers.
Given an integer n>=1 and an integerp>= 1 you have to write a program that determinesthe n th positive root of p. In this problem, given such integers nand 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 witheach integer on a line by itself. For all such pairs1<=n<= 200,1<=p<10¹⁰¹ and thereexists an integer k,1<=k<=10⁹ such thatkⁿ = 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

 

题意: k^n = p . 现在给出n ， p 求 k

 

解题思路:

                 1. 取对数运算: k^n = p <==> ln(k^n) = ln p <==> nlnk = lnp <==> lnk = (lnp)/n

                    <==>  lnk =ln(p^1/n) <==> k = p^(1/n).

                 2. 题目的精度直接pow计算值. long double是12个字节.

 

代码:

#include <cstdio>
#include <iostream>
#include <cmath>
using namespace std;

int main()
{
//   freopen("input.txt","r",stdin);
    double n ,p;
   while(scanf("%lf%lf",&n,&p) != EOF)
    {
      printf("%.0lf\n",pow(p,1.0/n));
    }
    return0;
}