Root of the Problem

sysu_strange

于 2015-08-26 21:42:26 发布

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分类专栏： ACM算法文章标签： sicily c++ 新手

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本博客介绍了一个算法，用于找到给定整数B的N次方根的近似值，范围在1到1,000,000之间，N从1到9。通过使用pow函数和条件判断，实现对输入值的处理。

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Description

Given positive integers B and N, find an integer A such that A^N is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that A^N may be less than, equal to, or greater than B.

Input

The input consists of one or more pairs of values for B and N. Each pair appears on a single line, delimited by a single space. A line specifying the value zero for both B and N marks the end of the input. The value of B will be in the range 1 to 1,000,000 (inclusive), and the value of N will be in the range 1 to 9 (inclusive).

Output

For each pair B and N in the input, output A as defined above on a line by itself.

// Root_of_the_Problem.cpp

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

int main (){
    int num , expet = 0 , lower = 0 , upper = 0 ;
    double p = 0 ;
    while (cin >> num >> p && p != 0 ){
        expet = pow( num , 1/p ) ;
        lower = num - pow( expet , p ) ;
        upper = pow( expet + 1 , p ) - num ;
        if( lower >= upper ) cout << expet + 1  << endl ;
        else cout << expet  << endl ;
    }
    return 0 ;
}