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最新推荐文章于 2020-06-12 21:55:45 发布

原创最新推荐文章于 2020-06-12 21:55:45 发布 · 303 阅读

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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.

Example Input:	Example Output:
4 3 5 3 27 3 750 5 1000 5 2000 5 3000 5 1000000 5 0 0	1 2 3 4 4 4 5 16

题意：

就是找到最接近b的a的n次方，就是输出a

分析：

就是找一个数的n次方小于等于n这个数加一大于n,做差，找最接近的数，输出

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long int b,n,i,x,y;
    while(cin>>b>>n&&n!=0&&b!=0)
    {
        for(i=1;i<=b;i++)
        {
            x=pow(i,n)-b;
            y=pow(i+1,n)-b;
            if(x<=0&&y>0)
            {
                if(fabs(x)<fabs(y))
                    cout<<i<<endl;
                else
                    cout<<i+1<<endl;
                break;
            }
        }
    }
    return 0;
}

感受：

这个题不知道怎么回事，老是不过，一直就是segmentation fault ,我感觉好像不能用数组，好像是开的太大了，反正就是交了10次才过