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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 01
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