ACM书中的题目 T-20

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Description
Given positive integers B and N, find an integer A such that AN is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that AN 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
题目简述:
输入正整数B　N，求一A，使得A的N次方最接近B
代码：

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

体会：
这个题目我当时不会做，看了别人的做法后稍微理解了一点，但还是要再看几遍

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