ACM程序设计 书中题目T

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

本题要输入B和N,找出整数A使A的N次方最接近B;

开始时我是用pow(B,1.0/N)，然后得出两个最接近的数，判断谁的N次方更接近B。代码如下：

#include <bits/stdc++.h>
using namespace std;
int main()
{
double a,b,sum;
int p[1];
while(cin>>a>>b)
{
if(a==0&&b==0)
break;
sum=pow(a,1.0/b);
p[0]=sum+1;p[1]=sum;
if(fabs(pow(p[0],b)-a)<fabs(pow(p[1],b)-a))
sum=p[0];
else
sum=p[1];
cout<<sum<<endl;
}
}

但测试正确提交错误，且找不出错误，于是换思路，从0开始扫描，提交正确，代码如下：

#include <bits/stdc++.h>
using namespace std;
int main()
{
int b,n,x,y,i;
while(cin>>b>>n)
{
if(b==0&&n==0)
break;
for(i=0;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;
}
}
}
}