Miaomiao's Geometry

　　HDU 4932  Bestcoder

Problem Description

There are N point on X-axis . Miaomiao would like to cover them ALL by using segments with same length.

There are 2
limits:

1.A point is convered if there is a segments T , the point is the
left end or the right end of T.
2.The length of the intersection of any two
segments equals zero.

For example , point 2 is convered by [2 , 4] and
not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3
, 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of
intersection doesn't equals zero), [1 , 3] and [3 , 4] are not(not the same
length).

Miaomiao wants to maximum the length of segements , please tell
her the maximum length of segments.

For your information , the point
can't coincidently at the same position.

Input

There are several test cases.
There is a number T (
T <= 50 ) on the first line which shows the number of test cases.
For each
test cases , there is a number N ( 3 <= N <= 50 ) on the first line.
On
the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the
position of each point.

Output

For each test cases , output a real number shows the answser. Please output three digit after the decimal point.

Sample Input

3

3

1 2 3

3

1 2 4

4

1 9 100 10

Sample Output

1.000

2.000

8.000

Hint

For the first sample , a legal answer is [1,2] [2,3] so the length is 1. For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2. For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.

注意本题的答案可能是某个相邻两点的距离差的一半。会有很多人考虑 不全面，所以适合作为cf和bf题目。

初次通过部分的数据的答案不会出现距离差的 一半，不然就不好hack了。

付一组数据

6

1 5 15 18 24

答案应该为5，而不是4

 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 #include<iostream>
 #include<math.h>

 using namespace std;

 int main()
 {
     int cas,i,n,right,left;
     double res,a[55],b[120];
     cin>>cas;
     while(cas--)
     {
         cin>>n;
         int j=0;
         for(i=0;i<n;i++)
         {
             cin>>a[i];
         }
         sort(a,a+n);
         for(i=1;i<n;i++)
         {
             b[j++]=a[i]-a[i-1];
             b[j++]=(a[i]-a[i-1]) /2 ;
         }
         sort(b,b+j);
         int flag=0;
         j=j-1;
         res=(double)b[j];

         while(1)
         {
             right =0; left=0;
             flag=0;
             for(i=1;i<n;i++)
             {
                 if(i==n-1) continue;
                 if(a[i]-res<a[i-1] && a[i]+res>a[i+1])
                 {
                     flag=1;
                     break;
                 }
                 if(a[i]-res>=a[i-1])
                 {
                     if(right==1)
                     {
                         if(a[i]-a[i-1]==res) {left=1; right=0; }
                         else if(a[i]-a[i-1]>=2*res) { left=1; right=0; }
                         else if(a[i]+res<=a[i+1]) { left=0; right=1; }
                         else flag=1;
                     }
                     else { left=1; right=0; }
                 }
                 else if(a[i]+res<=a[i+1]) {
                     right=1;
                     left=0;
                 }

             }
             if(flag==1) {
                 j--;
                 res=b[j];
             }
             else
             {
                 printf("%.3lf\n",res);
                 break;
             }
         }
     }
     return 0;
 }