HDU 4932 Miaomiao's Geometry 暴力

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Miaomiao's Geometry

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 438 Accepted Submission(s): 107


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
HintFor 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. 

Source BestCoder Round #4
有n個點在x坐標上，要求用一些線段將這些點全部覆蓋，要求這些線段不能重疊，且必須等長，x坐標上的點必須在線段的端點，求線段的最長長度。 一開始從第二個點到倒數第二個點找左右距離最大的值存到數組裡面，最後輸出此數組最小的即可，也就是求個最大最小。但是如果數據是這樣的：2 7 8 14 15的話，果斷WA，因為8和14共用一段區間，且此區間要大於2-7的5,。 正確解法應該是從大到小枚舉將這些數的差值以及差值的一半，如果符合要求，則就是答案。

//15MS	268K
#include
#include
using namespace std;
double s[57],array[107];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,k=0;
        scanf("%d",&n);
        for(int i=0;i=0;i--)
        {
            bool flag=1;
            double last=s[0],a=array[i];
            for(int j=1;j