Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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. 代码： 为什么要加后面那一段再判一下呢？删掉就错了呢？为什么呢？我也不知道啊，是吧？

1 #include 
         
           2 #include 
          
            3 #include 
           
             4 #include 
            
              5 using namespace std; 6  7 const double inf=1e9+7; 8  9 int n;10 double a[56];11 12 bool C(double x)13 {14     int i;15     double ab=a[1];16     for(i=2;i
             
              =x)19         {20             ab=a[i];21         }22         else23         {24             if(a[i+1]-a[i]
              
               x)32 {33 ab=a[i]+x;34 }35 }36 }37 return true;38 }39 40 int main()41 {42 int T;43 int i,j;44 scanf("%d",&T);45 while(T--)46 {47 scanf("%d",&n);48 for(i=1;i<=n;i++)49 scanf("%lf",&a[i]);50 sort(a+1,a+n+1);51 double lb=0,ub=inf;52 for(i=1;i<=100;i++)53 {54 double mid=(lb+ub)/2.0;55 if(C(mid))56 lb=mid;57 else58 ub=mid;59 }60 for(i=1;i
               
                lb)62 lb=a[i+1]-a[i];63 printf("%.3lf\n",lb);64 }65 return 0;66 }