Program B--codeforces 492B

Description

Vanya walks late at night along a straight street of length l, lit by n lanterns. Consider the coordinate system with the beginning of the street corresponding

to the point 0, and its end corresponding to the point l. Then the i-th lantern are at the point a_{I. The lantern lights All points} of the street is at the

Distance of In most D from it, where D was some positive number, common for all lanterns.

Vanya Wonders:what is the minimum light radius D Should the lanterns has to light the whole street?

Input

The first line contains the integers n, l (1≤ n ≤1000, 1≤ l ≤10^{9)-the number of lanterns and the length of the street respectively.}

The next line contains n integers a_{i (0≤ ai ≤ l). Multiple lanterns can is located at the same point. The lanterns is located at the ends of the street.}

Output

Print The minimum light radius D, needed to light the whole street. The answer would be a considered correct if its absolute or relative error doesn ' t exceed ^-9.

Sample Input

Input

7 15
15 5 3 7 9 14 0

Output

2.5000000000

Input

2 5
2 5

Output

2.0000000000




The main idea: Vanya walk in a long street, the street has n lights, this n can be in the random position on L, so the radius of light irradiation are the same, the minimum radius of the lamp to make the whole street has light.

Note: Any two lights must be adjacent.

Analysis: This problem can be solved directly with violence, but pay attention to two points. First, the n lamp includes the left and right two endpoints, so long as the two lamps are known to be the maximum distance (in sum),

Half of it is the minimum radius required;

If only one or two endpoints are included, the distance from the left end to the first light (in C) is compared to the last lamp to the right end (denoted by D),

Take a larger value to C, because C itself is the radius, so if C is twice times greater than or equal to the maximum distance of the two lights, output C, otherwise output sum of half.




The code is as follows:

#include<iostream>#include<cstdio>#include<algorithm>UsingNamespace Std;IntMain(){int n, lA[1005],cPQ, flag, ID;Double sum;While(scanf("%d%d", &n, &l)==2){C=0D=0P=0Q=0, Sum=0.0, flag=0;For(I=0; I<n; I++)scanf("%d", &a[I]);Sort(AA+n);For(I=0; I<n&&a[I]<=l; I++){If(A[0]==0&&a[n-1]==l) flag=1; Q=a[I]-p; P=a[I];If(q>=sum) sum=q;}If(!flag){If(A[0]!=0) c=a[0];If(A[n-1]!=l) d=l-A[n-1]; C=Max(cD);If(2*c>=sum)Printf("%.10LF\ n",(Double); else printf ( "%.10lf\n" ,sum /2);  else printf  ( "%.10lf\n  ",sum/2 return 0;< Span class= "Sh-cbracket" >             

Program B--codeforces 492B