Codeforces 492B B. Vanya and *s

Codeforces  492B   B. Vanya and *s

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86640#problem/B

题目：

Description

Vanya walks late at night along a straight street of length l, lit by n *s. 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 * is at the point a_i. The * lights all points of the street that are at the distance of at most d from it, where d is some positive number, common for all *s.

Vanya wonders: what is the minimum light radius d should the *s have to light the whole street?

Input

The first line contains two integers n, l (1 ≤ n ≤ 1000, 1 ≤ l ≤ 10⁹) — the number of *s and the length of the street respectively.

The next line contains n integers a_i (0 ≤ a_i ≤ l). Multiple *s can be located at the same point. The *s may be located at the ends of the street.

Output

Print the minimum light radius d, needed to light the whole street. The answer will be considered correct if its absolute or relative error doesn't exceed 10^- 9.

Sample Input

 

Input

7 15 15 5 3 7 9 14 0

Output

2.5000000000

Input

2 5 2 5

Output

2.0000000000

Sample Output

1 2

                

Hint

Consider the second sample. At d = 2 the first * will light the segment [0, 4] of the street, and the second * will light segment [3, 5]. Thus, the whole street will be lit.

题意：

N盏灯照亮长度为L的街道，使整条街全部被照亮。求灯的最少照亮半径。

分析：

1.将灯的位置排序，先求出相邻两盏灯之间的最大距离，最大距离是每盏灯所照的最大半径的两倍

2.再求第一盏灯到街首的距离和街尾到最后一盏灯的距离

3.比较三个距离的长度，最大值即为所求

代码：

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 using namespace std;
 ;

 int a[maxn];

 int max(int x,int y)
 {
     return x>y?x:y;
 }

 int main()
 {
     int n,l;
     scanf("%d%d",&n,&l);
     ;i<=n;i++)
     {
         scanf("%d",&a[i]);
     }
     sort(a+,a+n+);   //将灯的位置从小到大排序
     a[]=;  //街首
     a[n+]=l;  //街尾
     ;
     ]-a[],xx=a[n+]-a[n];  //首尾两盏灯到街首和街尾的距离
     ;j<n;j++)   //两盏灯之间的最大距离
     {
         a[j]=a[j+]-a[j];
         if(a[j]>m)
             m=a[j];
     }
     m=max(*b,max(m,*xx));   //求灯照射的最大直径
     double ans=m/2.0;
     printf("%.10lf\n",ans);
     ;
 }

昨天写题目的时候没有写max函数，提交后WA了，所以这次看完题目知道要比较大小，就先写了一个max函数。