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 ai. 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 ai (0 ≤ ai ≤ 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
7 15
15 5 3 7 9 14 0
2.5000000000
2 5
2 5
2.0000000000 题目大意：Vanya走在一条长为l的街上，这条街上有n盏灯，这n盏可以在l上的随意位置，所以灯照射的半径都相同，求灯的最小半径使得整条街都有光照。 注意：任意两盏灯必须要相邻才行。 分析：这题可以直接用暴力求解，不过要注意两点。首先n盏灯中包括左右两个端点，则只要知道其中两盏灯距离最大（用sum表示）就可以了， 它的一半就是要求的最小半径； 如果只包括其中一个端点或两个端点都不包括，先将左端点到第一盏的灯距离（用c表示）与最后一盏灯到右端点的距离（用d表示）进行比较， 取较大的值赋给c，由于c本身就是半径，所以如果c的2倍大于或等于距离最大的两盏灯，输出c，否则输出sum的一半。 代码如下：
#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
int main()
{
int n,l,a[1005],c,p,q,flag,i,d;
double sum;
while(scanf("%d%d",&n,&l)==2)
{
c=0,d=0,p=0,q=0,sum=0.0,flag=0;
for(i=0;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+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(c,d);
if(2*c>=sum)
printf("%.10lf\n",(double)c);
else
printf("%.10lf\n",sum/2); }
else
printf("%.10lf\n",sum/2); }
return 0;
}