codeforce492 B. Vanya and Lanterns

[align=center]B. Vanya and Lanterns[/align]
[align=center]time limit per test[/align]
[align=center]1 second[/align]
[align=center]memory limit per test[/align]
[align=center]256 megabytes[/align]
input
standard input

output
standard output

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 is at the point
ai. The lantern 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 lanterns.

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

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

The next line contains n integers
ai (0 ≤ ai ≤ l). Multiple lanterns can be
located at the same point. The lanterns 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 test(s)

Input

7 15
15 5 3 7 9 14 0

Output

2.5000000000

Input

2 5
2 5

Output

2.0000000000

Note
Consider the second sample. At d = 2 the first lantern will light the segment
[0, 4] of the street, and the second lantern will light segment
[3, 5]. Thus, the whole s
4000
treet will be lit.

题意：

有一条长为l的街道，把n盏灯放在路上，求使整个街道照明的最小半径。

思路：

求出两盏电灯之间的最大距离，除以二就是最小半径，但是考虑到路两边可能没有灯的情况，所以最后要把这个结果与a[0]与l-a[n-1]的最大值比较，其最大值，就是它的最小半径。

代码：

#include<cstdio>
#include<algorithm>
int const maxn=1000+5;
int a[maxn];
using namespace std;
int main()
{
int  n,l;
double a0,ax,d,ans=0.0;
while(scanf("%d%d",&n,&l)!=EOF)
{
for(int i=0; i<n; i++)
scanf("%d",&a[i]);
sort(a,a+n);
a0=(double)a[0];//算出左端点的最大半径
ax=(double)l-a[n-1];//算出右端点的最大半径
d=max(a0,ax);
for(int i=0; i<n; i++)
ans=max(ans,double(a[i+1]-a[i]));
ans/=2.0;//求出的直径除以二就是半径
ans=max(d,ans);//去取最大值就是结果
printf("%.10lf\n",ans);

}
return 0;
}