CF_492B Vanya and Lanterns

B. Vanya and Lanterns

time limit per test

1 second

memory limit per test

256 megabytes

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 1515 5 3 7 9 14 0

output

2.5000000000

input

2 52 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 street will be lit.

题意：

在一条路上放置路灯，求最小可能的路灯照射半径使整条路都等够被照亮。

题解：

很简单的一道题 ，数据量不大，简单枚举就可以A掉。从头至尾找两个路灯之间的最大的距离，是直径的长度，需要除以2，注意不要忘了两头。

代码实现：

#include <algorithm>#include <stdio.h>#include <stdlib.h>using namespace std;int a[1000];int main(){    int n,length;    double dis=-1;    scanf("%d%d",&n,&length);    for(int i=0;i<n;i++)        scanf("%d",&a[i]);    sort(a,a+n);    for(int i=1;i<n;i++)    {        dis=max(dis,(double)a[i]-a[i-1]);    }    dis/=2;    dis=max(dis,(double)a[0]-0);    dis=max(dis,(double)length-a[n-1]);    printf("%.10lf",dis);    return 0;}