B. VANYA AND LANTERNS

http://www.yyycode.cn/index.php/2020/05/17/b-vanya-and-lanterns/

anya 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 a i. 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 a i (0 ≤ a i ≤ 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.ExamplesinputCopy

7 15
15 5 3 7 9 14 0

outputCopy

2.5000000000

inputCopy

2 5
2 5

outputCopy

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：一条长度为 l 的街道，在这条街道上放置了n个相同的灯，设从一端位置为0，每个灯的位置在ai处，问灯的最小照射半径为多少时，才能满足整条街道都能被灯光照到。

思路：贪心，两个相邻的并且距离最远的路灯之间的距离的一半和最左端路灯到0的距离和l-最右端路灯的距离，三者取最大值。

要照亮的话只要距离最大的都能找亮，那么整条街都能照亮

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e5;
typedef long long LL;
LL a[maxn];
LL cnt[maxn];
bool cmp(LL c,LL d)
{
	return c>d;
}
int main(void)
{
	LL n,l;cin>>n>>l;
	for(LL i=1;i<=n;i++)
		cin>>a[i];
	sort(a+1,a+1+n);
	
	double dis1=a[1]-0;double dis2=l-a[n];
    
	for(LL i=1;i<n;i++)
    	cnt[i]=a[i+1]-a[i];
    sort(cnt+1,cnt+n,cmp);
    
	dis1=max(dis1,dis2);
	
	double dis=double(cnt[1]/2.0);
	
	dis=max(dis,dis1);
	
	printf("%.10f\n",dis);
return 0;
}

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