codeforce492 B. 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 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.

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 street will be lit.

題意：

有一條長爲l的街道，把n盞燈放在路上，求使整個街道照明的最小半徑。

思路：

求出兩盞電燈之間的最大距離，除以二就是最小半徑，但是考慮到路兩邊可能沒有燈的情況，所以最後要把這個結果與a[0]與l-a[n-1]的最大值比較，其最大值，就是它的最小半徑。

代碼：

#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];//算出左端點的最大半徑
        ax=(double)l-a[n-1];//算出右端點的最大半徑
        d=max(a0,ax);
        for(int i=0; i<n; i++)
            ans=max(ans,double(a[i+1]-a[i]));
        ans/=2.0;//求出的直徑除以二就是半徑
        ans=max(d,ans);//去取最大值就是結果
        printf("%.10lf\n",ans);

    }
    return 0;
}