2015 HUAS Summer Contest#3~B

最新推荐文章于 2015-08-21 19:57:00 发布

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最新推荐文章于 2015-08-21 19:57:00 发布

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原文链接：http://www.cnblogs.com/chenchunhui/p/4713197.html

本文探讨了一个关于路灯照亮整条街道的问题，即如何确定最小的照明半径来确保整个街道都能被照亮。文章详细介绍了求解此问题的方法，包括排序路灯位置、计算最大距离并取一半作为最小半径，以及处理街道两端没有路灯的情况。同时，文章提供了代码实现，并解释了输出格式要求。

Description

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 Input

Input

7 15
15 5 3 7 9 14 0

Output

2.5000000000

Input

2 5
2 5

Output

2.0000000000

Hint

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.

解题思路：题目的大概意思就是求最小能把整条街照亮的半径。但是需要注意的是有两种情况：第一种是街道的两端有路灯，那么这种情况下就只需考虑每两个路灯之间的距离，这样就只需要先将路灯排好序，然后求出最大的距离再除以2即可；第二种情况稍微复杂点，就是当街道两端没有路灯时，也要先运用第一种方法将最大距离的一般求出来，但是需要注意的是，这里还要将它与街道最左及最有右的路灯到街道两端的距离进行比较，取出最大值。最后还要注意输出格式，要记得保留10位小数。

程序代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
double d,mm,t;
long long a[1005],l;
int main()
{
    int n,i,p=0,q=0;
    scanf("%d%lld",&n,&l);
    for(i=0;i<n;i++)
    scanf("%lld",&a[i]);
    sort(a,a+n);
    for(i=0;i<n-1;i++)
    {
        t=a[i+1]-a[i];
        if(t>mm)  mm=t;

    }
    d=mm/2;
    p=a[0];
    q=l-a[n-1];
    if(p>=q&&p>d)  d=p;
    else if(q>d)  d=q;
    printf("%.10lf\n",d);
    return 0;
}