CodeForces - 492B Vanya and Lanterns (简单模拟)

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.
Examples
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.
问题链接： http://codeforces.com/problemset/problem/492/B
问题简述： 有一条长len的路，n个路灯，求每个路灯的照明半径最小是多少能照亮整条街
问题分析： 将路灯位置排序，找出各个路灯之间最短距离即可，（记得判断0跟len这两个点有无路灯的情况），输出用print("%.10lf"输出10位小数)
AC通过的C++语言程序如下：

#include <map>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <math.h>
#include <climits>
#include <queue>
#include <iomanip>
#include <vector>
#define ll long long
#include <stdio.h>
using namespace std;

double a[10000];
double b[10000];

int main()
{
    ios::sync_with_stdio(false);
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++)
        cin>>a[i];
    sort(a,a+n);
    //for(int i=0;i<n;i++)
    //    cout<<a[i]<<" ";
    //cout<<endl;
    for(int i=0;i-1<n;i++)
    {
        int j=i+1;
        b[i]=(a[j]-a[i])/2;
    }
    if(a[0]!=0)
        b[n]=a[0];
    else
        b[n]=a[0]/2;
    if(a[n-1]!=m)
        b[n+1]=m-a[n-1];
    else b[n+1]=(m-a[n-1])/2;
    double ans=0;
    for(int i=0;i<=n+1;i++)
        ans=max(ans,b[i]);
    printf("%.10lf",ans);
    return 0;
}