2019/3/31acm周三(三人)/CodeForces - 492B (模拟)

CodeForces - 492B

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.

题目大意：给出你灯的坐标，求等所照射的最小半径，使得能够照到全部;
思路：分两种情况(模拟)：
①：起点有灯柱，则可以计算每个灯柱之间最大的距离/2(double);
②：起点无灯柱，则多加一个从起点到第一个灯柱的距离，再与(其他距离(已除与2))进行对比，最长的则为所求ans;

#define ll long long
#include <stdio.h>
#include<algorithm>
#include<iostream>
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-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);//输出小数点后10位
    return 0;
}