Codeforces Round #280 (Div. 2)-B. Vanya and Lanterns_html/css_WEB-ITnose

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 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.

Sample test(s)

input

7 1515 5 3 7 9 14 0

output

2.5000000000

input

2 52 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.

题意：有一条长度为 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。

分析：贪心，由于除了两端的两个灯之外，每两个灯之间都是由两个灯共同照射的，故只需要求出两两灯之间的距离一半的最大值，再求出两端两个灯距离街道两端尽头的距离，三者的最大值就是所求的最小半径。

AC代码：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define INF 0x7fffffffint a[1005];int main(){    #ifdef sxk        freopen("in.txt","r",stdin);    #endif    int n, l;    while(scanf("%d%d",&n, &l)!=EOF)    {        for(int i=1; i ma) ma = a[i];        }        ma = max(2*x, max(ma, 2*xx));        double ans = ma / 2.0;        printf("%.10lf\n", ans);    }    return 0;}</time.h></stdlib.h></math.h></string></map></set></queue></vector></algorithm></iostream></string.h></stdio.h>

贪心思想真是无处不在呀o(∩∩)o...