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 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.
分析：一个长l的街道，在一些指定点上会有一些灯笼，问灯笼最小照亮半径为多少可以照亮整条街。
处理下，相邻元素最大值，注意一下0 or L处没有灯笼的情况。
#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <limits>#include <map>using namespace std;#define MIN(a, b) ((a) (b) ? (a) : (b))#define F(i, n) for(int (i)=0;(i)=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)::max();const double eps = 1e-10;const int INF = 0x7FFFFFFF;int a[1005];int main(){ int n,l; while(cin>>n>>l) { for(int i=0;i<n cin>>a[i]; } sort(a,a+n); int ma = -1; for(int i=1;i<n int tmp="a[i]" a ma="max(tmp,ma);" x="2*(a[0]" y="2*(l" double ans="(double)" printf> <br> <br> <p></p> <p> <br> </p> <br> <p></p> </n></n></map></limits></set></cmath></algorithm></cassert></stack></queue></vector></cstring></cstdlib></cstdio></iostream>
