Codeforces Round #280 (Div. 2) B_html/css_WEB-ITnose

题目：

B. Vanya and Lanterns

time limit per test

memory limit per test

input

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.

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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.

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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 MAX(a, b) ((a) > (b) ? (a) : (b))#define F(i, n) for(int (i)=0;(i)<(n);++(i))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MEM0(addr) memset((addr), 0, sizeof((addr)))#define PI 3.1415926535897932384626433832795#define HALF_PI 1.5707963267948966192313216916398#define MAXN 100000#define MAXM 10000#define MOD 1000000007typedef long long LL;const double maxdouble = numeric_limits<double>::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;i++)        {            cin>>a[i];        }        sort(a,a+n);        int ma = -1;        for(int i=1;i<n;i++)        {            int tmp = a[i] - a[i-1];            ma = max(tmp,ma);        }        int x = 2*(a[0] - 0);        int y = 2*(l - a[n-1]);        ma = max(ma,x);        ma = max(ma,y);        double ans = (double) ma/2;        printf("%.10lf\n",ans);    }}