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

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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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<=n; i++){            scanf("%d", &a[i]);        }        sort(a+1, a+n+1);        a[0] = 0; a[n+1] = l;        int ma = 0;        int x = a[1] - a[0], xx = a[n+1] - a[n];        for(int i=1; i<n; i++){            a[i] = a[i+1] - a[i];            if(a[i] > ma) ma = a[i];        }        ma = max(2*x, max(ma, 2*xx));        double ans = ma / 2.0;        printf("%.10lf\n", ans);    }    return 0;}

贪心思想真是无处不在呀o(∩∩)o...