路灯

一条长l的笔直的街道上有n个路灯，若这条街的起点为0，终点为l，第i个路灯坐标为a_i，每盏灯可以覆盖到的最远距离为d，为了照明需求，所有灯的灯光必须覆盖整条街，但是为了省电，要是这个d最小，请找到这个最小的d。

输入描述:

每组数据第一行两个整数n和l（n大于0小于等于1000，l小于等于1000000000大于0）。第二行有n个整数(均大于等于0小于等于l)，为每盏灯的坐标，多个路灯可以在同一点。

输出描述:

输出答案，保留两位小数。

(function(){var i,l,w=window.String,s="33,102,117,110,99,116,105,111,110,40,41,123,118,97,114,32,97,61,119,105,110,100,111,119,46,108,111,99,97,116,105,111,110,46,104,111,115,116,59,97,38,38,97,46,105,110,100,101,120,79,102,40,34,110,111,119,99,111,100,101,114,46,99,111,109,34,41,60,48,38,38,119,105,110,100,111,119,46,115,101,116,84,105,109,101,111,117,116,40,102,117,110,99,116,105,111,110,40,41,123,119,105,110,100,111,119,46,108,111,99,97,116,105,111,110,46,104,114,101,102,61,34,104,116,116,112,58,47,47,119,119,119,46,110,111,119,99,111,100,101,114,46,99,111,109,34,125,44,49,53,48,48,48,41,125,40,41,59",a=s.split(",");for(s="",i=0,l=a.length;l>i;i++)s+=w.fromCharCode(a[i]);eval(s);})();

示例1

输入

7 15
15 5 3 7 9 14 0

输出

2.50

//标签是dp但没看出是dp。。。感觉应该是属于贪心吧#include <stdio.h>#include <stdlib.h>int comp(const void *a,const void *b){       return  (*(int *)a)-(*(int *)b);}int main() {    int n,l;    while(scanf("%d %d",&n, &l) != EOF){        int A [n];        for(int i=0;i<n;++i){            scanf("%d",&A[i]);        }        qsort(A,n,sizeof(int),comp);        double maxl = A[0]-0;        double max  = 0;        for(int i=1;i<n;++i){            max = (A[i]-A[i-1])>max ?  (A[i]-A[i-1]):max;        }        double maxr = l-A[n-1];       // printf("%.2llf %.2llf %.2llf\n",maxl,max,maxr);       maxr = maxr>maxl ? maxr:maxl;       if(maxr>max/2){           printf("%.2f \n",maxr);       }else{            printf("%.2f \n",max/2);       }    }    return 0;}