Miaomiao's Geometry
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 191 Accepted Submission(s): 38
Problem Description There are N point on X-axis . Miaomiao would like to cover them ALL by using segments with same length.
There are 2 limits:
1.A point is convered if there is a segments T , the point is the left end or the right end of T.
2.The length of the intersection of any two segments equals zero.
For example , point 2 is convered by [2 , 4] and not convered by [1 , 3]. [1 , 2] and [2 , 3] are legal segments , [1 , 2] and [3 , 4] are legal segments , but [1 , 3] and [2 , 4] are not (the length of intersection doesn't equals zero), [1 , 3] and [3 , 4] are not(not the same length).
Miaomiao wants to maximum the length of segements , please tell her the maximum length of segments.
For your information , the point can't coincidently at the same position.
Input There are several test cases.
There is a number T ( T <= 50 ) on the first line which shows the number of test cases.
For each test cases , there is a number N ( 3 <= N <= 50 ) on the first line.
On the second line , there are N integers Ai (-1e9 <= Ai <= 1e9) shows the position of each point.
Output For each test cases , output a real number shows the answser. Please output three digit after the decimal point.
Sample Input
3
3
1 2 3
3
1 2 4
4
1 9 100 10
Sample Output
1.000
2.000
8.000
HintFor the first sample , a legal answer is [1,2] [2,3] so the length is 1.
For the second sample , a legal answer is [-1,1] [2,4] so the answer is 2.
For the thired sample , a legal answer is [-7,1] , [1,9] , [10,18] , [100,108] so the answer is 8.
Source BestCoder Round #4
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简直神奇,比赛的时候900多就21个过的...,自己当时没考虑到一条线段能覆盖两个点的情况,说到底还是自己太弱了,不够细心,还有就是自己太心急了,刚敲完就交了,导致罚时比较多,今后得慢慢改,注意到答案只能是距离或者距离的一半,依次枚举就行,对每个点只有两种选择,一种是选点左边的线段,一种是选右边的线段,当能选左边的时候一定要选左边的,否则选右边的,如果左右两边都不能选,那么这个线段肯定长了,假设当前枚举的距离为x,那么选左边的条件是A[j]-A[j-1]-vis[j]>=x||A[j]==A[j-1]+x,右边这种就是一条线段覆盖两个点的情况,vis[j]是上一个点对现在这个区间的影响.
代码如下:
#include
#include