Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2189 Accepted Submission(s): 774
Problem Description Hotaru Ichijou recently is addicated to math problems. Now she is playing with N-sequence.
Let's define N-sequence, which is composed with three parts and satisfied with the following condition:
1. the first part is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.
Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.
Input There are multiple test cases. The first line of input contains an integer T（T<=20）, indicating the number of test cases.
For each test case:
the first line of input contains a positive integer N（1<=N<=100000）, the length of a given sequence
the second line includes N non-negative integers ,each interger is no larger than
109 , descripting a sequence.
Output Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by a integer, the largest length of N-sequence.
We guarantee that the sum of all answers is less than 800000.
Sample Input
1
10
2 3 4 4 3 2 2 3 4 4
Sample Output
Case #1: 9
這題可以用Manacher算法做，因為題目要找的是三段（第一段和第二段對稱，第二段和第三段對稱），其實就是兩個連在一起的回文串，我們可以先用Manacher算法初始化各個點的p[i]值（即可以向右延伸的最大距離，包括本身，這時已經加入了-1代替算法中的'#',-2代替算法中的'$'），然後對於每個i，枚舉j（j屬於1~p[i]-1),如果i+j-p[i+j]+1<=i，那麼說明i，j可以分別作為第一、二段的點和第二、三段的點）。
這裡有個優化，因為枚舉時滿足條件的只有'#'（即'-1’），所以我們可以使i，j每次變化2.
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