HUST Monthly 2010.04.05
From: 20100405 12:00:00
To: 20100405 17:00:00
Now: 20170920 01:02:31
Status: Public
C  Dartboard
Time Limit: 2s
Memory Limit: 256MB
Submissions: 292 Solved: 118
 Description
 A dartboard is a disc divided into n segments, with each segment labeled with a distinct positive integer between 1 and n. That integer indicates the score for hitting the segment. To make the game more interesting, the arrangement of numbers is chosen in such a way that the risk is maximized. The risk is estimated based on the differences in scores of adjacent segments.
We're studying the following 'doublelayered' structure of segments in this problem:
i.e., n is always even, and we split the disc into two layers of n/2 parts along the circumference. We enumerate the segments in the following manner: the first segment is some outer segment, the second segment is the corresponding inner segment, the third segment is some adjacent outer segment, etc. An example of this enumeration is shown on the picture above.
The total risk is defined as the sum of squared differences between the scores of adjacent segments. If the value assigned to segment i is a_{i}, then the risk is:
R=∑_{i=1}^{n}(a_{i}a_{i+2})^{2}+∑_{i=1}^{n/2}(a_{2i1}a_{2i})^{2}
(we assume a_{n+1}=a_{1} and a_{n+2}=a_{2} in this formula).
You are to place the numbers from 1 through n into the segments in such a way that the total risk R is maximized.
 Input
 The input file contains an integer T means there are T cases followed.
The next T lines: there is an even integer n (6 ≤ n ≤ 100) in each line.
 Output
 Output the MAX(R);
 Sample Input

2
10
6
 Sample Output

461
87
 Hint
 When n equals to 10, “ 2 9 7 4 6 5 3 8 10 1”is one of the solutions. It makes R maximized(MAX(R)=461).
 Source
 Hust Monthly 10.04.05/木瓜仙人