"Displacement, reasoning" UVa 1315-creaz Tea Party

D secription

N participants of«crazy tea party»sit around the table. Each minute one pair of neighbors can change their places. Find the minimum time (in minutes) required-participants to sit-reverse order (so, left neighbors would bec ome right, and Right-left).

Input

The first line is the amount of tests. Each next line contains one integer n (1 <= n <= 32767)-The amount of crazy tea participants.

Output

For each number n of participants to crazy Tea Party print on the standard output, on a separate line, the minimum time re Quired for all participants to sit in reverse order.

Sample Input

3456

Sample Output

246

Test instructions: There are n people sitting in a circle. Adjacent can be exchanged, ask at least how many times to reach reverse order, that is, left and right adjacent to the number exchange.

Analysis: From the middle part, it is obvious 1~ (N/2) to the left interchange, (N/2) ~n to the right exchange. Like what:
n = 6 o'clock:
1 2 3 | 4 5 6
1 3 2 | 5 4 6
3 1 2 | 5 6 4 (above 3, 4 move to both sides respectively)
3 2 1 | 6 5 4 (2,5 move to both sides)
　　
The above completed reverse order transformation;

n = 7 o'clock:

1 2 3 4 | 5 6 7
...
4 1 2 3 | 6 7 5
...
4 3 1 2 | 7 6 5
4 3 2 1 | 7 6 5

So each number needs to be exchanged for (even) 0, ..., (N/2)-1, (N/2) -1,...,2,1,0
(odd) 0, ..., (N/2), (N/2) -1,...,2,1,0

Sum can be.

"Code": The code may be slightly cumbersome. I can simplify it.

1#include <iostream>2#include <cstdio>3 using namespacestd;4 intMain ()5 {6     intT scanf"%d", &T);7      while(t--)8     {9         intN, ans =0; scanf"%d", &n);Ten         if(n%2) One         { A             intnum = n/2; -Ans = (1+ (num-1)) * (num-1); -Ans + =num; the         } -         Else -         { -             intnum = n/2; +Ans = (1+ (num-1)) * (num-1); -         } +cout << ans <<Endl; A     } at     return 0; -}

"Displacement, reasoning" UVa 1315-creaz Tea Party