POJ Topic 1455Crazy Tea Party (mathematics)

Crazy Tea Party

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7164		Accepted: 4864

Description 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 required for all participants to sit in reverse order.

Sample Input

3
4
5
6

Sample Output

2
4
6

Source

Southeastern Europe 2003

Test instructions: n individuals surround a round table, each person can exchange positions next to each other, ask the last number of times after everyone's left and right opposite

Idea: http://m.blog.csdn.net/blog/u010893129/21127153

It is well known that the assumption is in the case of a row:

1 2 3 4 5 6 7

If the position is swapped:

7 6 5 4 3 2 1

Need to pass N (n-1)/2 times Exchange (bubbling method for Exchange)

Now the situation is a ring, how can I split the ring into two linear, then exchange, and then splicing together.

Examples are as follows:

1 2 3 4 5 6 7 (7 back 1)

If it is divided into 1 2 3, 4 5 6 72 parts, respectively, and then connected, is not satisfied with the condition.

After Exchange 3 2 1, 7 6 5 4

After the connection is: 3 2 1 7 6 5 4, in the loop, in fact, the position after the change of 7654321, is satisfied with the condition.

So the problem is solved:

Now the problem is that the length of the ring is assumed to be N, and if a screenshot is made, the number of exchanges is minimal.
Suppose that the ring of N is divided into a length of k, and the other length is a n-k segment.

Then the number of exchanges required is:

k* (k-1)/2

(n-k) * (n-k-1)/2

The sum is the total number of times that need to be exchanged.

The topic requires the least number of exchanges.

Then it's begging.

k* (k-1)/2+

(n-k) * (n-k-1)/2 minimum value.

It is easy to base the formula on the two-time function when the extremum is taken N/2. AC Code

#include <stdio.h>
#include <string.h>
int main ()
{
	int t;
	scanf ("%d", &t);
	while (t--)
	{
		int n;
		scanf ("%d", &n);
		int K=N/2;
		printf ("%d\n", (k* (k-1))/2+ (n-k) * (n-k-1)/2);
	}
}