Codey and Schedule

https://www.hackerrank.com/contests/codenection-2023-final-round-closed-category/challenges/cn-c12

Question

Codey is the chairman of a competition with n teams and the game cannot start without a schedule. Due to time constraints, Codey wants to arrange a schedule with a minimal number of rounds so that each of the n team will play against two different teams. Help Codey arrange the schedule!

Output m, which represents the number of rounds, and m pairs of integers $a_i$ and $b_i$ , which represent the $i$-th match up in the schedule. It is guaranteed that the answer exists.

Note that this problem uses a custom checker, so make sure your program compiles correctly and prints the output according to the format before submitting.

Input Format

The first line contains an integer n, which represents the number of teams.

Constraints

$$3 \le n \le 10^5$$

Output Format

Output an integer m, which represents the number of rounds, for the first line.

In the following m lines, output pairs of integers $a_i$ and $b_i$, where $a_i \ne b_i$, representing the $i$-th matchup in the schedule.

If there are multiple answers, you may output any of them.

It is guaranteed that the answer exists.

Sample Inputs:

Input 0

Input

5

Output

5
3 2
3 4
5 1
5 4
2 1

Explanation

There are 5 rounds in total.

In the first round, team 3 will play against team 2. In the second round, team 3 will play against team 4. In the third round, team 5 will play against team 1. In the fourth round, team 5 will play against team 4. In the fifth round, team 2 will play against team 1.

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Solution - Simplest Round Robin

Ignore all the descriptions, as they only to make you confuse about that.

Basically, all the question wants is, loop through array[n] vs array[n+1], then each team will guarantee that must have fought 2 different teams.

How it works? I draw the illustration, and you will understand:

Assume we have 2 identical arrays:
1 2 3 4 5
1 2 3 4 5
let's shift the second array to the left once:
1 2 3 4 5
2 3 4 5 1

Have you found the pattern? 
let's take team 1 as example, he will be facing 2 and 5;
take team 3 as example, he will be facing 4 and 5
...

And you will be find that, the amount of tries is exactly as same as the total team number.
This exactly solved the whole problem!

The only edge case we need to consider is explicitly print array[0] vs array[n] as array is not a cycle. That's it.

Now we found the procedure, now let's get into coding!

t = int(input().strip())
print(t)

for i in range(1, t+1):
    if(i >= t):
        print(t, "1")
    else:
        print(i, i+1)