Multiple Construction

The main trap is assuming we need to place the two copies of $x$ exactly $x$ apart. We don't. The statement only asks for a multiple of $x$, so $x$, $2x$, $3x$, etc. are all fine.

That extra freedom makes the construction almost stupidly simple.

Use this array:

$$n, n-1, n-2, \dots, 1, n, 1, 2, 3, \dots, n-1$$

So we print:

1. all numbers from $n$ down to $1$,
2. another copy of $n$,
3. all numbers from $1$ up to $n-1$.

This gives exactly $n+(1)+(n-1)=2n$ elements.

Checking occurrences

First, every number from $1$ to $n$ appears exactly twice:

• $n$ appears once at the start and once right after the decreasing block.
• every $x<n$ appears once in the decreasing block and once in the increasing suffix.

Now check the distances.

For $x=n$:

• first copy is at position $1$,
• second copy is at position $n+1$.

The distance is:

$$(n+1)-1=n$$

which is divisible by $n$.

For any $x<n$:

In the decreasing block $n,n-1,\dots,1$, the number $x$ is at position:

$$n-x+1$$

In the increasing suffix $1,2,\dots,n-1$, the number $x$ appears after the first $n+1$ positions, so its position is:

$$n+1+x$$

The distance is:

$$(n+1+x)-(n-x+1)=2x$$

And $2x$ is obviously divisible by $x$. No magic, no casework, no cursed backtracking. Just exploit the word “multiple”.

Complexity

For each test case we print $2n$ numbers, so the time complexity is $O(n)$ and memory usage is $O(1)$ besides output buffering.