Ladders and Queries

Write $s_x$ for a rung between bars $x$ and $x+1$. First place the following rungs in increasing height order:

$$s_1,s_2,\cdots,s_{N-1},s_{N-1},\cdots,s_2,s_1.$$

The second half reverses the first, so every starting bar initially leads to itself. This uses $2N-2$ type-$1$ operations.

For a target $(a,b)$ with $a=b$, simply print $2\ a$. Otherwise, if $a>1$, erase $s_{a-1}$ at height $a-1$ in the first half. If $b>1$, erase $s_{b-1}$ at height $2N-b$ in the second half. Print $2\ a$, then restore the erased rungs.

Erasing $s_{a-1}$ in the first half moves the path starting from $a$ onto the former path from bar $1$. Erasing $s_{b-1}$ in the second half leads that path to bar $b$. Hence, the query returns $b$. Restoring the rungs returns to the identity ladder for the next target.

The initial setup uses $2N-2$ type-$1$ operations, and each target uses at most four more. The total is at most $6N-2$, within the limit. The output length and running time are $O(N)$.

Proof

A rung $s_x$ swaps the two paths passing through bars $x$ and $x+1$ at that height. The initial rung sequence consists of a sequence followed by its inverse, so its overall mapping is the identity.

Removing $s_{a-1}$ from the first half swaps the paths from bars $1$ and $a$ in that identity mapping. Removing $s_{b-1}$ from the second half swaps the destination bars $1$ and $b$. With both removals, the path from starting bar $a$ first follows the former path from bar $1$ and then reaches destination bar $b$. If $a=1$ or $b=1$, the corresponding removal is unnecessary.

After each query, all removed rungs are restored. Thus, every query begins with the identity ladder and independently reaches its requested destination.

In subtask 1, output only the queries. In subtask 2, toggle the single rung between the two bars before and after each needed query. In subtask 3, temporarily erase a rung only in the second half. The full solution uses both halves.

Solution written by GPT5