Might as well start with B.

Let's say a subsequence is compact if all the adjacent numbers differ by 1.

After one operation, the permutation is partitionable into at most 2 compact subsequences.

After k operations, 2^k of them, and it turns out this is the IFF condition.

Let's look at the operations in reverse order.

We first sort all the compact subsequences: let's say there are M of them, S_1, ..., S_M.

One reverse operation allows you to pick a subset of them and place them at the end. When you select such subset, they have to form a suffix of S_1, ..., S_M by definition. It now splits into two independent part, [S_1, ..., S_i] [S_i+1, ..., S_M], and both part acts independently till the end.

Now set DP[x][l][r]: min cost to solve the interval [S_l, ..., S_r] using x last operations, 1 <= l <= r <= M. Transition is just iterating over all suffix.

Another way to look at it: for each operation, assign 0 or 1 to each number and look at the operations in reverse order. Then the initial order of any two elements is determined by which has 0 in the first operation where they have different numbers; if there's no such operation, then it's the order in $$$P$$$.

This is indeed equivalent to the intended solution. When you add $$$a_i$$$ and there are less than $$$k$$$ non-representable values, it means there are no more than $$$k$$$ segments before $$$a_i$$$. Therefore the number of segments increases by at most $$$k$$$ after adding $$$a_i$$$.

In general, after any number of steps there exist numbers $$$L, R$$$ such that we can be anywhere in the region $$$L \leq |x| + |y| \leq R$$$. If the next step is $$$d$$$, then the new boundaries are $$$R' = R + d$$$, $$$L' = \max(0, L - d, d - R)$$$. If we additionally put an upper bound $$$M$$$, then the first one simply becomes $$$R' = \min(M, R + d)$$$. Note that if $$$D$$$ is the largest step in the set, we have $$$M \geq D / 2$$$ to be able to make the step $$$D$$$. Also, if answer $$$M$$$ exists, it is at most $$$D$$$.

Respective to the upper bound $$$M$$$, say that a step $$$d$$$ is large if $$$d > M$$$, and small otherwise. Note that once $$$R = M$$$, is always stays that way (provided on every step $$$L \leq R$$$), thus any solution step sequence can be divided into two phases: before and after $$$R$$$ reaches $$$M$$$. With an exchange argument one can prove that the second phase should use steps by decreasing of magnitude. Further, in the second phase:

• after a large step $$$D$$$ in second phase we simply have $$$R' = M$$$, $$$L' = D - M$$$
• and after a small step $$$d$$$ we have $$$R = M$$$, $$$L' = \max(0, L - d)$$$.

Let $$$S$$$ be the sum of all small steps. Now there are a few cases:

• there is at most one large step in total. Then we just check that the largest step is at most the sum of all others.
• all large steps occur in the second phase. Let the sum of (small) steps in the first phase be $$$s_1$$$, and the smallest large step be $$$D$$$. We must have $$$s_1 \geq M$$$, $$$S - s_1 \geq D - M$$$.
• there is a large step $$$D_1$$$ at the end of the first phase, and the last large step in the second phase is $$$D_2$$$. Then, defining $$$s_1$$$ similarly, we have $$$s_1 \geq D_1 - M$$$, $$$s_2 \geq D_2 - M$$$.

Note that we can only consider $$$D, D_1, D_2$$$ above among the smallest two large steps available (check all combinations). Checking conditions requires finding the smallest representable sum $$$s \geq x$$$ of small steps. These can be maintained and queried with a bitset of reachable sums, along with gradually increasing $$$M$$$.

1) They count. (See the discussion here).

2) They also count. It must be a small bug and we will fix it.