Are we still in 20th century so that all tasks got ML=64MB? That caused me some problems in E and J :/

How to solve D? I got some ideas, but my local tests showed that even the simplest loop finding divisors of $$$n$$$ in $$$O(\sqrt{n})$$$ will time out xd

Only loosely related, but can somebody explain to me how changing the type of the board in E from vector<vector<bool>> to vector<vector<int16_t>> could have changed my MLE to OK?

B: If the string starts with a 'a', start with a swap. Now, assume the string starts with a 'b'.

If the string is all 'b', just copy $$$n-1$$$ times and merge $$$n$$$ times.

Otherwise, get lengths of continuous segments of the same letter, say these are $$$k_0, k_1, k_2, \dots, k_t$$$.

Then, for every $$$i = 0, \dots, t - 2$$$, copy $$$k_i$$$ times, merge $$$k_i - 1$$$ times, then swap if $$$k_i > 1$$$ and roll. After the $$$i$$$th operation, you have $$$i+1$$$ strings corresponding to the first $$$\sum_{j \leq i} k_j$$$ symbols at the start, and then a b or b a depending on the parity of i at the end.

For the last two $$$i$$$ you do one copy less. For $$$i = t-1$$$ you swap instead of swap and roll, and for $$$i = t$$$ you neither swap nor roll. Then in the end, merge until only one string is left.

This does around $$$3n - 2$$$ operations.

Is there an elegant solution to K? I don't have one so I'll leave my ugly approach instead.

Let's find a large cycle on the first board, and let $$$C=(C_1,C_2,\cdots,C_K)$$$ be this cycle. Then what we want next is a set of $$$K$$$ disjoint paths $$$P_1,P_2,\cdots,P_K$$$, where $$$P_{i,1}=C_i$$$ and union of these paths is precisely the set of cells on the first board. If we have such paths, the answer would be $$$P_1,second(rev(P_1)),second(P_2),rev(P-2),P_3,\cdots$$$. Here, $$$rev(P)$$$ means a reversed path and $$$second(P)$$$ means a path on the second board using the same positions.

How to get $$$C$$$: I repeated a certain pattern on the anti diagonal, which results in a cycle of length $$$O(N)$$$.

How to get $$$P_i$$$: I just run a DFS. When I choose the next vertex, I picked one with the minimum degree. Ties are broken randomly.

This algorithm sometimes succeeds, and sometimes fails because of randomness. Therefore I run several times to get a solution. In the worst case the running time didn't fit in the TL, so I pre-calculated for all $$$N$$$ a good seed which result in a solution.

My code

How to solve the problem I? I tried but failed qwq