The solution to $$$G$$$ is as follows. Note that the minimal period $$$a$$$ of the periodically repeating string $$$A$$$ must be a cyclic shift of the minimal period of the periodically repeating string $$$B.$$$ Also, note that $$$a$$$ must be the minimal period of the string $$$A + A,$$$ thus, it can be easily found by using prefix function. Now we need to find a multiset of strings from $$$S,$$$ such that $$$a$$$ is their concatenation's minimal period and their length is divisible by $$$|a|$$$. Let's build a directed graph on $$$n$$$ vertices $$$(n = |a|)$$$ numbered from $$$0$$$ to $$$n - 1.$$$ For each string $$$s$$$ from $$$S$$$ let's find all the positions $$$0 \leq pos \leq n - 1$$$ such that $$$s$$$ will be a substring of periodically repeating string $$$a$$$ and the position of its first symbol will be the same as some repeat's position $$$pos.$$$ For example, if $$$a = baababca$$$ and $$$s = ab,$$$ then $$$pos$$$ can be $$$2, 4,$$$ and $$$7.$$$ This part can be done using KMP algorithm with some heuristic optimizations. Then, for each $$$pos,$$$ add an edge in the graph from $$$pos$$$ to $$$(pos + |s|)$$$ $$$mod$$$ $$$n.$$$ Hence, we just need to find the length of the minimal cycle in this graph and since the graph's size is at most $$$500,$$$ it can be done by Floyd-Warshall's algorithm.

J is similar to ARC084 F, try to read its editorial.

No special cases, you just need $$$O(n^2)$$$ time and $$$O(n)$$$ memory.

In $$$E$$$ if you write numbers in base $$$7$$$, you need all powers of $$$7$$$ + numbers $$$2, 3 \cdots 6$$$ ($$$1$$$ is already included in the set and $$$0$$$ is not important). That is close to the answer (you have a few more numbers). But you can notice that $$$-1 = 6$$$ ($$$mod$$$ $$$7$$$), $$$-2 = 5$$$ and $$$-3 = 4$$$. So you do not need numbers $$$4, 5, 6$$$.

Let's ignore modulo. Let $$$steps[i][j][l]$$$ ($$$i \leq 18; j,l \leq 9$$$) be the minimum number of steps to come from $$$j \cdot 10^i + l$$$ to some $$$x$$$ such that $$$x \geq (j+1) \cdot 10^i$$$ and $$$where[i][j][l]$$$ be this $$$x$$$. Ignore states where $$$j=l=0$$$. For $$$i=1$$$ just simulate. To calculate these values for some greater $$$i$$$ you just need to know these values for $$$i-1$$$ (you perform $$$10$$$ such "jumps" for each state).

With these precalculations, you can move from any value to something not less than modulo in no more than $$$\log_{10}(mod) \cdot 10$$$ jumps. The problem is when for example you have to do $$$10^{18}$$$ steps and $$$mod=10^9$$$. To optimize it you can notice that each time after jumping over modulo you will go through some number which is less than $$$10$$$ (so you'll find a period quickly). Knowing the period you can skip several cycles and then just simulate till the end.

Worth mentioning: to simulate the process you should always try to make the biggest jump possible so that you won't run out of moves and won't exceed modulo too much. Also, notice that if $$$j$$$ would be greater than $$$9$$$ in the above definitions, then the simulation would go exactly the same if we replace $$$j$$$ with the greatest digit in $$$j$$$.

Which problem are you talking about?

I find the fact that the answer is on one of the axises (what is the right way to say this?) is pretty hard to prove.

Also L is pretty easy to write: https://ideone.com/NkhclY

So its more of a taste issue.

Maybe another reason could be the fact that the tests on the paper were very weak in general throughout this contest, and this problem in particular (which had some not-so-easy concepts like circles and rotations around a point) was particularly hard to reason about, especially due to the fact that it had no explanations and no figures whatsoever.

Ok, you were lucky and your code got accepted on the first try. My code, for example, gives out answer 3 instead of 2 on the sample. I understand that maybe at higher conding skills this may not happen, but at this point, with my solution not passing the sample, I am literally clueless as to what are the next steps, without a clear explaination of the sample test case or a figure at least.

I predict at least one hour of debugging on paper with a lot of poorly drawn circles and cartesian coordinate systems awaits me.

I believe people from November MW know why we put that name for the task :)

Generally official solution has a lot of cases too.

Just do 2D sparse table — $$$dp[x][y][z]$$$ is the MST taking the rectangle with corners $$$(x, y)$$$ and $$$(x + 2^{z} - 1, y + 2^{z} - 1)$$$. The restriction from the statement allows you to use $$$O(1)$$$ sparse table values answering a query, having $$$O(R \cdot C \cdot log(R) \cdot N) + O(Q \cdot N)$$$

Answer always exists. Let's make recursive function $$$rec(n, k, x, y)$$$ that gives construction for parametres described in the task where position of the first element is $$$y$$$. There are two options:

$$$k \geq n$$$ — then we put one element at position $$$y$$$ and call $$$rec(n-1, k - n + 1, x, y + x)$$$

$$$k < n $$$ — then we put one element at position $$$y$$$, $$$k$$$ elements at position $$$y + x$$$ and rest $$$n - k - 1$$$ elements at position $$$y + x - 1$$$ and stop with recursion (our $$$k$$$ will be $$$0$$$).

Note that second case will happen before $$$n$$$ becomes $$$0$$$.

Another solution — answer always exists even if we just divide all numbers in 4 groups — a 1s, b 2s, c (x + 1)s and then d numbers which are all at distance x from each other and from numbers from first 3 groups (i. e. 2x + 1, 3x + 1, ...). Also you will find answer very quickly despite seemingly cubic complexity

Most of the tasks are not from 2020 :)

I agree that we should make better memory limits for the problems, but we had convention of taking all data/constraints/limits from original setters and testers — we rewrote only needed stuffs before Baytdance round and fixed some bugs after noticing it during camp. Time limits are usually twice bigger than official solution in C++ (just for $$$B$$$ we put $$$5$$$ sec and official works for $$$2.9$$$ sec).