why problems A to C are Russian ?

How to prove that the order of the selected guests is not important in 907E - Party ?

Sorry, I think the Russian version analyzes this fact and so the English will. This comment has no point.

"There are a couple of corner cases:" Goes onto listing 9 of them.

The images in 906E — Reverses are broken.

For 906E, what is the algorithm to split a string into a minimum number of palindromes (presumably in linear time, given the constraints)?

I've found a description of an algorithm for deciding whether a string can be written as a sequence of even palindromes (P*) here, but it doesn't seem like it could be easily extended to find the minimum number of such palindromes.

Can anyone help me to understand why is it possible to use random approach in div2E?

Good tutorial! But in problem div.1 D, I wonder why holds? Can anyone explain it a little bit? Thanks a lot!

I think in the tutorial to Problem 907A — Masha and Bears, the point 4-->(Masha likes last car, so it's size is not more than 2·V3) must be (Masha likes last car, so it's size is not more than 2·Vm) . @veschii_nevstrui

What am i doing wrong here for D?

Does anybody know any link for proof for statements made in 906D?

Why Probelm D has tag "chinese remainder theorem"?
I think it should be replace with “Euler Theorem”

can someone explain solution of div 2 C shockers problem in easy way??with use of set or without it also?? relpy fast pls

For problem E, if we can reverse intersected substrings, is it an NPC problem or we can solve it in polynomial complexity?

Here's a randomized approach to 907D - Seating of Students that I think is easier to come up with than the editorial's & quite cleaner (although it is much less elegant):

1. Instead of dealing with a 2d grid, we can flatten out the grid. That is, instead of having the grid $$$[[3,2,4,5],[1,6,7,8]]$$$, we could have the grid $$$[3,2,4,5,1,6,7,8]$$$. Then, we know that something at position $$$i$$$ is adjacent to something at position $$$j$$$ iff $$$|i - j| = 1$$$ and neither $$$i$$$ nor $$$j$$$ is at the end (i.e. neither of them are $$$-1$$$ modulo $$$m$$$) OR if $$$|i - j| = m$$$.

2. We can incrementally build the grid. First, we take an arbitrary number and add it to a flattened grid. Say, we're building a $$$2 \times 4$$$ grid and our first number is $$$3$$$. Now, we can find out that $$$3$$$ can't be adjacent to adjacent to $$$1,4,5$$$ (using what we established in 1.) So once we chose our next arbitrary number, we have to make sure that the next number we chose can't be $$$1,4,5$$$. So if we do pick some of $$$1,4,5$$$, pick another number, and so forth, until you pick a valid number. Suppose we pick $$$6$$$. Now, we have a grid $$$[3,6]$$$. Pick another random number. If our new random number is either $$$1,4,5$$$ (since that's adjacent to $$$3$$$) or $$$4,5$$$ (since that's adjacent to $$$6$$$), then we re-pick until we get a valid number. In this way, we continue, until we have fully built our grid.

3. Just to formalize what's in $$$2.$$$ Suppose our incrementally build array is called ans. To check if the last element of ans violates any adjacency conditions, we just need to check two things (a) ans.size() % m != 1 % m && ans[ans.size() - 1] is not adjacent to ans[ans.size() - 2]. The first part is to account for the fact that just because two things are adjacent in the flattened grid, doesn't mean they're adjacent in the real grid (they usually are, but not when it's near the edge of the grid. (b) ans[ans.size() - 1] is not adjacent to ans[ans.size() - m - 1]. This basically accounts for vertical adjacencies.

Anyways, for those interested, here's the code: 137380117. I'd be interested if someone could hack this, since I know randomized solutions are prone to hacking. Also interested if someone could say the probability of using the aforementioned procedure, but not getting a valid grid. It does happen on the 10th test case, so to remedy this, I just applied the procedure 1000 times and took any valid result. Not sure how often this happens, though.