To me, the answer for K seems ambiguous as there can be considered to be 2018 sequences covering the whole circle. Hence, 2018*2019/2 is the answer that I was getting (even though I did not participate).

1164R — Divisible by 83 , Anser is 0, 0 mod 83 = ???? :)))

I think the answer to problem K should be $$$\frac{2018*2017}{2}+2018=2037171$$$
The whole circle may be counted 2018 times (because a sequence is an enumerated collection of numbers and there are 2018 possible starting points)
It can be seen that these 2018 sequences are pairwise distinct.

Why areas are 3 in B? I used affine transformations in B, to solve

Well, I for got the +1 in Problem K :'(

Is there a way to submit my answers now ? I couldn't take part of the contest and I would like to submit it ... Thanks :3

In some of these problems, you could take one particular instance and work out the answer for that, since you know there is only 1 answer to the problem (because of how the system works).

For example, for the 2018 integers written on a circle with sum 1. You can assume 2017 of them are positive and one of them is the negative sum of all the others +1. It's now easy to see that all sequences that do not contain the negative number are positive.

Can any one elaborate the solution of problem 1164I — maximum value please . How to transform the equality (a+b−1)2=ab+1 to obtain a2+b2=4−(a+b−2)2≤4 ??

Please make another mathforces contest.