It will allow all languages supported on Codeforces, atleast that was the case in previous years.

The online mirror will allow all languages supported by Codeforces. However, we can't guarantee that all problems will be solvable in each language.

Here is booklet of Bubble Cup 2016, it contain problems also: http://www.bubblecup.org/Content/Media/Booklet2016.pdf

Unfortunately, there will be no T-shirts this year. There are some custom laws that are complicating shipments from Serbia to Russia.
We are working on finding an alternative way to provide T-shirts, and if do find it, we'll update the blog post.

No.
From the blog post:
The contest will be unrated. The reason for this is because rules of this contest are not common for Codeforces. Also, it is only a mirror of an onsite competition.

Each team will be provided with one computer.

Yes, the registration page will be updated soon to allow team registration.

Radewoosh's team won, second place for Jagiellonian Armadillos from Poland and third place for Cheeks from Russia. From facebook.

Also me, but they kicked me from their team.

Mo's algorithm on tree

Yes. I did and solved problem C in my team, but I got stuck at this symbol, and I throwed two questions... (I spent ~15 minutes to read this and understand the symbol, but overall the contest was good and fun.)

One could calculate dynamics d[x] for numbers <= 9 * 200000 + C how fast we could reduce x to one digit. Call numbers with d[x] >= 3 bad. If sum of digits of A0 is not bad all is ok. Let's take nonzero digits in A and remove plus after them (merge two digits in one two-digit number), if the new sum will be good we will stop, otherwise continue this process. On every step A1 will increase on 9x, 1 ≤ x ≤ 9. If in A0 there are n digits we could perform at least n / 2 such steps. One could calculate another dynamics on "bad" numbers and see that it will not help only for 379 and 388. If sum of A0 digits is 379 or 388 we will merge on the first step 3 digits and then do the same staff.

The solution is

find the x s.t. and as it is prime, find modular inverse and calculate.

It is

I also doubted that, but it ran in 826ms somehow

I think There is an O(N^2 * K + N^3) solution (with preprocessing sorting-by-angle).

I was only a few seconds late :(

You should be able to submit now.

I have solution in O(n + m^2logl) so i think O(m^4) might be too slow.

How do you multiply two polynomials in m^4 :P?

The way interpreted the problem was essentially to calculate

where cp_i is the number of times i appear in the array c for all (x_i) s.t.

so we tried to multiply the polynomials

and sum over all the coefficients of x^L - 1y^M - k by multiplying them with the number of ways s_a + c_b + e_b = k via counting sort.

Could someone please help us with a better solution?

Solved it in O(m^3logL+N) using matrix multiplication.

A simple greedy of putting them all into single digits does not work. Say for example we have

(something) + 1 + 3 + 4 = 66 -> 6 + 6 = 12 (so impossible)

but if we do

(something) + 134 = 210 -> 2 + 1 + 0 = 3 (so possible)

I expect TC6 to be an example of this.

After reading the editorial for Problem , I am certain they literally copy-pasted the technique explanation from my blog and changed a couple of words here and there..

Unsurprisingly, that is how I solved the problem during the contest as well (Copy the code in the blog -> Change Inputs -> Submit -> Profit). Seems like even they submitted this.

EDIT:

Why am I being downvoted into oblivion? I didn't say anything untrue, did I? The editorial literally uses the same variable names (ST and EN) and copies the two cases verbatim from my blog. (the language is exactly the same)

I found a counterexample for ainta's solution of J.

5 5 9
1 3
2 3
4 3
5 3
2 2
3 2
4 2
1 1
5 1

is same as

5 5 5
1 2
2 2
4 2
5 2
3 1

and the answer is 2, but output is 3.

EDIT: