Since the announcement is in english I belive no. It is probably just a trap

Yes

l nas bt7fl 3lek leh ya 3m :"""""D

Good point. They are hosting contest on yandex site, but that is their cheap way to catch you off guard. They will change your cf ratings

Of course it is!

And I'm not sure if it is the first contest hosted on yandex site.

Финал конкурса пройдёт 19–20 мая по адресу: Российская Федерация, город Санкт-Петербург, Пискаревский проспект, д. 2, корпус 2, литер Щ. Каждый финалист должен подтвердить свое участие и представить все необходимые документы до 20 апреля. В противном случае он будет дисквалифицирован. Финальный раунд пройдет по правилам TCM/Time, длительность раунда будет между 120 и 180 минутами, объявлена непосредственно перед его началом.

Да, в этом году результат по каждому треку независимый.

if you send blind submission, then if you solve it correct, then the some amount of time is subtracted from your time .

One of the reasons to conduct a warm-up round is to make it possible for you to become familiar with TCM\Time rules.

Ok, so I switched my language to russian, and the time says 21:40. I still think the english version needs updating to clarify this though.

Yeah! Pls somebody clarify this ASAP. I'm giving the Yandex contest first time so don't know about how it works...

:) Looks like Yandex hired some good competitive programming engineers from Facebook.

brooo :D

I was able to persuade myself after staring at the screen that correct answer (except for n = 1) is Euler's totient function. No proof, but it worked.

A great hint for this problem:

Stern-Brocot tree

Simulate the process for small n and look up the sequence on OEIS.

The answer turns out to be phi(n). Can anyone explain why?

I was doing the same thing. Were you getting WA on test case 6?

I did the same and got WA at test 4 .

Checking whether all centers are collinear is enough. Probably you had a bug somewhere in your code.

I was just checking collinearity of the centers, because unless I am completely wrong, any line going through rectangle's center still splits it in two halves. But it didn't work.

Solution is check that all centers of mass* is on one line. But you should be carefull because centres can be same point

* for circle it is a centre, for rectangle it is { (x1 + x2 + x3 + x4) / 4, (y1 + y2 + y3 + y4) / 4 }

dp[i][mask1][mask2]

(Note that number of transitions (mask1, mask2) -> (mask2, mask3) for n = 7 is 18320)

I solved with DP and bitmasks.

Check all substrings of size 2 first, if palindromes are found, return lexicographically smallest one.

If not found, do same for size 3.

If not found, return -1.

You can show then for even n>=4, if palindrome exists, a palindrome of size 2 also exists.

And for odd n>=4, if palindrome exists, a palindrome of size 3 also exists.

Notice that you can first remove K - 1 sons of the root, which means you can remove their whole subtrees afterwards. The K-th removed son needs to be the last removed vertex overall — the situation for its subtree is the same as for the original subtree, so it can be solved recursively using some tree DP.

Besides that, there are some other sons of the root. These can't be removed at all, but we can remove K - 1 sons of each of them (with these sons' whole subtrees, just like above); the situation for the remaining sons' subtrees is the same again, so this can be done with another tree DP.

There are two tree DPs to compute in parallel: the max. number of removable vertices in the current subtree if the root needs to be removed last and if the root mustn't be removed.

Based on https://contest.yandex.com/algorithm2018/rules/ section 3 paragraph 3: yes

There are also qualification round

Don't fool yourself, you will definitely miss the ongoing rounds too and end up as a final round tester for the third time :)

Last year the editorial was published here in Codeforces, and in the schedule page, I think it'll be the same this year.

В меню выбрать "Соревнования", дальше "Яндекс.Алгоритм-2018".