Автор RussianCodeCup, 14 месяцев назад, ,

Всем привет!

Лучшие 200 участников первой квалификации уже в отборочном раунде, а остальные могут попытать свои силы во втором квалификацоинном раунде. Он состоится в это воскресенье, 16 апреля в 12-00 по Московскому времени . Лучшие 200 участников попадут в отборочный раунд, а остальные смогут попытать свои силы еще один раз, в третьем квалификационном раунде.

Желаем всем удачи на раунде и ждем на http://russiancodecup.ru!

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 » 14 месяцев назад, # | ← Rev. 2 →   0 If I qualified to the elimination round, how can I receive the certificate and I'm live in Egypt ?!
 » 14 месяцев назад, # | ← Rev. 7 →   +18 Has anybody received their shirts from last year?EDIT: oops... that was an unfortunate typo!
•  » » 14 месяцев назад, # ^ |   0 Don't be so critical!
•  » » » 14 месяцев назад, # ^ |   +18 Oh my gosh — thanks :P That wasn't intended!
•  » » 14 месяцев назад, # ^ |   +18 http://codeforces.com/blog/entry/51001#comment-349145by the way, the typo was really funny :D
•  » » » 14 месяцев назад, # ^ |   0 Thanks for the link! Did you end up with a reply?
•  » » » » 14 месяцев назад, # ^ |   +13 Yes. They said there was a problem with the post and they're going to send it again.
 » 14 месяцев назад, # |   +21 I qualified in the previous round. Can I take part in this round unofficially?
 » 14 месяцев назад, # |   +17 How to solve C?
•  » » 14 месяцев назад, # ^ |   0 Minimize the number of cycles.
•  » » 14 месяцев назад, # ^ |   0 The answer to the fully given array = N — number of cycles. So you collect all cycle segments (e.g. ? -> 1 -> 4 -> 5 -> ?) where ? is controlled by you. If you have one segment — you have no choice (just put one element you have). Otherwise you can connect all segments into a one cycle (e.g. end0 -> start1, end1 ->start2, ... endI->start0), thus maximizing the answer.
•  » » » 14 месяцев назад, # ^ |   +9 The edge here signifies (ind,arr(ind)) ???
•  » » » » 14 месяцев назад, # ^ |   0 Yes
 » 14 месяцев назад, # |   0 I have a feeling that E can be solved using MO's Algorithm but I don't know how xD So Can Someone Help Me ?
•  » » 14 месяцев назад, # ^ |   0 Yes, it does. I solved it by MO.
•  » » » 14 месяцев назад, # ^ |   0 Very Nice, So chemthan, Please explain your solution .
•  » » 14 месяцев назад, # ^ |   0 I tried to, but got TLE :(First, transform array into prefix sums + lift it up by 100000 to avoid negative values. So our goal is to find longest interval such that arr[l] = arr[r] inside the quered interval.We start MO's algorithm and check intervals. We maintain: for each value: a deque of indices of this value in our current interval (easy to add/remove index from any side); a multiset of maximal lengths for each value (that is multiset of all deque[val].front() - deque[val].back()). The multiset is also easy to update since we can compute the length for the value from the deque. The answer for current query is then the largest value in the multiset. At the end of block check, we can clear the multiset directly, and clean the deque's by collapsing the current interval.PS: I used map instead of multiset.
•  » » » 14 месяцев назад, # ^ | ← Rev. 3 →   +8 My solution is similar to hellman_'s one, except a trick to avoid log factor in the complexity :). Maintain two array a[], b[]. When adding/removing a segment of leng d we add/substract one from a[d / sqrt(N)] and b[d]. To find the maximal len, we need only O(sqrt(N)).
•  » » » » 14 месяцев назад, # ^ |   0 Nice trick!
 » 14 месяцев назад, # |   +42 Will the problems get added to Gym like Qualification Round 1?
•  » » 14 месяцев назад, # ^ |   0 Sure. You will be able to get it later in the trainings section
•  » » » 14 месяцев назад, # ^ | ← Rev. 2 →   +10 when?It hasn't been posted yet.
•  » » » » 14 месяцев назад, # ^ |   0