Привет.

В 19.08.2018 16:35 (Московское время), состоится общий рейтинговый раунд Codeforces Round #505. Этот раунд идет в паре с раундом #504.

Часть задач взята с Финала VK Cup 2018 (ashmelev, Errichto, lewin), часть придумана мной. Спасибо моим товарищам — Диме (_kun_), который, кстати, является координатором этого раунда, Коле (KAN), который позвал меня его проводить, а также Грише (gritukan) и Ильдару (300iq) за то, что они просто клевые.

Так же, как и в прошлом раунде, по инициативе компании ВКонтакте будут разыграны очки GP30.

Участники сортируются по очкам за оба тура (если участник не участвовал в одном из туров, очки, набранные за него, полагаются равными нулю); при равенстве очков как тай-брейк используется максимальное за оба тура время, прошедшее от начала тура до последней посылки, прошедшей претесты.

Напоминаю, что лучшие 10 по сумме очков GP30 получат фирменного плюшевого персика.

Надеюсь, что вам понравится. Удачи!

On Aug/18/2018 17:35 (Moscow time) Educational Codeforces Round 49 will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extented ACM ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given 7 problems and 2 hours to solve them.

The problems were invented and prepared by Vladimir Vovuh Petrov, Adilbek adedalic Dalabaev, Ivan BledDest Androsov, Mike MikeMirzayanov Mirzyanov and me.

Good luck to all participants!

Our friends at Harbour.Space also have a message for you:

Hi Codeforces!

Our Hello Barcelona Programming Bootcamp will be kicking off next month from Sept 26 — Oct 4, and we’d love to see you there!

The boot camp will once again feature the all-time greats Mike MikeMirzayanov Mirzayanov, Andrey andrewzta Stankevich, Michael Endagorion Tikhomirov, Gleb GlebsHP Evstropov, Artem VArtem Vasilyev, Ivan ifsmirnov Smirnov and other world renowned Russian coaches to train the participants.

You will be competing and learning side by side with some of the greatest teams in the world, while learning from the best coaches the ICPC has to offer.

Be sure to register before September 1st so everyone has time to get visas if needed, and of course for the Early Bird Discount of 5% or the Loyalty Discount* of 20% off registration for the boot camp!

*The loyalty discount is offered to universities and individual participants that took part in Hello Barcelona Bootcamps and Moscow Workshops ICPC.

Learn more about Barcelona ICPC Bootcamp

You can ask any questions by email: hello@harbour.space

Hi!

August 17, Aug/17/2018 17:35 (Moscow time), there will be a rated Codeforces round #504. Some of the problems will be from VK Cup 2018 finals, and PikMike and Vovuh have prepared other tasks for the full round.

The problems of this round are proposed, prepared and tested by: MikeMirzayanov, PikMike, Vovuh, Errichto, lewin, Endagorion, Um_nik, YakutovDmitriy, BudAlNik, izban, Belonogov, scanhex, 300iq, qoo2p5, Livace.

There will be prizes from VK social network in this round as well! The participants who took the first 30 places of this round and the round #505, also partially based on the tasks of VK Cup 2018 Finals, will get GP30 points.

Participants are sorted by sum of points for both rounds (if the participant did not participate in one of the rounds, the points scored for it are assumed to be equal to zero), with the maximum time for both rounds from the beginning of the round to the last submission that passed the pretests as tie-break.

The top 10 participants will receive a plush Persik!

There is no country nor language restriction, everyone can win a prize. One don't have to have participated in VK Cup to receive the prize. Exact selection algorithm will be announced before the start of the round.

Good luck!

Hello, codeforces!

All signs in the sky and on the ground indicate that you've enjoyed my first blog, so here is the second one. I've decided to choose a Polish task again, as there are plenty of interesting ones. This time we'll take a look at the "plot purchase" (you can submit here), which is a bit easier, but a few years ago I was very proud of myself when I solved it. The statement goes as follows:

You are given a square n × n grid (1 ≤ n ≤ 2000). In every cell, there is a number from the range [1, 2·10⁹]. You are also given an integer k (1 ≤ k ≤ 10⁹). A task is to find a subrectangle of this grid such that the sum of values in this subrectangle lies in the range [k, 2·k] (or report that there is no such subrectangle). Just print coordinates of its opposite corners.

If the problem would be more difficult, and we'd be asked to find a subrectangle with some given sum of values, then it'd be possible to find it in O(n³) time, just by fixing lower and upper sides and then using two pointers to find for every possible right side the proper left side (if existing). O(n³) is too much, so we have to use the fact that we don't have to achieve the exact sum, which looks really interesting, because many known problems become easier/possible to solve when we change the exact requirement to the range one (knapsack comes to my mind).

Let's firstly look at the sum of all numbers in the grid. If it's smaller than k, then we are sure that there is no proper rectangle (because all numbers in the input are positive). If it already lies in the desired range, then we can end and print the answer. If it's bigger than 2·k, then we need to try a smaller one. Here comes idea which most of the people would go through while solving this problem: let's divide this rectangle into two smaller ones. No matter how, possibly somewhere in the middle, because it looks the most natural (but probably it's a bit easier to implement cutting off rows/columns one by one). Also, the dimension which we'd decrease doesn't matter. After this cut, the sum of numbers in the rectangle with greater sum won't be less than the sum of all numbers divided by 2. So, if the sum was greater than 2·k, then it won't be smaller than k.

Sooo, is it the end? Can we just cut the rectangle as long as it has the sum greater than 2·k and then print what's left? Unfortunately, the answer is no. Let's consider a test case with only one 1 in the corner and 1000 in the rest of the grid. If k is equal to 1, then the only correct answer is to print only this corner. In our algorithm, we could lose it quickly if we'd stay in the wrong half (which is more likely as it has greater sum).

So we need some improvements. Firstly: all cells with values strictly greater than 2·k are definitely forbidden, so we can mark them as infinity. Among all rectangles with no forbidden cells we need to find one with the sum of elements not smaller than k — if we'd have it, then we'll run our algorithm on it. Here we'll again use the fact about only positive numbers: if some rectangle is a subrectangle of some bigger rectangle and the smaller one has sum not smaller than k — then, the bigger one also has such sum. This fact implies that it's enough to look only at rectangles maximal by inclusion, which turns out to be a well know task, but I'll describe it anyway.

We'll need some preprocessing. Firstly, prefix sums for each subrectangle with one corner in cell (1, 1). Secondly, for each cell, we know how far can we go down until we meet either a forbidden cell or an edge of the whole grid.

Let's fix an upper side of the rectangle that we want to find and build an array which indicates how far can we go to the down in each of the columns. Then, if we have some interval of columns corresponding to a rectangle, then it's optimal to go as far to the down as the minimum of values in the array (in this interval). It makes sense, cause it means exatly that when we know left, right and upper side, there is only one reasonable bottom side. Let's iterate over possible right ends of an interval and keep track of places on the left where interval's minimum changes. Why only these places? It's clear that if some interval [i, j] has this same minimum as interval [i - 1, j], then it makes no point to consider [i, j], so we are only interested in intervals whose minimums change after extending that interval to the left (we assume that before and after the array there are values equal to  - ∞). So we can keep these places using a stack. When new value comes, we pop out of the stack values greater than the new one and we consider intervals which start at them and end right before the column in which currently we are. It turns out that the number of considered intervals amortizes to O(n) with a fixed upper side of a rectangle (because we consider something only when we remove it from the stack), so in total there are O(n²) considered rectangles. Complexity is the same, cause when we find any rectangle with the sum not less than k, then we can run our partition-algorithm with the knowledge that for sure it'll find the desired rectangle.

What do you think about this task? Was it interesting or maybe too easy? Let me know what you think about it. See you next time! :D

Hello, codeforces!

The community wants so the community gets it! :D Here it is, my very first blog about tasks and algorithms. At the beginning I've decided to post my entries on codeforces, maybe I'll switch to something different if it becomes uncomfortable.

To pour the first blood I decided to choose a task from one of the old ONTAK camps. Task's name is "different words" (you can submit here). The statement goes as follows:

You are given n words (2 ≤ n ≤ 50 000), every of length exactly 5 characters. Each character can be a lowercase letter, an uppercase letter, a digit, a comma... basically, it can be any character with ASCII code between 48 and 122 (let's say that k is the number of possible characters). A task is to find all pairs of indexes of words which are . Two words are if they differ at all 5 corresponding positions. So for example words and are really different and words and are not, because in both of them the third character is . As there can be many such pairs (up to ), if there are more than 100 000 pairs, then the program should print that there are only 100 000 and print this number of pairs (arbitrary chosen).

Please, note that this task comes from the contest which took place a few years ago, so don't think about bitsets. :P

So, how can we attack this problem? At first, you may think about some meet in the middle. It turns out to be hard, even if you can come up with something with k³ in complexity, then it'll probably be multiplied by n. O(k⁵) is also too much. Inclusion-exclusion principle unfortunately also won't be helpful, as we want to find those pairs, not only count them, and working on sets of words won't be so effective.

The biggest problem is that k is big. If it'd be, let's say, up to 10, then it'd be solvable in some way, but I won't go deeply into this solution, cause it isn't interesting and k is of course bigger. But I've mentioned small k. Why couldn't we dream about even smaller k? If k would be up to 2 (so words would consist only of digits 0 and 1) then this task would be trivial. We'd just group words which are the same, and for each word its set of really different words would be the group with words where every zero is changed into one and vice versa. But again, k isn't small...

Buuuuut, we can imagine that it is! Let's say that we assume that characters with even ASCII characters are "new zeros" and characters with odd ASCII characters are "new ones". Then for sure if two words are really different with these assumptions, then they are really different without them, cause if they'd have this same character at some position, then this character would change into this same "new character". This allows us to find some pairs, unfortunately not all of them, but the way definitely looks encouragingly.

If we've found some of the pairs, maybe we should just try one more time? But how should we change characters now? Now comes an experience: if you have no idea what to do or you don't want to do some complicated construction, then just do something random! So we could randomly change every character into zero or one and then run our algorithm for k equal to 2 — match groups of opposite words. What's the probability for a fixed pair that we'd find it during one run of our algorithm? If we already know what we've assigned to each character of the first word, then on every corresponding position in the second word there should be a different character — for each position we have a chance, that this assigned character will be correct, so we have probability equal to that the whole word will match.

What's the number of needed runs? Looking at limits we can guess that it could be a few hundred. Let's calculate the probability of fail if we'd repeat algorithm 600 times. The probability that we wouldn't find some pair is equal to , the probability that we'd find it, of course, and finally, the probability that we'd find all of them (or some 100 000 of them) is equal to which is greater than 0.999, so it's definitely enough.

There is one more thing. Consider words and . They are really different. Let's say that we've assigned 0 to a. Then we have to assign 1 to b. Then we have to assign 0 to c. Then we have to assign 1 to a. So there is a problem... At first glance, it might look problematic, but it turns out that it's easy to fix it. We shouldn't just assign zeros and ones to characters. We should assign them to pairs (character, position). Then everything will go well.

Total complexity is something like O(n·600·log(n)), but we can get rid of the log factor by using hashmaps.

Sooooo, it's the end of my first blog, sorry if it's too long. I hope that you've found this task useful and interesting. What do you think about this format, what was missing, what was bad? It'd be great to see a feedback from you. See you next time! :D

Important day! Today we will know the winners of VK Cup 2018. The prize fund of the competition — numbers in binary notation:

• 1 place — 1048576 rubles
• 2 place — 524288 rubles
• 3 place — 262144 rubles
• 4-8 place — 131072 rubles

We are start at 10:40. Monitor the progress here.

Good luck to the competitors! At the evening I add the results in this post.

The first day of every year championship for programming VK Cup 2018!

Today VK Cup was opened by Alexander Konstatinov (VK) and Mikhail Mirzoyanov (CodeForces). After that was a non-formal part, there we looking for funny photos of our competitors and know some interesting facts about them.

Some teams from the top-20 of Round 3 have refused to participate in the final to save the chance for the next year. We invited the following by rating. So, this year's finalists, meet!

1. Нижний Магазин SU: V--o_o--V & LHiC
2. ИТМО 2.0: senek_k & demon1999
3. 120 Minutes Adventure: sinesight & SpyCheese
4. Кекс столичный: Melnik & hloya_ygrt
5. Z: egor_bb & Nikitosh
6. Saint Veronika: aytel & Constantius
7. Road to the Gucci store: Golovanov399 & I_hate_ACM
8. Группа внутренних автоморфизмов: MrKaStep & bixind
9. Keksik: Tinsane & kb.
10. Свист минигана: Sert & Alexandr_TS
11. Ананас: AllCatsAreBeautiful & arsijo
12. Pareidolia: Copymaster & Lilith
13. я и моя девушка: aid
14. Типичная вечеринка с бассейном: Au-Rikka & Seemann
15. Accepted, cats and dogs: Andrew_Makar & BigBag
16. Московский Вентиляторный Завод: Дудка и Трубник: mHuman & komendart
17. QuietGenius: [user:Semenar & [user:Vaterlou]
18. Влюбленные Мудаки: super_azbuka & Khusainov
19. R3tired: Z38
20. Вупсень и Пупсень: Ajosteen & adedalic

Trial round reviewed our system — trial round! The first problem of trial round was a quiz. The question was about Codeforces and VK. First team that solved it is Saint Veronika: Constantius, aytel. Сongratulate! :)

Yesterday finalists went to the Hermitage Museum and carting. Today we had a lunch on the ship.

Competitors now take part in Code Game Challenge. In 2018 this is football. At evening we will watch results. For winners, VK has very nice prizes.

Tomorrow we give a link to results and you can watch and chatter for the favorite teams ;) More about the competition you can watch at VK Cup group.

Hi!

Summer Informatics School (SIS / LKSH) is a summer school for students in grades 6-10. SIS is focused mainly (but not only) on students participating in the Olympiads in Informatics — from beginners to participants of international competitions. SIS is held in July and August, each of them is visited by about 200 students from all over Russia and abroad. The language of the camp is Russian. Additional information about SIS is posted at lksh.ru

Right now, the August branch of the Summer Computer School is running, and on August 11 the traditional team Olympiad is going to place. I am happy to present a rated round based on it!

The round will be rated for both divisions, will be held in Aug/11/2018 16:35 (Moscow time), in each division there will be 5 tasks and 2 hours to solve them.

The problems of this round and the SIS olympiad were authored and prepared by SIS's teachers: izban, achulkov2, Schemtschik, i_love_isaf27, senek_k, asokol, WreckingBall, burunduk2. Also, I would like to thank Dembel for his help with olympiad's organization.

Thanks to our problems testers: _meshanya_, burunduk2, gritukan, niyaznigmatul, manoprenko!

Thank you, MikeMirzayanov, for the codeforces and polygon systems!

Yes, we are aware, that this contest clashes with ProCon Junior on codechef. However, given the schedule of the SIS and the approaching VK cup finals we can't do anything with it, sorry for that =/

Good luck!

UPD: The round will have one interactive problem for both divisions. Please, read the post about interactive problems here: Interactive Problems: Guide for Participants.

Congratulations to winners!

Div1:

1. Marcin_smu
2. Radewoosh
3. Swistakk
4. Panole233
5. ko_osaga

Div2:

1. Onuz
2. aurelio
3. usachevd0
4. jebouin
5. etiennerossignol

Upd Thank you for participation! Due to vk-cup conduction rating recalculation will happen slightly later, than usually.

Upd The editorial is published!

You may also check the unofficial editorial, written by neal.

This round is in memory of Leopoldo Taravilse.

Leopoldo Taravilse (ltaravilse) was a distinguished member of the competitive programming community in Argentina and Latin America. As a contestant, he reached the ACM-ICPC World Finals in 2010 and 2012, and was active in CodeForces and TopCoder. After 2012 he became a problemsetter for the ACM-ICPC South American regionals and the Argentinian Programming Tournament, and later coach for a World Finalist team in 2016.

Leopoldo's passion was helping others realize their full potential in competitive programming. He taught at various competitive programming schools in Brazil, Cuba, Peru and Bolivia, and his role organizing the Argentinian Training Camp cannot be overstated. Under his wing, it grew from a small group of friends in 2010 to a major event featuring international sponsors and having hundreds of participants coming from all corners of Latin America. Through successive editions of the Training Camp, Leopoldo inspired students to learn and improve, to give back to the community, and to have fun doing it.

With his intelligence, warm heart, relentless work ethics and witty sense of humour Leopoldo touched countless lives. We will miss him as a friend, a mentor, a teammate, a drinking buddy, a role model, an overall great person.

Rest in peace my dear Leo.

You can see this blog: A round in the name of ltaravilse.

And you can donate for prizes of this round: In Memory of Leopoldo Taravilse.

Hello everyone!

I'm pleased to announce: Codeforces Round #502!

This round will take place on Aug/08/2018 17:05 (Moscow time). It is rated for all participants and everyone will have 8 problems and 2 hours 15 minutes to solve them.

The contest is created by Magolor. This is the first competition I proposed. I hope that you will enjoy it. The heroes are from a TV show that I like and do not have any relation to Leopoldo.

6 of the 8 problems are created by myself. Thanks to:

• ODT for inspiring me and discussing the problems with me.
• Anton arsijo Tsypko for examining my problems, helping me a lot on this contest and designing problems B and G.
• Max zubec Zub, Danya danya.smelskiy Smelskiy, Sofia Sonechko Melnyk, Stanislav StasyaCat Bezkorovainiy, Vitaliy kuviman Kudasov, Aleksandr Kostroma Ostanin, for testing the round and improving the problems and solutions.
• Mike MikeMirzayanov Mirzayanov for Codeforces, Polygon and the help he offered to me.
• Nikolay KAN Kalinin for helping with the contest and designing problem G.
• Ahmed ahmed_aly Aly for his initiative to host this round in memory of Leopoldo.

...Yet Another Chinese Round...?

Scoring distribution: 500-1000-1500-1500-2000-2500-3000-3250.

Prizes distribution: $153 for top 30, delivered using bitcoin or amazon gift card (each contestant chooses).

UPD1: Chinese Editorial is published. You can view it through This link. English Editorial well be published soon.

UPD2: Congratulation to winners!

1. Benq

2. 000000

3. mcfx

4. Swistakk

5. orbitingflea

6. ivan100sic

7. ko_osaga

8. dreamoon

9. fjzzq2002

10.step_by_step

UPD3: English Editorial is published.