Codeforces: summing up 2016 year <center> <img src="http://assets.codeforces.com/images/Happy-New-Year-HD-Wallpapers.jpg"/> </center> Somehow it turns out that for me, each coming the New Year means more and more. This is an excellent opportunity to take stock, to breathe fresh air and to go through a kind of catharsis. This is a rare opportunity to be with the family, dress up in Santa Claus and just have fun. Holidays give me a clear criterion that I do the right things. If during the holidays, I begin to think about it, to recall some deferred ideas, eager to implement something new. So I’m doing right things. My things. And now exactly the same as in several recent years: during New Year holidays I write code for Codeforces, trying new technology or simply I’m making minor changes in testlib. It is necessary to me: at this moment are no deadlines, do not press the obligation. Just now, at 3 am, I can not break away from being able to do something new. I would like to take a moment and realize that spun in ...

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Codeforces: results of 2014 <img src="http://assets.codeforces.com/images/statistics-2014/snowman-en.png" style="float:right; margin:0 1em 1em 1em;"/> Hello 2015! Hello Codeforces! It seems it is time to take stock. Frankly, I was almost feared to start summing statistics of 2014. In 2013 Codeforces showed rapid growth so that it would not be surprising to look bad on the background of 2013. Certainly not! I was pleasantly surprised by the statistics and reports! Just below is a list of major events and achievements of Codeforces over the year. For you, it's just a list, but please note &mdash; every item includes hard work of multi-day Codeforces team, writers of problems, the organizers of contests and tournaments, problem testers and volunteers. Yay! Together we have done all of this: * introduced Codeforces API * added (and sometimes improved) all Andrew [user:andrewzta,2015-01-03] Stankevich contests * Codeforces supported mode to work as iframe-widget, and Codeforces helped Google to run https...

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CROC 2016 — Elimination Hey Codeforces! The Elimination Round of the CROC 2016 Championship will take place on [Friday, March 18 at 16:35 UTC](http://www.timeanddate.com/worldclock/fixedtime.html?day=18&month=3&year=2016&hour=19&min=35&sec=0&p1=166). After [our last round](http://mirror.codeforces.com/blog/entry/21860), Yang Liu ([user:desert97,2016-03-17]), Michael Kural ([user:pi37,2016-03-17]) and I realized that we haven't had enough, so we joined forces with Kevin Sun ([user:ksun48,2016-03-17]) and Daniel Chiu ([user:waterfalls,2016-03-17]) to prepare another problem set for you guys. Our contest will be for combined divisions and consist of seven problems. And although only those who pass the Qualification Round can participate officially, the round will be open to and rated for all Codeforces users. As always, we'll be taking the tractor to Bovinia for some <s>farmland</s> algorithmic adventures with Farmer John, Bessie and her best friend Elsie! Before we begin, we'd like to thank [user:GlebsHP,2016-03...

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Codeforces: Results of 2013 Hello! Do you feel the breath of New Year? <img src="http://assets.codeforces.com/images/tree.png" style="float:right;margin: 0 1em 1em 1em;"/> The clock shows a day before New Year, so it is time to sum up the past. I do not have speech-writers for words in the style of "what recently seemed almost impossible, it becomes a fact of life" so I'll be short. Thank you all! Your unfading interest and constant help inspire Codeforces team for new achievements! Brilliant writers, invisible soldiers: testers, tireless and greedy for knowledge participants &mdash; all of you help Codeforces do better! I've dug into the records and made a partial list of the most notable major improvements of 2013. Some large improvements of our infrastructure are not in the list, as the community can not see them directly. * A simple way to add contest from [Polygon](https://polygon.codeforces.com/) to Codeforces::Gym. * Drafts embedded in text form fields in Codeforces and Polygon. * Multipl...

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CROC 2016: Programming Championship Hello, Codeforces! <img src="http://assets.codeforces.com/images/croc-eng-logo-200.png" style="float: left;margin: 0 1.5em;max-width: 100.0%;max-height: 100.0%;"> In the spring 2016 the company CROC together with Codeforces will hold the third Russian Open programming championship «CROC – 2016». The goal of this project, as before, &mdash; encourage of the most talented and innovative young specialists and stimulate interest to the information technology. As usual CROC company pays great attention to social projects, associated with the development of the IT-industry, to promotion of IT-professions and to the increasing interest in modern technology. CROC has rich experience of cooperation with universities, the organization of courses, seminars and competitions for pupils, students and experienced professionals. The registration to the championship will be open from March, 1st to March, 16. Anyone, regardless of nationality, place of residence and level of education, c...

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CROC 2016 — Qualification <img src="http://assets.codeforces.com/images/croc-champ-eng.png" style="float:left; margin:0 2em 2em 0;"/> The Qualification Round of the "CROC 2016" Championship will start on [March, 16, 09:00 (UTC)](http://www.timeanddate.com/worldclock/fixedtime.html?day=16&month=3&year=2016&hour=12&min=0&sec=0&p1=166). Please, read about the Championship by the link http://mirror.codeforces.com/blog/entry/43229 You need to participate in the Qualification Round to make it to Round 1. All contestants who gain a positive score will advance to the Round 1. At the Qualification Round you will find a few problems, roughly ordered by the increasing complexity. During the Qualification Round the problems are judged only on pretests and system testing will take place after the end of the Qualification Round (round continues for 48 hours). The pretests do not cover all possible cases of input data, test your programs carefully! The Qualification Round has no hacks or decreasing values of the pr...

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Croc Champ 2012 — Round 2 Hello everyone! Today is the second round of the [Open Moscow Programming Championship By CROC](http://mirror.codeforces.com/croc2012) will be. Start is planned at [19:00](http://timeanddate.com/worldclock/fixedtime.html?day=20&month=4&year=2012&hour=19&min=0&sec=0&p1=166). Competition will happen by usual rules of Codeforces, with hacks and score falling in process of time. All contestants who passed score no less than contectant of the 300-th place in the Round 1, can participate in the Round 2. Every other contestants can patricipate the Round 2 out of competition. Specially for contestants of the second division we prepared more easy unofficial problemset. The official problemset and the unnoficial one have some common tasks. **Round will be rated for all participants.** Some number of problems are waiting you. They are roughly ordered by the increasing complexity. Score distribution is standard for both divisions (500-1000-1500-2000-2500). Don't forget that during contest your...

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Croc Champ 2012 — Qualification Round Hi everybody! Let me remind you that on the [3rd of April, at 08:00](http://timeanddate.com/worldclock/fixedtime.html?day=3&month=4&year=2012&hour=12&min=0&sec=0&p1=166) the Qualification Round of the [Open Moscow Programming Championship By CROC](http://mirror.codeforces.com/croc2012) will start. You need to participate in the Qualification Round to make it to Round 1. Contestants who gain a score equal to the 1000-th place finisher score or greater will advance to the Round 1 (also you need to gain positive score). At the Qualification Round you will find a few simple problems, roughly ordered by the increasing complexity. During the Qualification Round the problems are judged only on pretests and system testing will take place after the end of the Qualification Round (round continues for 24 hours). The pretests do not cover all possible cases of input data, test your programs carefully! The Qualification Round has no hacks or decreasing values of the problems. The round...

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ALL CodeForces Rounds Tutorial <html> <body> <br /><br /> >>>>>>>**Hi I just Say Really Sorry For Delay in UPD This Blog...!**<br /><br /><br /> <u>This is not Complete</u> , but i try to find <b>all Tutorial codeforces round</b> and posted here , thanks in advance :D <br /><br />This post will <b> <span class="Apple-style-span" style="border-collapse: collapse;color: rgb(255,0,0);font-family: verdana , arial , sans-serif;font-size: 12.0px;"> UPD</span></span></b> after each contest . <span>If anyone have or writes any editorial that's</span> <b> <span> <span class="Apple-style-span" style="border-collapse: collapse;color: rgb(255,0,0);font-family: verdana , arial , sans-serif;font-size: 12.0px;"> not in here</span></span></b> , Plz post here for ALL . <br /><br /><b>UPD </b> &nbsp;: Table1 updated : in Future!!! <br /><br /><b>UPD </b>&nbsp;: Table2 updated : CF # 118 ...to... CF # 169 added .<br /><br /> </p><hr /><br /></p> <div align="justify"> <br /><br /><b><s...

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The Final Round of CROC 2016 and Codeforces Round #347 <br> Hello Codeforces! The Final Round of CROC 2016 will be held today at [17:15](http://timeanddate.com/worldclock/fixedtime.html?day=15&month=4&year=2016&hour=17&min=15&sec=0&p1=166) Moscow time, where the best 50 participants from the Elimination Round will be competing for the valueable prices, as well as for their personal entertainment. Everyone else will be able to participate in Codeforces Round #347 tomorrow at [19:35](http://timeanddate.com/worldclock/fixedtime.html?day=16&month=4&year=2016&hour=19&min=35&sec=0&p1=166) Moscow time that will feature an almost identical problem set. It is going to be a usual **unrated** round, separate for each division. The problem set was prepared by Evgeny Vihrov ([user:gen,2016-04-15]), the one and only coordinator of Codeforces Gleb Evstropov ([user:GlebsHP,2016-04-15]) and me. I would also like to thank Mike Mirzayanov ([user:MikeMirzay...

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Editorial for CROC 2016 Elimination Round [problem:645A] --- **Idea:** [user:yummy,2016-03-18] One solution is to just brute force and use DFS to try all the possibilities. Alternatively, note that two puzzles can be changed to each other if and only if the $A$, $B$, and $C$ have the same orientation—clockwise or counterclockwise—in the puzzle. A third option, since the number of possibilities is so small, is to simply classify all of the $4!=24$ configurations by hand. <spoiler summary="Code"> ~~~~~ #include <bits/stdc++.h> using namespace std; string b, b1, b2, e, e1, e2; int main(){ cin >> b1 >> b2 >> e1 >> e2; swap(b2[0], b2[1]); swap(e2[0], e2[1]); b = b1 + b2; e = e1 + e2; b.erase(b.find('X'), 1); e.erase(e.find('X'), 1); if((b + b).find(e) != string::npos){ cout << "YESn"; } else { cout << "NOn"; } } ~~~~~ </spoiler> [problem:645B] --- **Idea: [user:ksun48,2016-03-18]** Loosely speaking, we’re trying t...

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Croc Champ 2013 — Qualification Round <img src="http://assets.codeforces.com/images/start.jpg" style="float:right;margin: 0 0.5em 0.5em 0.5em"/> Hi everybody! Let me remind you that on the [12th of April, at 20:00](http://timeanddate.com/worldclock/fixedtime.html?day=13&month=4&year=2013&hour=0&min=0&sec=0&p1=166) the Qualification Round of the [All-Russian Programming Championship CROC-2013](http://mirror.codeforces.com/croc2013) will start. You need to participate in the Qualification Round to make it to Round 1. Contestants who gain a score equal to the 2000-th place finisher score or greater will advance to the Round 1 (also you need to gain positive score). At the Qualification Round you will find a few problems, roughly ordered by the increasing complexity. During the Qualification Round the problems are judged only on pretests and system testing will take place after the end of the Qualification Round (round continues for 48 hours). The pretests do not cover all possible cases of input data, test your pr...

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CROC-MBTU 2012, Elimination Round Good day! Elimination round of the CROC Collegiate Programming Competition for the MSTU Bauman students is going to be here soon. The official championship participants will take part in the competition, all others will be out of the competition in the table of results. The contest is held by the well-known rules ACM-ICPC. Competition duration is 2:00 hours. Supported programming languages are C/C++, Pascal, Java, C#, Python, Ruby, PHP, Haskell, Scala, OCaml, and D. This Round will be the rated for the Div-2 participants, regardless of whether one is the competition championship party or not. For Div-1 participants this round is unrated. Please note that the official participants of the competition are not need to register for this competition they will be registered automatically (with advance registration at [website] (http://www.crocok.ru/championship/)). Good luck to everyone! **UPD.** Competition is completed, thank you very much for your participation! I hope th...

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CROC-MBTU 2012, Elimination Round — Editorial Hello, It's my first Blog post and I decided to dedicate it to [CROC-MBTU 2012, Elimination Round](http://mirror.codeforces.com/contest/245/). I tried to do my best and explain ideas completely. Bear in mind that solutions aren't unique and they're only my ideas. At the end of each explanation, there exists a link to **C++** implementation of problem. I hope this editorial will be useful. Any constructive criticism is appreciated. Please report any bug and/or mistake you encounter here in comments. #### [A. System Administrator](http://mirror.codeforces.com/contest/245/problem/A) Let's define following variables: $A_Reached$ : Number of packets that reached $A$. $A_Lost$ : Number of packets which didn't reach $A$. $B_Reached$ : Number of packets that reached $B$. $B_Lost$ : Number of packets which didn't reach $B$. We iterate over input and update each variable accordingly. Now answer for server $A$ is `LIVE` if and only if $A_Reached \ge A_Lost$ otherwise it's `DEAD`. Also a...

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Open Moscow Programming Championship By CROC [CROC](http://www.croc.ru/eng/about/index.php) invites talented and innovative programmers (at least 18 years old) to participate in the Open Moscow Programming Championship. The championship is held on the basis of the Codeforces project. Saratov State University involved into jury and the technical committee. The official championship language is Russian, but the problems will be in Russian and English. The championship will consist of qualification, two online rounds (on April 3-4, 6 and 20) and the final round that will take place on April, 27 in Moscow in the CROC Headquarters. The championship finals will also include a AI-game round. The championship winners will be awarded with valuable prizes: the first place winner gets 100000 rubles, the second and third place winners get Apple MacBook Pro. The contestant who wins the first place in the AI-game round will be awarded with an Apple MacBook Air. To participate in the championship, you must register on [http://w...

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Codeforces: the Results of 2012 Happy New Year! The New Year is in a two-day distance from us and it's good time to summarize the ending year. In a nutshell, this year's been highly productive for the project. Codeforces has set up large championships, rounds followed one after another, we've launched the Codeforces::Gym project, we've added numerous improvements/fixes/innovations. Here's a short list of our deeds: <img src="http://assets.codeforces.com/2012/2012-codeforces-google-trends-en.png" style="float:right;margin:0 1em 1em 1em;"/> - when a user votes for a post/comment, his/her authority is taken into consideration, we've introduced anti-bot and froud-detection heuristics - the Codeforces::Gym has been opened and has come through many improvements - the Codeforces Markup, based on Markdown has been introduced, accompanied with special Codeforces tags - Codeforces helped VK.com to carry out a amplitudinous VK Cup 2012 championship which had around 6000 participants! - the CROC company has used the...

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The CROC All-Russian Open Programming Championship Results Last Friday, on May 17, CROC finishes the All-Russian Open Programming Championship. The CROC company (together with Codeforces) holds a competition for the best coders of our country for the second year in a row. Informational technologies are a dynamic habitat where the best minds dwell and sometimes innovative ideas come up at a great speed! This time the organizers stated that the championship is all-Russian but they were surprised to see the championship interest extending beyond the national border. <center> <img src="http://lh3.googleusercontent.com/-bHHNR_A6z6k/UZpYlF2ElYI/AAAAAAABDVs/OlIAHRayouI/s560/90520.jpg"/> </center> <i>The interest towards the competition was such that we had many participants from Russia, former Sovyet republics and even far away countries. On this championship, coders solve sports programming problems: to win, they need to come up with effective algorithms, code them quickly, consider all non-typical situations. Besides, there is the additi...

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CROC Champ 2012 — Round 3 (Editorial) This is the editorial for [contest:183] <h3>Remarks</h3> One day, I applied to write another round in Codeforces. To my surprise, I was offered to write CROC instead. I'm quite surprised they offer me (an unknown programmer from a notorious country) to write for such an important match! So I happily oblige --- kudos for the open culture nurtured by Codeforces! I like most the problems of this final round. I think all of them are beautiful in some aspects :). Oh, and all of them emits such concise solutions ;). <h3><a href="http://mirror.codeforces.com/contest/183/problem/A">Problem A</a></h3> Author: [user:dolphinigle,2012-04-25]<br/> Ad Hoc I supppose... This problem is equivalent to calculating the number of reachable locations from point (0, 0) by doing the allowed moves. The naive solution is to process the information one by one from first to last, and to keep track on all the reachable positions. Claim: At any moment in time, all reachable locations as described abo...

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CROC Champ 2012 — Round 3 (Final) [contest:183] **Final Update:** Congratulations to the prize winners! <ul> <li>First place: [user:neex.emil,2012-04-28] --- 100000 rubles</li> <li>Runner Up: [user:Dmitry_Egorov,2012-04-28] --- Apple MacBook Pro 15</li> <li>2nd Runner Up: [user:vepifanov,2012-04-28] --- Apple MacBook Pro 13</li> </ul> Videos for the onsite round (posted by [user:ivan.popelyshev,2012-04-28]): [1](http://www.youtube.com/watch?v=QIRrD2zET3Q) [2](http://www.youtube.com/watch?v=eKEguEMPDDc) [3](http://www.youtube.com/watch?v=I97YzgF-3a4) [4](http://www.youtube.com/watch?v=E-mM5WBri5g) [5](http://www.youtube.com/watch?v=-YvmZvW9MiM) [6](http://www.youtube.com/watch?v=C2T1b37rWxQ) Congratulations to the winners of the mixed group (onsite & offsite)! <br/>[user:ivan.metelsky,2012-04-27] <br/>[user:WJMZBMR,2012-04-27] <br/>[user:neex.emil,2012-04-27] [user:rng_58,2012-04-27] deserves honorable mention for the only competitor to correctly submit both D and E :) [Editorial](http://co...

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Croc Champ 2013 — Round 2 — Editorial Here you can find the solutions for the problems from the past round. You can ask questions in the comments below. **[A Div 2](http://mirror.codeforces.com/contest/299/problem/A)** If $a_i$ divide $a_j$ than $a_i \le a_j$. So the number which will divide every other number should be less than or equal to every other number, so the only possible candidate &mdash; it's the minimum in the array. So just check whether all elements are divisible by the minimal one. **[B Div 2](http://mirror.codeforces.com/contest/299/problem/B)** Easy to see, that Ksusha is unable to complete her journey if there is a sequence of consecutive # with length more than k. **[C Div 2](http://mirror.codeforces.com/contest/299/problem/C) / [A Div 1](http://mirror.codeforces.com/contest/293/problem/A)** The first observation &mdash; we don't care about the actual strings, all information we need &mdash; number of pairs {0,0}, {0,1}, {1,0}, {1,1}. Count that and then just follow the greedy algorithm, for the first player: try...

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CROC announces start of All-Russian Programming Championship CROC-2013! <div style="background-color: #EAF4FF; padding: 2px;"><i>Due to many requests</i> we decided to expand the Championship the target audience, inviting to participate not only the Russians. According to the updated rules, we invite to take part (like last year), everyone! However, please note that all finalists should to confirm invitation until May 2.</div> <img src="http://www.croc.ru/bitrix/templates/croc_redesign/images/en/logo.gif" style="float:left;margin:0 1em;"/> We are growing and spreading! All-Russian Programming Championship CROC-2013 is followed after the Open Moscow Championship and the Championship for the Bauman MSTU Students. The CROC company, with the help of Codeforces and Saratov State U, sets up a competition for the top talented programmers of the country. If you are a university student or a young specialist over 18 (<s>and the citizen of Russian Federation</s>), we invite you to participate. The first prize in the Championship is 100000 rubles, the secon...

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Croc Champ 2012 — Round 1 — Editorial **A.** Naive solution in $O(n)$ (some simulation all of n rounds) gets TL. Let's speed up this solution. Let's consider rounds on segments [1..mk], [mk+1..2mk], [2mk+1..3mk] and so on. You can see that results of games on these segments will repeat. So you can simulate over exactly one segment and then take into consideration them [n/(mk)] times ([x] is rounding down). Remainder of n%(mk) last rounds you can simulate separately. So you have $O(mk)$ solution that easily fits into time limits. Authors are ~Ripatti,2012-04-06 and ~Gerald,2012-04-06 [cut] . **B.** In this problem you should build bipartite graph with n+m vertices. All vertices of the firts part correspond to the rows of the table, vertices of the second part correspond to the columns of the one. Edge connects vertices iff some regular column is placed in the cross of corresponding row and column. Firstly death ray covers only last row of the table that corresponds to one of vertices of our graph. When you are ch...

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Croc Champ 2013 — Final Round — Editorial [Problem A. Morning run](http://mirror.codeforces.com/contest/308/problem/A) -------------------------------------------------------------------- We were asked to find the expected value of meetings between the runners. How to do that? As the first step, expected value is lineal, so we can split the initial problems into the different ones: find the expected value of meetings between the fixed pair of runners. We will solve these problems. To do that we need to make some observations: 1. Let $x$ be the distance between the two runners and they run face to face for infinite amount of time (probability of that obviously equals to $0.5\cdot 0.5 = 0.25$). Then the first meeting will happen at time $\frac{x}{2}$, the next one &mdash; $\frac{x}{2} + \frac{l}{2}$, the next &mdash; $\frac{x}{2} + \frac{l}{2} + \frac{l}{2}$ and so on. 2. Let us assume that every run ran for $l$ time units. Then if two runners meet &mdash; they meet exactly two times. The probability of the meeting equals to ...

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Croc Champ 2012 — Round 1 Good day, Codeforces! Today at [19:00](http://timeanddate.com/worldclock/fixedtime.html?day=6&month=4&year=2012&hour=19&min=0&sec=0&p1=166) the Round 1 of the [ Open Moscow Programming Championship By CROC](http://mirror.codeforces.com/croc2012) will start. This is usual two-hour Codeforces round with hacks and decreasing values of problems. Everybody, who passed Qualification Round and registered for today’s round, has advanced to Round 1. The remain participants are allowed to participate in this Round out of competition. Round will be rated for everyone as usual. Contestants who gain a score equal to the 300-th place finisher score or greater will advance to the Round 2 (also you need to gain positive score). You will find a few problems, roughly ordered by the increasing complexity. Score distribution is standard (500-1000-1500-2000-2500). During the Round the problems are judged only on pretests and system testing will take place after the end of the Round. The pretests do not c...

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CROC Champ 2013 — Round 2 Hello everyone! Round 2 of [All-Russian Programming Championship CROC-2013](http://mirror.codeforces.com/croc2013) will take place today. Round was prepared by [user:sdya,2013-04-22], [user:Seyaua,2013-04-22], [user:Gerald,2013-04-22] and traditionally, problem statements were translated to English by [user:Delinur,2013-04-22]. Good news for people, who didn't qualify to this round &mdash; today everyone can participate out the competition. Additionally, round will be rated for both official participants and out of competition participants. Remind some facts about the official participants: - All the participants should be 18+ years old - The championship finals are going to take place on May, 16-17 in Moscow in the CROC office (50 participants) - The CROC company pays for the accomodation in Moscow during the finals - For Russian citizens: the travel expenses around Russia will be covered, the transport expenses outside Russia can be covered possibly partially but you need...

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Editorial for CROC 2016 Finals and Codeforces Round #347 [problem:664A] -------------- Author of the idea &mdash; [user:GlebsHP,2016-04-17] We examine two cases: 1. $a = b$ &mdash; the segment consists of a single number, hence the answer is $a$. 2. $a < b$ &mdash; we have $gcd(a, a + 1, a + 2, \dots, b) = gcd(gcd(a, a + 1), a + 2, \dots, b) = gcd(1, a + 2, \dots, b) = 1$. <spoiler summary="Code"> ~~~~~ #include <bits/stdc++.h> using namespace std; int main() { string a, b; cin >> a >> b; if (a == b) cout << a; else cout << 1; return 0; } ~~~~~ </spoiler> [problem:663A] -------------- Author of the idea &mdash; [user:gen,2016-04-17] First we check whether any solution exists at all. For that purpose, we calculate the number of positive (the first one and any other with the $+$ sign) and negative elements (with the $-$ sign) in the sum. Let them be $pos$ and $neg$, respectively. Then the minimum value of the sum that can be possibly obtained is equal to $min = (1\,\cdot\,pos -...

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Croc Champ 2012 — Round 2 — Editorial **A div2.** Required row is row that have only one star inside. Requred column is comumn that also have only one star inside. So, you can iterate over all rows/columns, calculate number of stars inside them and find the answer. Authors are ~MikeMirzayanov,2012-04-20,, ~Gerald,2012-04-20 [cut]. **B div2.** Naive solution $O(n^3)$ (where you check all triplets) doesn't fin into time limits. You can see that for every two points from triplet (for example, for A and C) you can find place of the third point. So you can find number of requred triplets just inerate over all pairs of points and check middle point between points in every pair. How to fast check position? You can add 1000 for all coordinates (you can see that this operation doesn't change the answer) and mark them in the 2-dimensional boolean array 2001х2001. So you can check every position for point in $O(1)$ time. Obtained solution works in $O(n^2)$. Authors are ~MikeMirzayanov,2012-04-20,, ~Gerald,2012-04...

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Croc Champ 2013 — Round 1 Good day) Soon is coming Round 1 of the [All-Russian Programming Championship CROC-2013](http://mirror.codeforces.com/croc2013). Contestants who gain a score equal to the 400-th place finisher score or greater will advance to the Round 2 (also you need to gain positive score). Round will be held by usual [Codeforces rules](http://mirror.codeforces.com/blog/entry/456) (with hacks and decreasing values of the problems). During the round the problems are judged only on pretests and system testing will take place after the end of the contest. The pretests do not cover all possible cases of input data, test your programs carefully. Before the end of the round it is strictly forbidden to publish the problem statements/solutions/any thoughts and ideas about them elsewhere. It is forbidden to talk about the problems, discuss the statements, solutions and so on. The problems were prepared by the group of authors: Pavel Kholkin ([user:HolkinPV,2013-04-15]), Gerald Agapov ([user:Gerald,2013-04-15])...

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КРОК: Чемпионат по программированию <img style="float:right; margin: 0 1em 0.5em 1em;" src="http://assets.codeforces.com/images/croc-article-2012/1.png"/> В апреле 2012 года компания КРОК на базе проекта Codeforces провела свой первый открытый чемпионат по программированию, в котором приняли участие около двух тысяч человек из различных городов России. Цель этого проекта --- поощрение самых талантливых и прогрессивных молодых специалистов и стимулирование интереса к сфере информационных технологий. Компания КРОК активно участвует в различных социальных проектах. Особое внимание уделяется проектам, связанным с развитием ИТ-отрасли и продвижением ИТ-профессии. У КРОК имеется богатый опыт сотрудничества с ВУЗами, организации студенческих олимпиад, открытых семинаров и лекций. Этой весной «по мотивам» мировых чемпионатов был организован первый открытый чемпионат КРОК по программированию. Проект оказался успешным, и потому проведение подобных соревнований может стать новой традицией для компании. Стоит отметить, что ...

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Make a Flying Drone together with CROC Being recently in the CROC office, I've heard about another interesting contest. It's called "Flying Robots". I am far from robots, but the subject is interesting and popular. I am pleased to publish a small announcement. ### About contest September 3, 2012. CROC, one of the five largest IT companies in Russia, calls for Flying Drone contest applications. The contest is a part of the '20 Ways to Brighten Up Your Life' project. The first prize will be **1 million** Rubles. Contestants shall make a pilotless and autonomous drone which must take off from a starting point, overcome an obstacle, land within a certain zone, take off again and then come back to the starting point. Please submit your application till November 15, 2012. For contest details please visit [http://robots.croc.ru/](http://robots.croc.ru/) (in Russian). Results announcement and award ceremony will take place in August 2013. ### About CROC CROC is a leader in IT infrastructure design and implement...

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