By lperovskaya, 6 years ago, translation, ,

June 27 is an important date – that’s when the test round of Yandex.Algorithm kicks off. Don’t miss your chance to register, so you can be in the running for a place in the final round, which takes place on August 21-23, 2013 in St. Petersburg. The final round venue is truly unusual for such an IT event – the palace of Grand Duke Vladimir, built in 1870, will welcome finalists with all its historical charm.

In the final round, the best participants will be left one-on-one with the testing system – no internet connection or prewritten code, only properly set-up computers and IDEs. We will provide text coverage of the round.

If you want to be part of the action, don’t leave your registration to the last minute – remember, the test round starts tomorrow.

More good news: we decided to almost double the number of T-shirts. The best 75 non-finalists among participants of the elimination stage, another 75 random participants of the elimination stage who successfully submit a correct solution to at least one problem and, of course, all finalists will receive an exclusive Yandex.Algorithm T-shirt.

UPD: The test round is available for participants now: http://algorithm.contest.yandex.com/contest/306.

Current standings are open for everyone: http://algorithm.contest.yandex.com/contest/306/standings.

UPD2: Official results are avaliable on the official website

UPD3: Round archive is avaliable at the official website

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 » 6 years ago, # |   +10 is this contest for Russian only ?
•  » » 6 years ago, # ^ |   +6 Not at all, statements and the official web site are avaliable in English as well. You can switch the language using button on the bottom of any page of official website or during the registration process.
 » 6 years ago, # |   +2 I think I'm encountering 404…
 » 6 years ago, # |   +3 I am very glad to see that organizers decided to provide more T-shirts as prizes. I think it will greatly enhance the enthusiasm of contestants.
 » 6 years ago, # | ← Rev. 2 →   +5 registration page is in russian .......how .i can register and there is no link..4 english ver
•  » » 6 years ago, # ^ |   +5
•  » » » 6 years ago, # ^ |   +7 thanks...
 » 6 years ago, # |   +6 Where is the contest link?
•  » » 6 years ago, # ^ |   0
•  » » » 6 years ago, # ^ |   0 You mean contest will be held on this link??
•  » » » » 6 years ago, # ^ | ← Rev. 3 →   -8 I believe it will be held here on Codeforces but I'm not 100% sure because it doesn't mention anything on their website.
•  » » » » » 6 years ago, # ^ | ← Rev. 2 →   -8 It's been updated here's the link. http://algorithm.contest.yandex.com/contest/306/
•  » » » » » » 6 years ago, # ^ |   -8 thank you!
 » 6 years ago, # | ← Rev. 2 →   +1 2337 submission for problem A only 109 got Accepted !!!most people had precision errorseven I used a user-defined function for comparing that didn't work long double aabs(long double a){ if(a>0){ return a; } return -a; } long double EP=0.0000001; bool is_equal(long double a,long double b){ if(aabs(a-b)
•  » » 6 years ago, # ^ |   +1 Maybe someone could comment how to solve A? Especially those, who got it right at the first attempt.
•  » » » 6 years ago, # ^ | ← Rev. 3 →   +5 , so you can just compute fractions with integer numerators and denominators and find the fraction with the most occurrences in the multiset. (I've got it right at the second attempt, but only because I used stdin/stdout instead of files in my first submission).
•  » » » » 6 years ago, # ^ |   +3 do you mean R=(abc)/(4*S) or R=(4*abc)/(S)
•  » » » » » 6 years ago, # ^ |   +2 Of course , it was a small typo.
•  » » » » » » 6 years ago, # ^ |   +3 you didn't fix the other fraction in your comment :)
•  » » » » » 6 years ago, # ^ |   0 It does not matter for solving :)
•  » » » » » » 6 years ago, # ^ |   0 actually yes it's like you multiplied all fractions with 16
•  » » » » 6 years ago, # ^ | ← Rev. 3 →   0 a and b and c themselves may not be integers how can we save the numerator in integer variable ?Edit: oops got it now , also S may not be integer but 16*S is always an integer
•  » » » » » 6 years ago, # ^ | ← Rev. 2 →   0 a2 = (x2 - x1)2 + (y2 - y1)2, and so on. (2S)2 = ((x1 - x0)·(y2 - y0) + (x2 - x0)·(y1 - y0))2.
•  » » » » 6 years ago, # ^ |   0 But numerator's maximum value can be  (1400)^6 which will cause overflow even for unsigned long long int. How to store the value?
•  » » » » » 6 years ago, # ^ |   0 2^63 is around 9*10^181400^6 is around 7*10^18
 » 6 years ago, # |   0 Can i have the test case 9 for problem A?
•  » » 6 years ago, # ^ |   0 Tests are usually not disclosed in this type of competitions. It decreases motivation to prove your theoretical solution and test/debug your code.
•  » » 6 years ago, # ^ |   +21 We decided to publish the competition archive, you can find it at the competition site
 » 6 years ago, # |   +7 how to solve D?
•  » » 6 years ago, # ^ |   -11 div1 or div2 ? XDonly joking
•  » » 6 years ago, # ^ |   0 I was thinking about some kind of Maximum Knapsack with an array that keeps track of objects used( in this case numbers used) to get to the number, but I didn't complete it on time, and I am not sure if the idea is valid. A problem was the huge n.
•  » » 6 years ago, # ^ |   0 Seems that my code will describe my solution much better than me.
 » 6 years ago, # |   +1 Will there be an editorial for this contest?
•  » » 6 years ago, # ^ |   +1 Yes, there will be.
 » 6 years ago, # |   +16 HiI'm advanced, but can I participate in the Qualification round?
•  » » 6 years ago, # ^ |   0 Yes, you can
 » 6 years ago, # |   +34 seems awesome but what is Yandex
•  » » 6 years ago, # ^ |   +9 It's the 4th largest search engine in the world, surpassing Bing recently by the number of searches and succesfully competing with Google in Russia for many years. You can read more here and here and here
 » 6 years ago, # |   +16 Problem E: submitted a program that outputs nothing. Accepted.
•  » » 6 years ago, # ^ |   0 By some odd reason after the test round checker type was set wrongly for this problem. Fixed, rejudged all upsolving submissions.