We’d like to invite you to take part in the qualification round of the Yandex.Algorithm competition that will start on May 25, 2014, at midnight. You need to start your qualification within 24 hours from the beginning of the qualification round. Qualification will last 100 minutes.
To advance to the next stage of Yandex.Algorithm you will need to solve at least one problem during your qualification.
Participants, already advanced to the elimination stage from warm-up round, are also welcome on the qualification.
Good luck!
I have a question:
From the Rule, "If a problem is solved with a “blind” submission, the total penalty time is decreased by n seconds, where n = ((number_of_contestants_who_did_not_solve_this_problem) × (contest_length_in_seconds)) ÷ (total_number_of_contestants)."
Who will be counted as "number_of_contestants_who_did_not_solve_this_problem" and "total_number_of_contestants"? I wonder those who are eligible for the round but not participate in the round at all will be counted or not.
It includes all people who have made a submission (even get 0 point in the end).
You can confirm this by calculate on this standing.
thanks!
can someone explain how to solve the problems ?
I hope there are editorial after contest has ended
This webpage has a redirect loop :(
Try opening the page in an incognito window if you r using google chrome
thanks Shawkat
Will the coming rounds be held and rated on Codeforces? I mean there was a previous round on CF in 2011...
sadly I missed it :( :(
Actually, you have about 5 hours to start it
BIG Thanks I thought it's saturday not sunday ,Thanks again
Ah! it needs further registration, I thought codeforce handles are enough. Too bad!
Is there any way to see other participants solution after the contest?
Why cant we resubmit blind submissions?
Because it is no more blind. You have already been told that what you initially thought was incorrect, and you have been provided a chance to correct it.
Who told me?:D
What?
It does not say after the submission if it has passed the tests or not (only sample tests are checked).
What's the point of blind submission now?
Do you want to check that your submission passes the tests from the problem statement? That is something you can do manually.
while regular submissions are being tested immediately.
No and this is why the result of blind submission is useless.I think I didn't got you correctly. Were you asking why yo can resubmit blindly during the contest? I was thinking about upsolving due to some reason. If your question is about the contest then I think the reason is to add more drama, you should be sure about first time you submit it.
In the final standings, the + indicates normal submission, while a tick indicates blind?
Yes.
Is there any editorial for the contest ??
Of course not, its not even over yet.
I don't understand, the qualification round is not finished ??
It's a "virtual" contest. You can start your participation in any moment between it's start and finish time. At the end, results of the participation of all participants will be merged and we will have final standings.
got it now, thank you :D
Contest already finished.
Actually, no. What's over is the time window when one can start a virtual contest. A participant can start at 23:59 MSK and still have 100 full minutes to compete.
The real end time is 01:40 MSK.
)
I put together a short editorial in case someone needs it: http://www.rivsoft.net/blog/yandex-algorithm-2014-qualification-round/
tourist solved all problems in 41 minutes! That is insane!!
They were mostly implementation problems, though.
Damn, why no any reminding email :(!? I missed such a good contest, because I missed qualification round :(.
Did you participate in Warm Up round?
No, I didn't have time to (lot of work to studies) and since it was "Warm Up" I assumed that problems presumably won't be amusing. Were reminding e-mails sent to Warm Up participants only? If so, that isn't a very clever idea.
It is pity, because Warm Up round was also an opportunity to qualify for main rounds (in the same way as qualification round — by solving at least one problem).
Oh, I didn't know :P.
Seriously, why no email reminder? I missed it too, it was a really good contest and I wanted to participate :( :(
I believe the statement of problem F was incorrect.
Currently the statement says: "The sequence starts with an opening round bracket “(”, ends with a closing round bracket “)” and holds only English letters (one or more) in between the brackets. Examples: (blush), (facepalm), (SmileSmileSmile)."
However, there was no "(one or more)" in the beginning of the contest. Without "(one or more)", I think "()" satisfies this condition (and I guess this is the reason of so many failures).
We considered to retest all failed on test 4 submissions before fix with "one or more" was added with understanding when () is a smiley. Retesting will be done soon.
Update: Retesting complete; all solutions failed on test 4 before statement fix were rejudged.
But anyway, i am not sure if there is no difference between "holds only English letters" and "does not contain any characters, other than English letters" (something like that must be used for set of English words + empty set). For the first version of statement more likely string must be non-empty
Thank you!
C# solution for F:
In java:
out.println(in.readLine().replaceAll("(\\([a-zA-Z]+\\)|[^a-zA-Z0-9 ()]\\))", "<S>$1</S>"));
in perl:
s/(([a-zA-Z]+)|[^ a-zA-Z0-9()]))/<S>$1<\/S>/g, print while (<>)
What was the approach to solve problem E?
Change the sign of all elements in odd positions. Now you need to find the segment with the largest sum (in absolute value). If you calculate partial sums (from the start of the array to the current position), then the sum of any segment is the difference of two partial sums (corresponding to the segment's start and end). So all you need to do is go through the array once, keeping track of the current sum, and the minimum/maximum value of this sum so far.
The rules say that the participants with the highest cumulative scores will advance to the finals. Does this mean that we have to participate in all 3 rounds? They are scheduled in such a way that in any time zone at least one contest will be in the middle of a night.
It doesn't — you can participate however you want. Still, the more you participate, the better your chances, since once excellent result gives you much more than 3 good ones.
That's good to know. The rules say it uses total points, number of solved problems and total penalty, but doesn't say which formula is used, so I can't tell how important it is to participate in all 3 rounds. Also, I thought the first round was yesterday, guess I'll wait for it today then as its the only one that happens at night for me.