Yandex.Algorithm 2013: qualification round
By lperovskaya, 11 years ago, In English

Hi!

Qualification round of the Yandex.Algorithm competition that will start on July 8, 2013, at 7 p.m. This round is virtual, so you need to start your qualification within 24 hours from the beginning of the qualification round. Qualification will last 100 minutes even if you will start at 6:59 p.m. Please, do not discuss the problems until 8:40 p.m. July 9 — someone can be still solving the problems.

The 2,000 best contestants who have solved at least one problem will advance to the elimination stage. Already advanced from test round participants will have a chance to work with the TCM/Time.

Good luck! Almost 200 t-shirts and more than half a million rubles are waiting for you!

Registration

Link to the contest

UPD: Archive is at the competition site.

  • Vote: I like it
  • +62
  • Vote: I do not like it

»
11 years ago, # |
  Vote: I like it -14 Vote: I do not like it

Qualification will last 100 minutes even if you will start at 6:59 p.m. Please, do not discuss the problems until 8:40 p.m. July 9 why same as start 4:00 p.m someone can be still solving the problems. You need not type did not :D

→ Reply
»
11 years ago, # |
  Vote: I like it -10 Vote: I do not like it

Is pascal allowed? because this is the only programming language that i can use.

→ Reply
»
11 years ago, # |
  Vote: I like it +12 Vote: I do not like it

Maybe I missed something, but how can you prevent someone from cheating, like, creating two accounts, opening the problems with one account, solving them, and then submitting them with the other account later?

→ Reply
  • »
    »
    11 years ago, # ^ |
      Vote: I like it -11 Vote: I do not like it

    This does not work, actually. The quota is 2k, you can write at most 5 different solution during the contest, so you can not make a significant impact on results.

    → Reply
  • »
    »
    11 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    We have a system that finds similar solutions and presents suspicious pairs of solutions to the judge.

    → Reply
    • »
      »
      »
      11 years ago, # ^ |
        Vote: I like it +28 Vote: I do not like it

      No, I mean using the first account to read the problems in order to gain time advantage.

      → Reply
      • »
        »
        »
        »
        11 years ago, # ^ |
          Vote: I like it +2 Vote: I do not like it

        so true. Also, if you do bad in your allocated 100mins, you can just register a new account and try again. so basically, the qualification round is 24hrs.

        → Reply
        • »
          »
          »
          »
          »
          11 years ago, # ^ |
            Vote: I like it +63 Vote: I do not like it

          Basically, most participants do not think about any cheating, cause that makes no sense — if u are not able to qualify to 2000 in fair manner, your chances to go to final are equal to 0.

          Also, I suppose cheaters will be banned for good and all.

          → Reply
          • »
            »
            »
            »
            »
            »
            11 years ago, # ^ |
              Vote: I like it +40 Vote: I do not like it

            most participants do not think about any cheating

            Not sure if you have ever organized online contests, but from my experience I can say that there will as many cheaters as rules will allow. And for this competition virtual contest seems like a poor idea. Why not make it 24 hour long contest like google does?

            → Reply
      • »
        »
        »
        »
        11 years ago, # ^ |
          Vote: I like it +45 Vote: I do not like it

        Just don't do it — it's that easy. Almost as easy as to go into the store and don't steal anything.

        → Reply
»
11 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Good luck to everybody!

→ Reply
»
11 years ago, # |
← Rev. 2   Vote: I like it +8 Vote: I do not like it

a

→ Reply
»
11 years ago, # |
  Vote: I like it +8 Vote: I do not like it

How is the contest system supposed to work:

  1. Should someone who advanced from test round be able to participate in this round?
  2. Should everyone be able to see the current standings?
→ Reply
»
11 years ago, # |
  Vote: I like it -20 Vote: I do not like it

Yandex has a foul contest scheduling. Totally insane.

→ Reply
  • »
    »
    11 years ago, # ^ |
    ← Rev. 2   Vote: I like it -10 Vote: I do not like it

    i support yandex! its a good encouragement for all.

    (mistook in last comment)

    → Reply
  • »
    »
    11 years ago, # ^ |
      Vote: I like it -25 Vote: I do not like it

    They exposed themselves. Go home yandex, u r drunk!!

    → Reply
  • »
    »
    11 years ago, # ^ |
      Vote: I like it -16 Vote: I do not like it

    By unliking this comments you guys are proving nothing went wrong. Huh!! Just for a t-shirt the insanity scale is rising above. Go on, we are being amused by this as more effectively as u unlike more.

    → Reply
»
11 years ago, # |
← Rev. 3   Vote: I like it +11 Vote: I do not like it

How many problems are there in the contest? It would be very kind if you answer and only after that put a minus on this comment, because there are already 3 people who could answer me but they didn't.

→ Reply
»
11 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Could you enable us to see the test cases, like in codeforces? If not, please put them for download (but ofcourse after the contest ;) ). It would be very great if you did that! :) Thanks!

→ Reply
»
11 years ago, # |
  Vote: I like it 0 Vote: I do not like it

i think my solution on problem B was better than one int the site; we know that k>=4 . if k>=4 and k^2=p1*p2+1 than k must be even and k-1 and k+1 must be primes. at first i use eratostenes sieve and than iterate (k=4;k<=n; k+=2;) if(k-1 and k+1 prime)print k;

please coment it's time complexity.

→ Reply
  • »
    »
    11 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    why is this if k>=4 and k^2=p1*p2+1 than k must be even and k-1 and k+1 must be primes,I dont understand the theorem you used.can you please explain...

    → Reply
    • »
      »
      »
      11 years ago, # ^ |
      ← Rev. 2   Vote: I like it +3 Vote: I do not like it

      IMHO, k^2=p1*p2+1 => k^2-1=p1*p2 => (k-1)*(k+1)=p1*p2 => (k-1)=p1, (k+1)=p2. But it's wrong. UPD1: I was wrong.

      → Reply
      • »
        »
        »
        »
        11 years ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        It isn't wrong. p1,p2 are prime so
        k-1 = p1, k+1=p2 OR k-1=1, k + 1 = p1*p2 = 3 but that's impossible

        → Reply
      • »
        »
        »
        »
        11 years ago, # ^ |
          Vote: I like it +6 Vote: I do not like it

        It's correct.
        There are 4 cases:
        1) k — 1 = 1, k + 1 = p1 * p2 => k = 2 , p1 * p2 = 3 — impossible
        2) k — 1 = p1 * p2 , k + 1 = 1 — impossible
        3) k — 1 = p1 , k + 1 = p2
        4) k — 1 = p2 , k + 1 = p1 — equal to 3)

        → Reply
        • »
          »
          »
          »
          »
          11 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          You are right. I was wrong.

          → Reply
»
11 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Will you make editorial as the test round you make a very good one? :)

I want to understand the C (Board Game), I have lot of try (I know the nim..) but my last try got TLE, so I download the solutions, but I didn't get it. The solution is very short, exists some short explanatation? (Or just should realize this pattern?)

UPD: Sorry my comment, I found the analysis .

→ Reply