PraveenDhinwa's blog

Invitation for CodeChef SnackDown Round 1A
By PraveenDhinwa, 10 years ago, In English

CodeChef invites all the programming enthusiasts to take part in the CodeChef SnackDown Round 1A. The contest is a part of CodeChef SnackDown, a multi-round onsite programming contest and we want you all to be a part of it. It is a team contest open to students and professionals alike. The contest will have 3 problems and will be of 2 hours duration. You surely will enjoy cracking them.

All you need to take part in this contest is a CodeChef username. If you do not have one, please register here to participate.

Also, it is a team contest, so you need to register your teams first to take part in the contest. To register your team, register/login through CodeChef and go to: http://snackdown.codechef.com/register/

Date of Contest: 23rd May 2015 Time: 14:30:00 hrs to 16:30:00 hrs IST Check out the local time in your City/Time Zone here.

Contest Details

Prizes
For the current round, Along with top ranked participants, top 10 fastest successful submissions will get cool CodeChef SnackDown Merchandise. Please see details of the prizes of the entire competition here. You can also see schedule and other details on the Snackdown page

Top 500 teams in the round will advance to next Elimination Round. Others will have two more chances to qualify by participating in Round 1B and Round 1C.

Please also note that we will also provide Vietnamese translations in all Snackdown rounds. A big thanks to I_love_Hoang_Yen (Thanh Trung Nguyen) for making the proposal of including Vietnamese translations and providing big help in finding out translators.

We hope to see you all participating in the contest. Good Luck! Hope to see you participating.

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10 years ago, # |
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Please note that start time of the contest is unusual. It will be 9 AM UTC instead of usual 4 PM UTC.

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10 years ago, # |
  Vote: I like it +17 Vote: I do not like it

6001 teams, sure hang website :|

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10 years ago, # |
← Rev. 2   Vote: I like it +7 Vote: I do not like it

Annnd, Codechef is down.

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10 years ago, # |
  Vote: I like it +20 Vote: I do not like it

The problems are blank .. Are they open to interpretation to test your creativity ?

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10 years ago, # |
  Vote: I like it +4 Vote: I do not like it

Typical CodeChef.

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    10 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    I saw the third problem CHEFVOTE. Couldn't read the complete problem as I mistakenly clicked F5. It might be something about graph as I saw the words "bidrectional relation". There were 500 test cases. This is unfair. There might be many who must have gained access to the problems.

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10 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Is it something wrong with registration system? Because I've got the following message: "You do not belong to any team who has registered for SNCK151A.". But I can see my team profile:http://www.codechef.com/teams/view/dmytro

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    10 years ago, # ^ |
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    You have registered after 1:30 pm when the registration for the round was closed, due to which you can't not participate in this round. Please participate in Round 1 B. We are really sorry for inconvenience.

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10 years ago, # |
  Vote: I like it +11 Vote: I do not like it

Do we need separate registration to 1A round? Because I can't submit anything at all...

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10 years ago, # |
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problem links? UPD- got it.

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10 years ago, # |
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What is the solution for DINING? The best I could come up with was a twisted,unproven two part Min-Cost-Max-Flow with log of probabilities as edge weights for first part and difference of log of probabilities for second.

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    10 years ago, # ^ |
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    That's it. You need to have 1 edge with cost -infty from source to each day vertex and K-1 with cost 0 to fulfill condition on 1 meal every day

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10 years ago, # |
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How to solve HistJunk?

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    10 years ago, # ^ |
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    2 vertex to connect with all your vertices and 2 vertices to connect with first 2.

    and also you need to do test with n=3

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    10 years ago, # ^ |
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    Here is my solution.

    Case 1: if n  ≤  2, then answer is 0 0

    Case 2: if n  ≥  4, then use following construction, add n+1,n+2 to all n nodes and n+3 to n+1 and n+4 to n+2. Notice that now d(v) = 2 for all n vertices and 3 or greater for the 4 new vertices we added.

    Case 3: if n = 3, then the above construction adds too many edges. So if n = 3 and m = 3 i.e. cycle then 0 0 is answer else current graph is a tree with 2 leaves. Add 4,5 to 1 leaf and 6,7 to another. Now join 4 and 6 and join 5 and 7.

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      10 years ago, # ^ |
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      Oh, stupid me. Thanks

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      10 years ago, # ^ |
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      Also with n=3 you can do this: if 2 and 3 are leaves, then add next edges: 1-4,2-5,3-5

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      10 years ago, # ^ |
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      How isn't d(v)=1 for n+1 and n+2 as u can travel to any junction from them using one road only? Edit : Got it.. all distances are 1 except for the n+1 to n+2 and n+4 one and so on

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    10 years ago, # ^ |
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    Добавляем 4 вершины и 2n + 2 ребер. Соединяем все вершины с n + 1 и с n + 2. А также n + 1 с n + 3 и n + 2 с n + 4. Тогда для v — старой вершины d(v) = 2, а для новых 3 или 4. Рассматриваем отдельно случаи при n ≤ 3.

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10 years ago, # |
  Vote: I like it +3 Vote: I do not like it

Tasks were pretty hard.I could solve only the first problem...

Where I can find who passed in elimination round ?

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