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
- Contest link: http://www.codechef.com/SNCK2015
- Problem Setters: CherryTree (Sergey Kulik) and PraveenDhinwa (Praveen Dhinwa)
- Problem Testers: kevinsogo (Kevin Charles Atienza) and Piyush Kumar
- Editorialist: kevinsogo (Kevin Charles Atienza)
- Translators: CherryTree (Sergey Kulik) (Russian), MinakoKojima (Minako Kojima) (Mandarin), ptnk8554517 (Tran Phan Anh Khoa) and duykhanhuet (Khanh Duy Nguyen) (Vietnamese).
- Contest Admin: PraveenDhinwa (Praveen Dhinwa)
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.
Please note that start time of the contest is unusual. It will be 9 AM UTC instead of usual 4 PM UTC.
6001 teams, sure hang website :|
Annnd, Codechef is down.
The problems are blank .. Are they open to interpretation to test your creativity ?
Typical CodeChef.
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.
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
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.
Do we need separate registration to 1A round? Because I can't submit anything at all...
problem links? UPD- got it.
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.
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
Ah. That is a nice method.
How to solve HistJunk?
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
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.
Oh, stupid me. Thanks
Also with n=3 you can do this: if 2 and 3 are leaves, then add next edges: 1-4,2-5,3-5
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
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.
Tasks were pretty hard.I could solve only the first problem...
Where I can find who passed in elimination round ?
http://www.codechef.com/rankings/SNCK151A
I know results, but rules said maximum top 100 non Indian teams go to elimination round. I don't see standings for non Indian teams...