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unstoppable_N's blog

CoDeft 2.0, organised by CP Club, AIT Pune
By unstoppable_N, history, 2 years ago, In English

Hello Codeforces,
Competitive Programming Club, AIT Pune is excited to invite you to CoDeft 2.0 which will take place on Monday, April 18, 2022 at 21:00 IST. This contest will be an ICPC styled contest to provide an exciting learning platform to showcase your CP skills.

Many thanks to all the people who made this contest possible:

This event will be held in 2 stages:

Qualifiers

Finals

  • Date: Saturday, April 30, 2022
  • Participation: Individual
  • Mode: Onsite, Top 100 participants from online qualifiers will be called for onsite finals at AIT, Pune.
  • Participants reporting for onsite finals will have three events CodeRed, Bug-Off and Retracer to take part.

Prizes and Goodies

Qualifiers

  • Top 25 Indian participants will receive goodies, delivered to their addresses.
  • 5 Random Indian participants from the top 300 will receive goodies, delivered to their addresses.
  • 5 Random 1st and 2nd-year students from AIT will receive goodies.

Finals

  • All the onsite finalists will receive a CoDeft T-shirt + Goodies.

  • Event Winner Runner Up
    CodeRed 10,000/- INR 6,000/- INR
    Retracer 5,000/- INR 3,000/- INR
    Bug-Off 5,000/- INR 3,000/- INR

  • Goodies for the top 7 participants on the leaderboard in all the 3 events separately.

Don't forget to register to be eligible for goodies and prizes — Link
Join Discord Server for regular updates — Link

codeft-codeforces

Good luck, see you on the leaderboard!

UPD : Contest starts in 15 mins. All the best!!

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2 years ago, # |
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Auto comment: topic has been updated by unstoppable_N (previous revision, new revision, compare).

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2 years ago, # |
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Where will I get the link for the contest?

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    2 years ago, # ^ |
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    You can find the contest link in the blog. Also, don't forget to register before taking part in the contest.

    Contest Link: Link
    Registration Link: Click

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2 years ago, # |
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Auto comment: topic has been updated by unstoppable_N (previous revision, new revision, compare).

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

Reminder : Online Qualifier starts in 15 mins. Hoping to see you all in the leader-board.

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2 years ago, # |
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tfw when u are doing multiset.erase(variable) for 45 minutes

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2 years ago, # |
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Was F related to matrix exponentiation?

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    2 years ago, # ^ |
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    Yes, just you have to take care that if you have no moves, the number of ways will be 0, as stated in the announcement....

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      2 years ago, # ^ |
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      Hey, why is my solution for F giving WA? I have handled the case as told in the announcements.

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        2 years ago, # ^ |
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        I have gone through your solution, it was stated in the announcement that when the moves will be 0, the answer will be 0. So when the home node is not connected to any other node , you can not perform any move so the answer will be 0, but your code is giving 1.

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      2 years ago, # ^ |
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      when was this announced? am I the only one who missed it? very unusual corner case.. (´。＿。｀)

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        2 years ago, # ^ |
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        Infact, initially in constraints, K was > 0. They changed it later.

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2 years ago, # |
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Why is the form not accepting responses?

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2 years ago, # |
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Was D related to some kind of dp?

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    2 years ago, # ^ |
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    Yes....there are five states x, y, dx, dy and turn(alice or bob)

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2 years ago, # |
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In country planning, we just had to find the diameter of each component, sort them in reverse order and then had to make 4 cases right? cnt(no of components)= n, 1,2 and >2 for cnt == n ans=1, for cnt == 1 ans = (d1+1)/2, for cnt == 2 ans = (d1+1)/2 + (d2+1)/2+1, for cnt > 2 ans = max((d1+1)/2 + (d2+1)/2+1,(d2+1)/2 + (d3+1)/2+2). where d1, d2 and d3 are the largest diameters(d1>d2>d3)

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

    When the count of the components is n then
    ans=0 if n==1
    ans=1 if n==2
    ans=2 if n>=3 (because the graph can be made like a star graph) What is Star Graph

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      2 years ago, # ^ |
      ← Rev. 4   Vote: I like it +8 Vote: I do not like it

      Here n is no. of nodes or no. of components?

      If number of components is 1 and we have more than 2 nodes isn't the component just a tree which already has defined edges? how can we convert it to star graph.

      And if you meant n is count of centers of the components, I did the same case work as you but we need to take account of the diameters of largest components as well right for end points of each component?

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        2 years ago, # ^ |
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        When the number of the components is n, it means there is no edge (m=0), then the answer can be greater than 1 , but in your previous comment you mentioned that is the number of components is n , the answer would be 1.

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          2 years ago, # ^ |
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          Sorry I misread your previous comment. Got my mistake. Thanks!

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2 years ago, # |
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Looks like E was this problem verbatim and F was this but with implementation issues in the intended solution. I wonder how the sole person who got AC on F solved it, since most people were getting multiple WAs (it was somewhat suspicious too).

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    2 years ago, # ^ |
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    I studied a few codes and found that everyone missed this corner case, please have a look at this comment Comment Link

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      2 years ago, # ^ |
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      Still ambiguous. I had a couple of submissions where I ignored self-loops, and it really doesn't make sense to ignore self-loops for the home vertex but not for the remaining vertex.

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