Utkarsh.25dec's blog

By Utkarsh.25dec, history, 3 weeks ago, In English

We invite you to participate in CodeChef’s Starters 13 this Sunday, September 26th, rated for both division 2 & 3 participants.

Time: 4 PM — 7 PM IST.

Joining me on the problem setting panel are:

The video editorials of the problems will be available on our YouTube channel as soon as the contest ends. Subscribe to get notifications about our new editorials.

Also, if you have some original and engaging problem ideas, and you’re interested in them being used in CodeChef's contests, you can share them here.

Hope to see you participating.

Good Luck!

 
 
 
 
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3 weeks ago, # |
  Vote: I like it +22 Vote: I do not like it

I'm really excited to participate in it. Wish everyone a good luck!

Happy cheating coding.

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3 weeks ago, # |
Rev. 2   Vote: I like it +5 Vote: I do not like it

Wishing negative delta for cheaters

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    3 weeks ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    This is not possible in codechef :)

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3 weeks ago, # |
  Vote: I like it +17 Vote: I do not like it

please add plagiarism system like codeforce in codechef there are not many so many cheater in codechef.

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3 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

Hey, Can You coordinate Timings of Codechef Contest and ABC ? I want to give both(sed)

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3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Great! Good luck everyone!

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3 weeks ago, # |
Rev. 2   Vote: I like it +7 Vote: I do not like it

Maybe its not the first time a CodeChef Starter is clashing with Atcoder Beginner contest.. It would be great if codechef admins set the time so that these won't happen again as I wanna participate in both rounds -_-

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    3 weeks ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    The Starters held on the last Sunday of the month has always been from 4PM IST to 7pm and is announced a month in advance. We will coordinate with AtCoder to avoid clashes.

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      3 weeks ago, # ^ |
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      when will be the editorial will added with the problems ?

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3 weeks ago, # |
  Vote: I like it +17 Vote: I do not like it

i think the codechef admins should think upon decreasing duration to 2hour , this will not only reduce cheating but will also reduce clashes!

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3 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

Can the contest be reschduled so that it doesn't clash with Atcoder contest ?

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3 weeks ago, # |
  Vote: I like it +8 Vote: I do not like it

I wonder what explanation do you guys have for this

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3 weeks ago, # |
  Vote: I like it +2 Vote: I do not like it

Are there prizes?

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3 weeks ago, # |
  Vote: I like it +16 Vote: I do not like it

Why did you leak the solution of prefix as a substring in youtube?

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3 weeks ago, # |
Rev. 2   Vote: I like it +31 Vote: I do not like it
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3 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

What is Expected time Complexity Of (MAKE ARRAY) ? As My solution is Nlog(n) . Still Giving TLE in last 3 test cases. If Possible write down your Approach

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3 weeks ago, # |
  Vote: I like it +9 Vote: I do not like it

Due to an unfortunate manual error from our end, the video editorial for SSTRPREF was made public for a while during the contest. Due to this, we are forced to make this problem non-scorable. We will recalculate the rankings and ratings accordingly soon. We apologise to the 6 users who are affected by this issue.

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3 weeks ago, # |
Rev. 3   Vote: I like it +1 Vote: I do not like it

can anyone explain a solution of Interesting Pairs ?

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    3 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Complete Editorial is yet to be posted.

    For now you can go this hint:

    Represent each value by (i , A[i]) .

    reduce (i , A[i]) to the lowest form — (a , b).

    a represents that the value needs a factor a, and b represents that it can provide b.

    m[a][b] — no of indices for which a is available, but it requires b.

    Now, if I encounter (a , b) — ans += m[a][b’] for all b’ which are factors of b.

    And m[b’][a]++ , for all b’ which are factors of b.

    Complexity — (N * factors of N * log) = O(N ^(4/3) * log (N^(4/3))