CodeChef_admin's blog

Invitation to CodeChef February Lunchtime (Rated for all) — Today
By CodeChef_admin, history, 2 years ago, In English

Update: The contest has been postponed to tomorrow — 1st March. Apologies for the inconvenience.

We invite you to participate in CodeChef’s February Lunchtime, this Monday, 28th February, rated for all.

Time: 8:00 PM — 11:00 PM IST

Joining me on the problem setting panel are:

Note that this Lunchtime will not have Subtasks.

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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Comments Comments (40)
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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Now there are 4 divisions, is the prize structure changed or still we get 750rs if come in the top 200 for the first time in div 2,3,4?

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2 years ago, # |
  Vote: I like it +6 Vote: I do not like it

There seems to be some issue with AWS. We are postponing the contest by 30 minutes. Apologies for the inconvenience.

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2 years ago, # |
  Vote: I like it -41 Vote: I do not like it

20:20 IST: There still seems to be some lingering issues with AWS. Hence to be safe, we are postponing the contest to 20:00 IST tomorrow — 1st March, Tuesday.

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    2 years ago, # ^ |
      Vote: I like it +34 Vote: I do not like it

    February Lunchtime happening in March. Who's cooking?

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2 years ago, # |
  Vote: I like it -12 Vote: I do not like it

Contest is Postponed to tomorrow....

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2 years ago, # |
  Vote: I like it +35 Vote: I do not like it

Codechef back to being codechef....

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can't participate tomorrow. Do I have to do something to unregister?

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    2 years ago, # ^ |
      Vote: I like it +2 Vote: I do not like it

    I don't think Codechef even has a registration button anywhere. Unless something has changed recently you won't be counted as having participated in the contest unless you make a submission.

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2 years ago, # |
← Rev. 3   Vote: I like it +21 Vote: I do not like it

DELETED

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

CodeChef_admin You should inform us 8: p.m itself. You don't know how much time we have wasted for this. Codechef back to being codechef...

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

24 hours delay! Thanks for the amusing maneuvering.

I intended to participate in tomorrow's Kotlin contest at CF. So, it seems that I won't be able to participate in the postponed Lunchtime contest.

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

So will I give february lunchtime or march lunchtime tomorrow? Lol.

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

Hey listen please! First of all there are only 2 3 contests for div2 and only 2 for div1 and that also are conducted like a gully-cricket match. If you are taking all of us lightly so will we. If you knew this, you could have informed this earlier. We are being informed when timer stops. Codechef is not a small college level website that is not checked/ monitored regularly. You must have checked this earlier, What were you doing then?? I can't tolerate now. I am boycotting your website for this gully-level management.

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    2 years ago, # ^ |
    ← Rev. 3   Vote: I like it +32 Vote: I do not like it

    I do not think that we should blame CodeChef for trying to secure the proper operation of the web site during the contest.

    Temporary issues in Amazon Web Services (AWS) can happen at any time. Internet issue happened right before few CF contests before, and forced the CF team to postpone the contest shortly before the start time.

    No big deal! Take a deep breath, and have a little smile.

    Best wishes

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      2 years ago, # ^ |
        Vote: I like it -18 Vote: I do not like it

      Ok

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

      Their is a big difference between "sometimes"(as you said w.r.t. codeforces) and "often"(what codechef do).

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        2 years ago, # ^ |
        ← Rev. 4   Vote: I like it 0 Vote: I do not like it

        I agree with you. CodeChef team should be able to let us know how many times these internet issues happened during the past few years, even without comparing this number to other competitive programming internet platforms. The team should be able also to endeavor for improving the reliability of the on-line services during short-time contests to world-level standard.

        Just keep in mind that the contest is a competitive programming sport event, no more no less.

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2 years ago, # |
  Vote: I like it +15 Vote: I do not like it
Everytime when I feel Codechef is good:
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2 years ago, # |
  Vote: I like it +19 Vote: I do not like it

Everyone be like to codechef :

Meme
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2 years ago, # |
  Vote: I like it +88 Vote: I do not like it
Meme
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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

How will prizes be given if there are multiple users with same points?

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    There is a tie-breaker, so prizes will be given depending on that:

    "Those who achieve the score first will be placed higher in the ranklist in case of a tie."

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

Its strange that there were no sub-tasks!!!

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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

(VISEMALL) Do the following conditions (any one) guarantee existence of hamiltonian path in $$$[L, R]$$$?

  1. $$$L = R$$$

  2. $$$X =L$$$ and all elements in $$$[L, R]$$$ are same.

  3. There is at least one pair of unequal elements in $$$[L, X]$$$.

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    No, it's not enough.
    L=1,R=5, X=2
    00110
    2->5->3->4->1.

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Not sure here, but according to me, these should be the conditions(assuming we have a 0 at x) : let's denote by i and j the leftmost and rightmost occurrences of 0

    1)if L = x, and all elements in [L,R] are same

    2)if i = x, then there should be at least one '1' in [L,j], otherwise there should be at least one '1' in [i,j]

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2 years ago, # |
  Vote: I like it +5 Vote: I do not like it

does someone have some good proof to COOKPERM, I just assumed that this was a good place to use gcd.

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    2 years ago, # ^ |
    ← Rev. 3   Vote: I like it 0 Vote: I do not like it

    Lets denote cycle from first array by $$$(1,2,...,n-1,0)$$$ and cycle from second array by by $$$(1,2,3,....,m-1,0)$$$, Let $$$g=gcd(n,m)$$$.

    We can see that on the ith step we will be at $$$(i \%n,i\%m)$$$. We can now see that start from $$$(0,0)$$$ we will reach back to $$$(0,0)$$$ only when $$$i$$$ is a multiple of $$$g$$$ because on each step $$$ (i \%n) \% g = i \%g$$$ and $$$ (i \%m) \% g = i \%g$$$.

    This gives the lower bound. We can also see that cycles starting from $$$(j,0) \forall 0 \leq j \lt g$$$ are disjoint.

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2 years ago, # |
  Vote: I like it +14 Vote: I do not like it

Few Hints regarding Problem Magical Modulo, would be very much appreciated.

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    2 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it
    Hint 1
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      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Can you please link your submission . I tried something similar and got TLE.

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

    Let us try to limit the possible values of X(the modulus number).sum=initial sum of the array, p_sum= sum of the permutation from 1 to N. If we are able to convert the initial array to a permutation by applying some modulo operation X, then this condition should hold true sum%X==p_sum%X, right? or in other words X*floor(sum/X)+residue=sum and X*floor(p_sum/X)+residue=p_sum. If we simplify it more, X*(floor(sum/X)-floor(p_sum/X))=sum-p_sum, from the above equation we can clearly see that X should be a factor of sum-p_sum.So it will be sufficient for us to consider only such X, such that X divides (sum-p_sum) After that, you just need some minor optimizations.

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      2 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Cool, that's a much better way to go about it. Never thought about it like that.

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    2 years ago, # ^ |
    ← Rev. 4   Vote: I like it +3 Vote: I do not like it
    Hint 1
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2 years ago, # |
  Vote: I like it 0 Vote: I do not like it

A request to code chef to please let us see the test cases after the contest.

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2 years ago, # |
  Vote: I like it +1 Vote: I do not like it

CodeChef_admin Please look at the cheating happening in the codechef starters . The submissions for travelling in the graph increased exponentially . My rank changed from 150 to 900 in just a last hour. I request you to do plagiarism checks so that those who worked hardly their efforts doesn't go in vain.

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