lucifer2709's blog

Invitation to Alkhwarizm 2017
By lucifer2709, history, 8 years ago, In English

The much awaited signature event of IIIT Allahabad's Tech Fest Aparoksha 2017 — Alkhwarizm is here to endeavour with a challenging problem set.

It will be a 5 hour individual contest with algorithmic problem-set of diverse nature to be hosted on Codechef. Contest link is here.

Prize money worth INR 30k for Indian winners and other exciting goodies for Global winners. CODE ON !!!

So be ready to have a nail biting experience on March 24th, 2017 — 9 pm to 2 am IST.

Register here to be eligible for prizes.

It will be first External rated contest that will be hosted on Codechef.

Problem Setters — m17, lucifer2709, shivamg_isc, hybrid

For any updates follow our event page.

Update

The contest has ended, Thank you all for participating. Hope you liked the problem set.

Following are winners of Alkhwarizm 2017

  1. adkroxx
  2. harshil
  3. ajinkya1p3

Congratulations to the winners !!!

For now we have provided small editorials for 8 problems. You can find it here. Detailed editorials will be published soon.

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8 years ago, # |
  Vote: I like it +44 Vote: I do not like it

I still haven't recieved any cashprize from last year's contest.

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

    Hi Rajat.

    I apologize to you on behalf of Aparoksha 16 organizing team. There was a serious blunder last time.

    I am seriously sorry for your bad experience of Alkhwarizm last year. Though I assure you that the same blunder won't happen again. The prize money will be rewarded for sure this time.

    I hope you have a great contest tomorrow. :)

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    8 years ago, # ^ |
      Vote: I like it +51 Vote: I do not like it

    You won't have the same problem this year coz I am competing. Ez

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8 years ago, # |
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Will there be any editorial ?

Also how to solve Charles and the BRUTAL subsequences ?

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    8 years ago, # ^ |
    ← Rev. 2   Vote: I like it +2 Vote: I do not like it
    • Let dp1[i]-> No. of subsequences if the last chosen string is i and it has greater length than the previous one.
    • Let dp2[i]-> No. of subsequences if the last chosen string is i and it has lesser length than the previous one.
    • Then we have: dp1[i]= Summation (dp2[j]) j is 1.2...i-1 && s[i] and s[j] have common prefix, dp2[i]= Summation (dp1[j]) j is 1.2...i-1 && s[i] and s[j] have common prefix
    • The check for second condition could be implemented using segment or fenwick tree.
    • Figuring out base cases and answer is easy.
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    8 years ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    The solution hints to some problems can be found here

    Detailed editorials will be posted soon.

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

When will the editorials come??

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

    The solution hints to some problems can be found here

    Detailed editorials will be posted soon.

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