Блог пользователя lucifer2709

Автор lucifer2709, история, 7 лет назад, По-английски

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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7 лет назад, # |
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I still haven't recieved any cashprize from last year's contest.

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    7 лет назад, # ^ |
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    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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    7 лет назад, # ^ |
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    You won't have the same problem this year coz I am competing. Ez

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7 лет назад, # |
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Will there be any editorial ?

Also how to solve Charles and the BRUTAL subsequences ?

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    7 лет назад, # ^ |
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    • 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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    7 лет назад, # ^ |
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    The solution hints to some problems can be found here

    Detailed editorials will be posted soon.

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7 лет назад, # |
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When will the editorials come??