Блог пользователя lighters

[HackerRank] CodeAgon on August 15th
Автор lighters, история, 9 лет назад, По-английски

Hello CodeForces Community!

NOTE This contest has prizes and job opportunities for Indian participants.

I am glad to share that CodeNation's CodeAgon is scheduled on 15th-August-2015 at 14:30 UTC

Go ahead and register now at https://www.hackerrank.com/codeagon to show off your coding chops.

Prizes worth 2 lakh rupees will be given and you'll get to connect with CodeNation for a career opportunity.

Contest will be unrated, scoring is 20 — 30 — 50 — 80 — 100. Tiebreaker is person to reach the score first.

Editorials will be live soon after the contest ends.

Good luck and have fun.

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9 лет назад, # |
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I think the contest is rated :)

And why I must fill this form for participating? I don't know what the half of this things mean.

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9 лет назад, # |
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what is the duration of contest ?

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9 лет назад, # |
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Please recheck the tie breaking in the leader board part. I think that the leader board is not showing the first person to reach the score. People who reached the score later than me are going ahead of me.

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9 лет назад, # |
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Problem E was nice. It wasn't too difficult, rather it was one of those "feels good to solve" problem. There are at least 5 different methods to get accepted in time.

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    9 лет назад, # ^ |
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    what did you try?I tried seg tree + two pointers and solved the problem with multiplication not with big numbers and with complexity O(number of digits),but I only got 83 points. I think using big numbers with base very big would be practically better :-?

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      9 лет назад, # ^ |
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      1. Prime factorization followed by sliding window.
      2. Prime factorization followed by binary search.
      3. Sparse table followed by binary search with complexity O(N * log(N)^2 * log(K)). I believe it can pass with some optimizations.
      4. Method 3 can be reduced by a factor of log(N) using modified binary search.
      5. Sqrt(n) decomposition. Might require some optimizations to pass the time limit though.
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        7 лет назад, # ^ |
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        I know it's been 3 years since 2015 ! But, can you tell me how to solve it using prime factorisation followed by sliding window ?

        Here's the link to the question.

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          7 лет назад, # ^ |
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          Fix l=0 and find the shortest r such that the product of the interval l-r divides k. Now move from l=0 to l=1 and decrease the count of the prime factors corresponding to the value at l=0. If the interval from l=1 to r does not divide k, keep incrementing r until it does. Pretty basic stuff if you are familiar with sliding window.

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            7 лет назад, # ^ |
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            i think what you have described as "sliding window" is considered as "two-pointer" for general audience .

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