amnesiac_dusk's blog

By amnesiac_dusk, history, 3 years ago, In English

We invite you to participate in CodeChef’s November Cook-Off, this Sunday, 22nd November, from 9:30 pm to 12:00 am IST. 2.5 hours, 5 problems.

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.

Joining us on the problem setting panel are:

Prizes: The top 10 Indian and top 10 Global participants will receive CodeChef laddus, with which the winners can claim cool CodeChef goodies. Know more here.

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.

Good luck and have fun!

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3 years ago, # |
  Vote: I like it +38 Vote: I do not like it

Gentle Reminder: Contest starts in ~150 minutes.

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3 years ago, # |
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Unbalanced contest :(

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3 years ago, # |
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Could have had a better difficulty gradient (C was much tougher than B, which made the contest speed forces), but other than that, nice contest.

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3 years ago, # |
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Uh is it just me or did the contest length change to 3 hours?

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    3 years ago, # ^ |
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    Yeah

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      3 years ago, # ^ |
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      Ah, I reloaded the page a few minutes later and saw the announcement. I do find the addition of a problem during the contest to be rather sketchy though.

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    3 years ago, # ^ |
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    This is actually really bad management of codechef, no announcement whatsoever. Everywhere its still written 2.5 hour.

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3 years ago, # |
  Vote: I like it +79 Vote: I do not like it

Div-1 was perfectly balanced as all things should be. Thanks for the round.

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3 years ago, # |
  Vote: I like it +46 Vote: I do not like it

:P the Wikipedia page that was linked in the problem statement for Set Queries had the solution.

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3 years ago, # |
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How to solve "Hiring Workers" ?

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    3 years ago, # ^ |
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    We can see that the lcm of all groups should be equal to x so we can first divide the number x into coprime factors such as x=(p1^c1)*(p2^c2)... where p1,p2.. are prime numbers now if k is greater than or equal to the the number of prime factors we assign the remaining groups as 1.

    If k is less than the number of prime factors we can check by brute force as x cannot have more than 7 distinct prime factors and even if k is 6 we have to check 6^7 cases.

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      3 years ago, # ^ |
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      How in the world will someone know the upper cap is 7! You must be a genius to figure that out. Kudos.

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        3 years ago, # ^ |
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        Just multiply the smallest 8 prime numbers.

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          3 years ago, # ^ |
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          Thank you, that's really neat. I spent about 2 hours and I couldn't figure that out. I was trying to come up with a some greedy approach. This proves I'm dumb!

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      3 years ago, # ^ |
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      There's a way to do it using bitmask dp too (by iterating over submasks), in like P^2*3^P time or something (don't exactly remember the complexity), where P is the number of prime factors of X and P<=7.

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3 years ago, # |
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Video editorials for 4 problems have been uploaded here. The other 3 videos will be uploaded over the next day or two.

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3 years ago, # |
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For HIRINGWO can someone tell whether there was another approach other than brute force when $$$k$$$ was less than number of distinct prime factors? that was like $$$6^{8}$$$.

Also for KDIAMS do we make graph with diameter 2 only no?

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    3 years ago, # ^ |
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    "Also for KDIAMS do we make graph with diameter 2 only no" ==> you can also have a complete graph which will have diameter as 1.

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3 years ago, # |
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In hiring workers problem, I find a pair of least sum and having LCM x and then added k-2(k-2 groups having only 1 worker) to sum of pair. what went wrong in my approach? Somebody please help

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

Nice problem set but too many observation related problems :(

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3 years ago, # |
Rev. 3   Vote: I like it +10 Vote: I do not like it

https://youtu.be/nTAQi-7fd1E made a video on Hiring workers. Give it a watch if you are stuck.