Блог пользователя antontrygubO_o

Автор antontrygubO_o, 4 года назад, По-английски

We invite you to participate in CodeChef’s May Cook-Off, this Sunday, 1st May, Rated for All.

Time: 8:00 PM — 11:00 PM IST

Note that from this month, Cook-Off and Lunchtimes will be held on Sundays instead of on Saturdays.

Joining me on the problem setting panel are:

A couple of extra words about the contest:

  • All Div-1 problems have been set by me.

  • The contest is harder than usually, so the duration is changed to 3 hours instead of 2.5 hours

  • I hope everyone will enjoy the contest!

Also, announcing Scholarship for CodeChef Certification in Data Structure & Algorithms — More than 100 Indian participants in Divisions 1, 2, and 3 will win scholarships for the CodeChef Certification exam (discounted prices). Scholarship criteria can be found on the respective contest pages.

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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Note that this is the new schedule:

  • First Sunday of the month: Cook-Off [Rated for all]
  • Second Weekend of the month: The first Long Challenge (Starts Friday & Ends Monday) [Rated for Divs 3 and 4]
  • Third Sunday of the month: LunchTime [Rated for all]
  • Fourth weekend of the month: The second Long Challenge (Starts Friday & Ends Monday) [Rated for Divs 3 and 4]
  • Starters will be there on all Wednesdays of the month [Typically rated for Divs 2, 3 and 4. Sometimes for 3 & 4].
  • All short contests at 8 PM IST (+5:30 GMT)
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Adhoc OP

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Div-2 Constructive_chef

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Maximum value can be found in MAXMINCRCDIF by using ore's thereom+binary search.How to construct the answer though?

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MAXMINCRCDIF is so similar to the problem I created in the past. So I think I must solve it in the contest.

Finally, I do it :)

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Any idea why this solution works for B?

let first array B be x1.....x1 x2.......x2 x3......x3 xn....xn

where x1,x2 xn are elements sorted by frequency

then let second array C be reverse of this. Find the first index j from back such that there exists some subarray B[j.....k] which is a permutation C[j......k]

the array C is then = B[n-1-j....n-1]+B[0.......n-1-j-1]

Check if Array C satisfies the conditions.

I was able to come up to this pattern after observing quite a few cases but im unable to prove it

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    4 года назад, скрыть # ^ |
     
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    Bro did you solve the triple inversion problem. I am not able to understand the editorial or anyone's solution. Can you please tell me how did you solve that?

    And for this question you can just rotate the array from mid that's it just one edge case in this when there are only 2 elements and their frequency is more than 1 like 1 1 2 2 or 1 1 1 2 2 2 like wise they will fail coz when you rotate 1 1 1 2 2 2 2 2 2 1 1 1 at mid their is a subarray which is a permutation so it will fail.

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Can I play with the big boys in Div1 now, antontrygubO_o?