piyush_pransukhka's blog

By piyush_pransukhka, history, 16 months ago, In English

Hii Codeforces!

We invite you to participate in CodeChef’s Starters 83, aka International Coding Marathon, in collaboration with Technex '23, IIT (BHU), this Wednesday, 29th March, rated for all divisions.

Time : 8:00 PM — 10:30 PM IST

Note that the duration is 2.5 hours. Read about the recent CodeChef changes here.

Joining us on the problem setting panel are

ICM is the flagship CP event of Byte the Bits, the set of programming events organized under Technex, the annual techno-management fest of the Indian Institute of Technology (BHU), Varanasi.

Prizes

Note: Prizes are only for Indian participants (combined ranklist for divisions 1 and 2).

  • Winner: $$$10,000$$$ INR
  • $$$1^{st}$$$ runner up: $$$6,000$$$ INR
  • $$$2^{nd}$$$ runner up: $$$4,000$$$ INR
  • Top $$$25$$$ contestants: Goodies and coupons

Please fill out this form to avail them.

Some of our previous rounds:

Written editorials will be available for all on discuss.codechef.com. Pro users can find the editorials directly on the problem pages after the contest.

The video editorials of the problems will be available for all users for 1 day as soon as the contest ends, after which they will be available only to Pro users.

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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16 months ago, # |
  Vote: I like it +49 Vote: I do not like it

As admin and one of the setter, I can assure you problems are really good, Its worth to spend your 2.5 hours of day.

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    16 months ago, # ^ |
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    May I ask how the difficulty is roughly divided?

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      16 months ago, # ^ |
        Vote: I like it +10 Vote: I do not like it

      It will be like usual starters, for div 2 and 1 all of the problems will be common to have common ranklist for prizes.

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16 months ago, # |
  Vote: I like it +7 Vote: I do not like it

Nice to see CodeChef become independent from shitacademy. Hopefully it can find new sponsors and regain its past glory soon.

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16 months ago, # |
Rev. 2   Vote: I like it +24 Vote: I do not like it

Great!

From now on, it would be helpful if you could announce whether it will be rated up to Div.1 as soon as possible.

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    16 months ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    Sure, we'll try to do that.

    Just to set the right expectation — even though some of the Starters will be rated for all, for now they will not be of the same standards that Cook-Offs and Lunchtimes were. As mentioned here, it will be less challenging for the highest rated participants. But we do hope and believe it'll still be a good experience! :)

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16 months ago, # |
  Vote: I like it +5 Vote: I do not like it

Again Trouble ??

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16 months ago, # |
  Vote: I like it +8 Vote: I do not like it

Cannot get access to the site.

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16 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Nice contest! What's the intended solution for Mex Segments ?

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    16 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Let F(M, L) = Number of subarrays whose mex is >= M and length is <= L. Answer to a query [M1, M2], [L1, L2] is F(M1, L2) — F(M1, L1 — 1) — (F(M2 + 1, L2) — F(M2 + 1, L1 — 1))

    Computing F(M, L): F(M, L) is the same as "How many subarrays of length at most L completely contain the minimal range whose MEX is M". Let's say the minimal range is [l_M, r_M) (prefix minimum and maximum on the sequence of positions of values in p). The number of subarrays counted in F(M, L) whose length is r — l is 1. length r — l + 1 is 2, then 3 and so on until say k when the closer of the boundaries of the permutation is touched. Then it remains k until the length touches the other boundary and then reduces one by one until 1 when the length equals n. F(M, L) is the sum of a prefix of this sequence which can be carefully computed in O(1).

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16 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

Fun run with good problems. Will be fun to upsolve the remaining ones!

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16 months ago, # |
  Vote: I like it +16 Vote: I do not like it

Cool problems. Enjoyed the contest.

PS
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16 months ago, # |
  Vote: I like it 0 Vote: I do not like it

CodeChef please address the problem of plagiarism during contests ?? All it took me is to look at some random submissions and I could easily make out the similarity in the code!!

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    16 months ago, # ^ |
      Vote: I like it +19 Vote: I do not like it

    No, we do not plan on doing anything about plag during the contest. We only penalize them after the contest, and it takes many days. You can see the updates till last week's contest here, where about 1000 participants were caught.

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      16 months ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      Thank you for the clarification!! Your efforts are much appreciated !!

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16 months ago, # |
  Vote: I like it +18 Vote: I do not like it

Weak Tests for XOR_ORDER problem.

My solution just checking for a > b > c or a < b < c and giving -1 for all other case passes.

Can someone hack?

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    16 months ago, # ^ |
      Vote: I like it +38 Vote: I do not like it

    Just printing -1 works lol.

    Not Weak Tests but Incorrect Checker.

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      16 months ago, # ^ |
        Vote: I like it +8 Vote: I do not like it

      No wonder it got so many AC submissions in such less time.

      If they will re-judge almost everyone will fail.

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16 months ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

Beauty Sum is another usage of my blog!

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16 months ago, # |
Rev. 3   Vote: I like it 0 Vote: I do not like it

Can problem Saber Slays be solved only using segment trees without any dp. just keeping track of the minimum start time and corresponding ending time and maximum value for ranges of nodes of segment tree. then there will be few cases for merging. my merge function is of something this sort:

pair<pair<int, int>, int> p1 = tree[2 * treeidx]; // left child
pair<pair<int, int>, int> p2 = tree[2 * treeidx + 1]; // right child

tree[treeidx].second = max(tree[2 * treeidx].second, tree[2 * treeidx + 1].second); // for storing maximum

// if maximum of both range same then answer will be max+1 or if any of the range start with max+1//

if (p1.second == p2.second or tree[2 * treeidx].first.first == tree[treeidx].second + 1 or tree[2 * treeidx + 1].first.first == tree[treeidx].second + 1)
		tree[treeidx].first = {p1.second + 1, p1.second + 1};
	else if (tree[2 * treeidx + 1].second == tree[treeidx].second)
		tree[treeidx].first = tree[2 * treeidx + 1].first;
	else if (p1.first.second > p2.first.first)
		tree[treeidx].first = tree[2 * treeidx].first;
	else if (p1.first.second == p2.first.first)
		tree[treeidx].first = {tree[2 * treeidx].first.first, tree[2 * treeidx + 1].first.second};
	else
		tree[treeidx].first = {p1.second + 1, p1.second + 1};

Can someone check this as I don't know why i am getting wrong answer?

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16 months ago, # |
  Vote: I like it -33 Vote: I do not like it

In XOR_ORDER, the checker had an issue due to which some WA solutions have gotten an AC. This has been fixed for Practice. Since most affected participants did not use this 'loop hole' intentionally, we believe it is not fair to rejudge the submissions now, and are leaving it as it is. Apologies for the error.

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    16 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Its definitely fair to rejudge the submissions as either they just tried something random or they were just guessing the solution so it is not fair for someone who tried to think and code the correct solution.

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      16 months ago, # ^ |
        Vote: I like it +26 Vote: I do not like it

      It is not like system testing on CF, since the rules of CF and CC are inherently different. CC has full feedback, and the participants know this. Hence when someone gets an AC, there is no need to think more about it and try to fix any issues which the pretests might not have caught, which they would do in a CF contest.

      Yes, the current announcement is unfair towards people who coded the correct solution during the contest, but that does not make it fair to turn an AC into a WA post-contest for a person who would have gotten an AC in the next try, if they had gotten a WA during the contest. It is a case of lesser of the two evils.

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16 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Nice Problemset!