Vladosiya's blog

By Vladosiya, history, 2 weeks ago, translation, In English

Hello! Codeforces Round 925 (Div. 3) will start at Feb/13/2024 17:35 (Moscow time). You will be offered 6-8 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have a rating of 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ICPC). Thus, solutions will be judged on preliminary tests during the round, and after the round, it will be a 12-hour phase of open hacks.

You will be given 7 problems and 2 hours and 15 minutes to solve them.

Note that the penalty for the wrong submission in this round is 10 minutes.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a trusted participant of the third division, you must:

  • take part in at least five rated rounds (and solve at least one problem in each of them)
  • do not have a point of 1900 or higher in the rating.

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.

Problems have been created and written by our team: myav, Gornak40, senjougaharin and Vladosiya.

We would like to thank:

  1. MikeMirzayanov for help with ideas and Polygon and Codeforces platforms.

  2. nigus for red testing.

  3. vladmart, Gheal, KseniaShk for yellow testing.

  4. htetgm for purple testing.

  5. natalina, SashaT9, lockdown, bma20006 for blue testing.

  6. RedDreams for cyan testing.

  7. the_Incharge, Aurora__, Longqiang for green testing

Good luck!

P.S. Happy Valentine's Day!

UPD: Let's continue streak of announces with photo of the authors :)

UPD2: Editorial

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2 weeks ago, # |
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Wow, you put a lot of effort to make a lot of quality div3. Thanks for a lot good div3 round to help beginner to advances in CP.

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$$$\mathtt{Thanks \ for \ Legendary \ Div.3 \ Rounds , Looking \ forward \ to \ join \ the \ authoring \ forces}$$$

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Yeahhhh!!! Div 3 Letsssss Gooooo

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Div 3 are my favourite. Lots to learn.

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the round number and date is wrong please fix

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the round and date are wrong

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speedforces let's go

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Thursday, October 12, 2023?

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I hope this is my last round before cyan

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Thank WHO for red testing 😳

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My first unrated Div3 I mog everyone

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Excited!! Hoping to reach pupil this time :)

UPD : FINALLY REACHED :)

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don't make your gf sad, prepare gift after contest

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    2 weeks ago, # ^ |
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    cp and gf? nah man something doesn't feel right. Just don't touch grass and stay home I guess

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I had no idea authors could be so pretty. I'm blushing (,,>﹏<,,)

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    2 weeks ago, # ^ |
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    mmm, why?

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      I was hoping the authors would see it and feel happy.

      Spoiler
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        2 weeks ago, # ^ |
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        No, I mean why you thought that authors couldn't look pretty. As far as I'm concerned they're all human beings, just like the rest of us

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          2 weeks ago, # ^ |
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          If I became an author, I would be a good counter-example :)

          But you are correct. Everyone is beautiful. That's why its great to have pictures of the authors in the announcement.

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so sweet post. Happy Valentine's Day. Do spend time with ur loved ones, instead of attending contests

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Can you pin social media handle (Instagram) of girls in image, i need tips for CP from them :)

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I am the one in the button right on Valentine's day :⁠,⁠-⁠)

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I don't get it,why are authors making a 'C' with their hand?

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    2 weeks ago, # ^ |
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    May be because Mike also did the same on his picture (to hold the burger)?

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    2 weeks ago, # ^ |
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    It's not the letter "C", it's half a heart (we tried, ahahah) We wish the community a happy Valentine's Day!

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      2 weeks ago, # ^ |
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      Got it,that makes sense,i need to go out and meet some real people.

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Thanks for the contest but why y'all throwing up gang signs?

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    2 weeks ago, # ^ |
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    Bro failed to get some contribution XD

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      Lol, at +116, even 100 upvotes on this comment won't change my contribution at all

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Valentine's Day ? Oh man, i am CPer

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Wow, this is great.

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Looking forward to it!

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My first unrated contest

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They are so similar

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why are the names of the authors written in reverse

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I really hope to become Pupil in this contest.

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Hopefully my last rated div3

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Wish everyone will have a good score!

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why the solution will be out 5 minutes before the contest ends?

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Hope positve delta for all rated participant.

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Finna reach 1000 elo ;))

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I'm really looking forward to the Codeforces contest taking place tonight, on the fourth day of the Lunar New Year. It's an exciting opportunity to challenge myself and improve my problem-solving skills!

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NOOOO THE CONTEST END 10 MINUTES AFTER MY OLYMPICS(school)

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Good luck to everyone.I hope to become expert on this round.

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Hoping to solve atleast 2 problems this contest

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Iam ready to solve just one problem As usual

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Thank you

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Hope to reach expert again

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hope to be specialist,,if i could solve E, my day will be great otherwise it gonna worse,,,

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I give up more than an hour trying to figure out B and its just not working

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    2 weeks ago, # ^ |
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    same here... i dont know what are the mistakes i am doing as A,B,C all the three questions are very easy

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Disgusting problem F

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    2 weeks ago, # ^ |
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    It's not, F is basically "Detect cycle in a directed graph".

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    2 weeks ago, # ^ |
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    you can do "F" just by making first line as 1, 2, 3, ... n, and these ai changes in next k — 1 lines too.

    Then for each line just check order is maintained (i.e. curr element is less than last)

    As an edge case: ensure "1" is correct in all.

    submission

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I absolutely clutched problem E with 10sec remaining!!

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Got the right idea for F about an hour after the contest started but spent the next hour debugging double-hash. :(

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    Hope no one tries to hack me after I write this comment. :(

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please hint for C

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    2 weeks ago, # ^ |
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    Try to fix equal element, either equal element can be first element of the array or last. Then find the range i and j

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    You MUST choose x as the first OR the last element.

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    three possible ranges you will select from k to n , from 0 to n-k , or from i+k to n-k. k is the index is where there's no more duplicates from the beginning or end from the array, like [1,1,1,2,3,4,5] k is at index were ai is equal 2 , so we can make the range we select [4,7] , same from the back. i+k to n-k is when the duplicated at front is the same from the back . sorry for bad explaining

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      2 weeks ago, # ^ |
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      i did that but test case number 2 gives WA 246226405

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        Same problem, everything seems right still tc2 shows wrong, 112th answer being wrong by 1.

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    2 weeks ago, # ^ |
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    You can look for the longest segment of equal elements from the start, let's call it a, and the longest segment of equal elements from the end, call that b. Now, there are two cases:

    (1)If the first and last elements are equal, you can choose the contiguous segment of elements not in either a or b. (notice, if len(a) + len(b) > n, the answer is zero. Else the answer is n — len(a) — len(b))

    (2)If the first and last elements are not equal, you choose the equal element as the element that comes in the longer segment. (notice, the answer in this case is n — max(len(a), len(b)))

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The CloudFlare feature is really a shit ,can't submit the solution even after doing it

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i am annoyed i couldn't get F i didn't think it will be cycle detection at all.

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Problem D is nice, thanks for the contest. But how to solve F?

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I feel like idea of F was simple but the implementation was tedious. Also, loved G!

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Any hints for G?

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    2 weeks ago, # ^ |
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    Try reducing this problem to a stars and bars problem Try to think what are the prerequisites if we want to place the 4th block at any place same goes for the 3rd Think which blocks are the limiting factors and which we can take care of irrespective of their quantity Have a Great solve!!

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Maybe it's just me, BUT F seemed way easier than both D and E, although didn't spend much time on E after I saw F, which I had just the right idea for. Overall very fun Contest.

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Quick hints for all the problems before the Editorial comes out.

A: Just brute force on each index using 3 for loops.

B: As the water can't flow backward, therefore the amount of water in any prefix can only decrease or stay constant.

C: Check if all equal, else count prefix equal length and suffix equal length.

D: What if we store the remainders in pair<int, int>?

E: Replace each number by the count of its trailing zeroes, now play the game greedily.

F: We can easily create the order from the first 2 arrays given (if k = 1, then its YES). Just make sure to handle one edge case [1,2,3,5,4] and [2,1,3,5,4], basically those in which there are 2 possible orders)

G: Pieces 1 and 2 can only appear alternatively, and the pieces 3 and 4 will always fit in the between their occurences. Also, pieces 3 and 4 can be repeated, while 1 and 2 can't. Try fixing Piece 1 or 2 as the first piece in the chain, do you get some general pattern?


Spoiler
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      Easier way to solve F without using graph

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      Yeah this subproblem looks way more easy and familiar, I was just hasty and went with my first thought.

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    In F, I think the simplest way to code is model it as a graph and use topo sort. I think that makes implementation really easy with template :)

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      F feels like AtCoder style problem, doesn't it? :)

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        I think it's more like the first few problems of a regional.

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    can up please tell what's wrong with this

    code
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    I did F after constructing the graph and checking if the number of SCC is equal to $$$n$$$:)

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      can someone define the statement in a prettier manner? I can't understand that. I know how to do cycle detection but the statement itself is unclear.

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F — "Detect cycle in a directed graph."

Submission- 246203312

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can anyone share any hint for problem D?

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How to combine pieces in the second test case of problem G (one sample way would be enough, please)?

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    121212121244 or 121244121212

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      Hmm, but there are only 2 pieces of type 2 and single piece of type 1: c = [1 2 5 10]

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        2 weeks ago, # ^ |
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        Sorry I misread it as second last test case, for the 2nd tc, here is one,

        333332124444444444 or 333332444444444412

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          Thanks! Now I see why I could not solve the problem :)

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I just wanna say thank you for making this contest, I was able to solve till F( can't believe I did this )

I think I have very unique solution for problem D: Problem D Solution

I used modular operations for storing single value in map

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    I did not get the logic behind your code. int v = mod*a + b ; int fv = mod*fx + fy ; why did you do this?

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      2 weeks ago, # ^ |
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      For pair $$$(a,b)$$$ we interested in pair $$$((x-a)\mod{x}, b)$$$

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      2 weeks ago, # ^ |
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      this is a trick,

      for representing two number in single number,

      lets a , b are the numbers i want to encode it into one number, then i will choose mod > max( a , b )

      then encoded_number = mod*a + b

      when you want,

      to get b = encoded_number%mod ;

      to get a = encoded_number/mod ;

      so with one encoded_number i have stored two values

      EDIT: fixed formatting

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just 3 again

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Code editor is not working, keeps running and no result, makes the whole experience shit. Cant run test cases.

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    2 weeks ago, # ^ |
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    you just calculating binomial coefficient, how is this related to solving G?

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      Because the only thing that came to my mind was:

      • place a 3 on this group (between 1 and 2),
      • or move to the next group

      which led me to the formula: $$$dp[i][j] = dp[i - 1][j] + dp[i][j - 1]$$$, and I didn't realize that it was effectively just comb(i + j, j), so that's what ChatGPT told me.

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Here is how you can choose the next type of element to add in problem G.

Two nice observations:

  • This condition must be satisfied: $$$\left|c_1 - c_2\right| \le 1$$$.
  • Think about the problem without elements of types $$$3$$$ and $$$4$$$, then try to insert them in the generated chains.
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When I saw other's solutions,I found the problem E very easy. However,I think F is easier than E because I can get the idea to solve it without thinking.(if you know something about graph)

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    I agree. I knew immediately I could use Khan's algorithm to solve F right after I read it. But the strategy for E took me some time. (and I solved F before E)

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I was wondering whether this contest gives any rating? I am 401 rated. I have participated and solved 2 problems but haven't got any rating. Can anyone tell me if i'll get any rating?

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Does the following work for F?

Rearranging any of the participants except the last one cannot change the last participant from the permutation. Therefore, we can find the real permutation by checking which one has the last participant only once. Now, that we have obtained the real permutation, we can compare all of the k inputs to the real one and check.

This doesn't work when k == 2 but in that case we can simply check subsequences ignoring the first two numbers.

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    2 weeks ago, # ^ |
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    Better Idea:

    Simply Ignore the first Position of Every observation. Take an edge pointing from v[i] to v[i+1] (1 <= i <= n-2) for Every participant observation.

    Check for the cycle in the graph, If the cycle exists then NO else YES.

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      2 weeks ago, # ^ |
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      How to check if a cycle exists? I tried doing it using DFS but it exceeded the time limit.

      246185460

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          Thank you!

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            2 weeks ago, # ^ |
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            TLE coz of MAX maybe, You are unnecessarily assigning 2e5+7 values to the array, even when N = 1.

            Just imagine for the test case where t = 10^4 and N for each case belongs to {1,2,3,4,......2e5}, even if k = 1.

            As this test case fits under the given constraints, You will be doing 2e5*(1e5)/2 roughly around 1e10. That's y.

            I think so, I can be wrong.

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              2 weeks ago, # ^ |
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              Yeah, I think that was the problem. I just resubmitted resetting only n values and it got AC. Just missed it during the contest :(

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    246262272 Maybe hashing was unnecessary, also who knows if this might get hacked.

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Loved the beautiful Stars and Bars problem (G) Got to the solution but didn't submit earlier as I was unsure if it would be rated for me Nonetheless Here's my solution in case any one wants a reference: https://codeforces.com/contest/1931/submission/246256160 :)

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As an expert I failed D: TLE on test 9. Here is my code: 246198274. Could anyone help me with debugging? I think this is O(nlogn).

Edit: This is O(y). Forget it

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How to solve E? Didn't get any idea at all :(

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    1) Ans "yes" if number of digits in the last number >= m + 1. Anna in each step tries to reverse number with maximum zeroes on the end. Sasha in each step does concatenation number with maximum zeroes on the end with any number without zeroes on the end (number without zeroes on the end is guaranteed to exist, since it either exists initially, or Anna will make it her first move)

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    it is optimal for both of them to the choose the numbers which have most no. of zeros at the end

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    multiset<pair<trailing_zeroes, digits>> and just model a game

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    you can simulate it as follow: for anna reversing a numbers is equivalent to deleting the trailing zeros so she should take the number which have the most trailing zeros every turn for sasha she will also choose the number with most trailing zeros and concat it with any number (it doesn't matter but we can choose the second most number having trailing zeros for implementation purpose) by doing this it save the trailing zeros in the largest number so anna can't remove them by reversing the number thats it u can keep track of the numbers having most number of trailing zeros using a priority queue and update it each turn accordingly

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This is my first AKed competition, and the problems were wonderful and quality, thanks! I like G best, it's an interesting counting problem.

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FUN QUESTIONS, Extremely challenging but really good questions.

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2 weeks ago, # |
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Can someone please help me in Problem C?

I am not getting why it gives me wrong answer on test case 2 (112th answer in tc2 is wrong it says, and I am not able to see what that case is because after certain lines the input of test case just shows "...")

Here is the submission —

246248784

Thanks!

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    2 weeks ago, # ^ |
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    I dont know python so I apologize if I am wrong but if first and last element are equal then the ans should be 'n-(s+e)'

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      2 weeks ago, # ^ |
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      yes so in python ".strip()" will remove all occurences of the passed argument from the front AND the back ONLY(not in the middle). So that part works alright.

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        2 weeks ago, # ^ |
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        Have you considered striping the last part too?

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2 weeks ago, # |
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I think you should reduce the time limit of this problem. this submission -> https://codeforces.com/contest/1931/submission/246247517 runs in 1 second. this submission should not get AC because # of testcase is 10000 and this submission is executing a for loop of 200005 every test case. that's more than 2e9 operations.

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2 weeks ago, # |
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!What is this submission code
is he know the testcase?

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    2 weeks ago, # ^ |
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    it's hacked

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      2 weeks ago, # ^ |
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      Yes but how did he AC in the first place?

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        2 weeks ago, # ^ |
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        testcases were week

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          2 weeks ago, # ^ |
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          Oh no, we didn't make tests with all possible t value! What a weak testset!

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2 weeks ago, # |
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After successfully wasting 1.5 hour on B i smashed binary search in B

can someone hack it?

246247536

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    2 weeks ago, # ^ |
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    why would you even think of binary search . you already know that the good value is sum/n . that's the only way you can make tha array the same

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      2 weeks ago, # ^ |
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      yeah i am dumb. do not know why that did not come into my mind

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        2 weeks ago, # ^ |
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        same here I thought of every thing in this life except that the only possible number is sum/n by the way I wanted to do binary search but I didn't know how to handle the good function but I learnt that from you now I mean the 1 , -1 and 0

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2 weeks ago, # |
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Video Editorial for problem B (Audio : Hindi) TUTORIAL VIDEO LINK

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2 weeks ago, # |
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Is this hackable B ? 246258951

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    2 weeks ago, # ^ |
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    I don't think so.. I used a similar approach so.. I hope it is not

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2 weeks ago, # |
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Thanks to the authors for an interesting div 3 without clay and with tasks that are fun to solve. After this div and the Cognitive Technologies Olympiad, I realized that the name Gornak in the list of authors => good Olympiad/div.

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2 weeks ago, # |
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Can G be solved using inclusion exclusion ?

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2 weeks ago, # |
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May I ask why I passed 3 questions but didn't receive a rating in the end? Or who should I seek help from? I really want this rating, thank you.

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    2 weeks ago, # ^ |
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    Rating appears after 1 to 24 hours from the end of a contest, and if there is a hacking phase, like this contest, then it may last up to 24 hours or less.

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2 weeks ago, # |
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in Problem E , i wrote code which was absolutely right, the only thing i missed was, if number>=10^m then number of digits should be >=(m+1) not >=m (it was pretty close :) )

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2 weeks ago, # |
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I submitted E when there were 10 seconds for the contest to finish and my submission got vanished...

why so ?

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2 weeks ago, # |
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In Problem C anti hash test case can be created as many unordered map solutions have been accepted.

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2 weeks ago, # |
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How to solve Problem G?

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2 weeks ago, # |
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wow that was very cool

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2 weeks ago, # |
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A different way to do F:

  • Each element (after the 0th element in the kth list) can only be it's current position or its current position+1

  • Store both possibilities in a vector

  • Remove possibilities for elements when they're no longer possible (i.e. the current element's position in the kth list conflicts with the elements position in a previous list)

  • The answer is no if an element has no possibilities, or if its only possibility conflicts with another elements only possibility

  • Otherwise the answer is yes

Code: https://codeforces.com/contest/1931/submission/246250979

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2 weeks ago, # |
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I am getting wrong on test case 3 for problem E

My approach is " The first set contains numbers with leading zeros, while the second set comprises numbers without leading zeros. Within a while loop, the first element of the first set is removed, and its length is added to the second set. Additionally, the first value of both sets is combined and inserted into the second set. The game iterates until the first set is empty. Afterward, the sum of the lengths in the second set is calculated. If this value exceeds a specified threshold, Sasha wins; otherwise, Anna wins.

can anyone help me find what's wrong in this?Solution link

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2 weeks ago, # |
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really enjoyed this contest, particularly problem D. start to see some common patterns on int-mod problem..

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2 weeks ago, # |
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my respect to Aleksander Gornak

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2 weeks ago, # |
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I was getting runtime error (STATUS_ACCESS_VIOLATION) in E, during contest , but after contest when i submitted same solution it got accepted . runtime error during contest code link.
accepted code after contest. Thanks

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    2 weeks ago, # ^ |
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    The problem is this for loop here: for(ll i=vect.size()-1;i>=0;i-=2).

    In C++, the return type of std::vector::size is an unsigned integer. When vect is empty, then vect.size() - 1 becomes 0 - 1 which then overflows to UINT_MAX. This causes out of bounds error which is why your program got runtime error. (It seems that this is changed in C++20, which is why your other code got accepted)

    To fix this, you just have to convert vect.size() into any signed integer type. For example for(ll i=(ll)vect.size()-1;i>=0;i-=2). New code

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2 weeks ago, # |
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One of the best rounds I've ever participated. Thanks to all the writers and testers for offering us such a round.

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2 weeks ago, # |
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The best div3 round contest I've ever participated!Thank you!

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2 weeks ago, # |
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when will the rating come out?

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    2 weeks ago, # ^ |
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    you need to wait that the system testing ends first (is around 80% right now). after that eventually the rating will be updated.

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      2 weeks ago, # ^ |
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      Oh,Thank you.This is my first time to attend CF.(I find out my G turned out to WA,It used to be AC)

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        2 weeks ago, # ^ |
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        I saw it as a AC, when the system testing is running and your AC code is still on queue, if you had a WA before the AC, you will see as if you get a WA. So congrats on your first contest participation in CF was really good!

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        2 weeks ago, # ^ |
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        First CF got G AC,congratulations!

        I'm waiting the rating, too.

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        2 weeks ago, # ^ |
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        Thank you for your encourages!

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2 weeks ago, # |
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Can anyone tell me where have I made a mistake in these two codes? Only difference I have made from my side is making the capital N,small n? Basically I had initialised all visited array elements to zero in reset function. Thankyou in advance.

1st submission 246198873

2nd submission 246233252

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    2 weeks ago, # ^ |
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    I don't understand why exactly gives you another answer (actually I'm more curious of how didn't get you RTE hahaha weird stuff)

    But the reason why the change works is because in the WA submission, you have the following code:

    vi g[N];
    void reset(){
    	fr(i,0,N){
    		g[i].clear();
    		vis[i]=0;
    		vis1[i]=0;
    	}
    }
    

    and fr is #define fr(i,a,b) for(int (i) = (a); i <= (b); (i)++)

    So, the cycle in reset() will access to the position $$$N$$$. But the arrays g, vis and vis1 are of size $$$N$$$, I change you code putting a -1 and now it gets TLE, that makes a lot of sense, because $$$t = 10^4$$$ and $$$N=2e5+69$$$, so you will do around of $$$2\cdot 10*9$$$ operations just to reset.

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2 weeks ago, # |
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In Question 4 this code exceeds time limit on test case 33 cant understand why its o(n) approach can someone please help me?

include<bits/stdc++.h>

using namespace std;

define ll long long

int main() {

int t;
cin >> t;
while(t--)
{
    ll n, x, y;
    cin >> n >> x >> y;
    unordered_map<ll, ll> dp;
    ll a[n];
    ll ans = 0;
    for(int i = 0; i < n; i++)
    {
        cin >> a[i];
        ll f = (x - a[i] % x) % x;
        ll l = a[i] % y;
        ans += dp[f*y+l];
        dp[(a[i] % x)*y+a[i] % y] += 1;
    }
    cout << ans << '\n';
}
return 0;

}

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    2 weeks ago, # ^ |
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    Same happened with me many times... :(

    Its because of unordered_map if you use simple map it will pass....

    For more: https://codeforces.com/blog/entry/62393

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    2 weeks ago, # ^ |
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    Worst case time complexity of unordered map is o(n) while normal map is of o(logn) that's why it is giving Tle. I got AC with your solution by just changing unordered map to normal map

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2 weeks ago, # |
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But may I ask why I still haven't received a rating? It should have been a long time since the hack phase ended. I really want the rating for div3 this time, thanks.

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    2 weeks ago, # ^ |
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    You must wait for some more hours after the system test. Codeforces needs time to calculate rating changes. Just chill

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2 weeks ago, # |
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Why there is no rating change? My rating is less than 1600

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    2 weeks ago, # ^ |
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    It takes a while after the system test for codeforces to remove cheaters and recalculate the rating changes.

    By the way, have we met somehow before?

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2 weeks ago, # |
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can anyone just tell me which are valid patterns in G? I can't find any for 1 2 5 10

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    2 weeks ago, # ^ |
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    One of the valid sequences is:

    2 4 4 4 4 4 4 4 4 4 4 1 3 3 3 3 3 2

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      2 weeks ago, # ^ |
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      Thank you... I didn’t notice self loops.

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2 weeks ago, # |
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when it will be rated?

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2 weeks ago, # |
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All my submissions have been blocked due to one answer(246237190) for question 1931F being very similar to the ones given by two other people.

One of them is from another guy studying in my institute who happens to study from the same materials and has given similar contests, hence the answers are very similar. Please review it and I hope my submissions are considered.

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    2 weeks ago, # ^ |
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    This clearly is a very repeated question, since many people have the same solution Please review it once.

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      2 weeks ago, # ^ |
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      It is very clear that You have copied this code 246237190 from somewhere, you didn't even put the effort to indent it properly. I think it is justifiable to get blocked.

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        2 weeks ago, # ^ |
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        So your justification of me getting blocked is the code not being indented

        i was in a hurry, thats it

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        2 weeks ago, # ^ |
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        You know what Leave it, you and I both know how much ever I fight there isn’t going to be any change I better focus on future contests, even if they unjustly block me again I will create new account and start again

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2 weeks ago, # |
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Key observation of problem E!
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I am new to code forces this is my first contest, I have solved two questions still didn't get any rating. Did I miss anything?

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Hi, I participated in 925 and it wasn't rated for me, does anyone know why? I am < 1600 (878).

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2 weeks ago, # |
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Give me my rating pls((

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Finally blue. Was a great experience.

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Hi I attended this contest and I solved problems A & B and got D wrong. And i'm not able to find the constest under my contests and my rating afterwards didn't although I am under 1900. Can you please explain why my rating did not improve?

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2 weeks ago, # |
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Become a pupil from this contest.

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I am sorry for using a code snippet that is used by lot of other people. This made me into this trouble of making code solution identical to other users. I regret using this snippet now. I shall prevent myself from further using this snippet anymore. Sorry CF ...

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2 weeks ago, # |
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why aren't the ratings changing ? Its been 2 days

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2 weeks ago, # |
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Hello respected judges,

My code was accidentally matched with the other participants. I have no idea about this rule of the CodeForces of contests. I didn't know this before. I had no selfish intention to do this cheat of any type. That participant is my college friend, and we discussed not exactly but similar type question before this about how to solve it. But I didn't know how we both accidentally write up the same code.

So, please can you give me my rating back, in future I will take care of it and not publish my code publicly anywhere.

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    2 weeks ago, # ^ |
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    You mean your rating reduced?

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      2 weeks ago, # ^ |
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      No, my rating has not increased. My solved question which matches with others was skipped means not considered for evaluations.