Newtech66's blog

Codeforces Round #819 (Div. 1 + Div. 2) and Grimoire of Code Annual Contest 2022
By Newtech66, 15 months ago, In English

Hello Codeforces!

Grimoire of Code, the official Competitive Programming club of IIT Kharagpur, is happy to invite everyone to take part in Codeforces Round 819 (Div. 1 + Div. 2) and Grimoire of Code Annual Contest 2022 which will take place on Sep/06/2022 17:35 (Moscow time). This round will be rated for everyone.

You will be given 8 problems and 2 hours 15 minutes to solve them.

The problems have been authored and prepared by Grimoire of Code members Anubhav anubhavdhar Dhar, Debajyoti little_angel Dasgupta, and Mainak Newtech66 Roy.

We'd like to thank all the people who made this round possible:

We hope everyone will enjoy the contest.

See you on the leaderboard!

About Grimoire of Code:

Grimoire of Code is the official Competitive Programming club of IIT Kharagpur created by and for competitive programming enthusiasts. We promote competitive programming culture in our college, and provide a forum for interested minds to discuss their thoughts and ideas. We also conduct mock coding rounds for placements and internships in Indian colleges.

You can check out our Facebook page here.

You can check out the problems from last year's Annual Contest here.

Updates

UPD1: The contest duration has been extended to 2 hours 15 minutes.

UPD2: Score distribution: $$$500-1000-1500-2000-2250-2750-3250-3500$$$

UPD3: Congratulations to the winners!

  1. Benq
  2. tourist
  3. jiangly
  4. Maksim1744
  5. kotatsugame
  6. tatyam
  7. TLEwpdus
  8. gyh20
  9. turmax
  10. Radewoosh

UPD4: Editorial (editorial for F will be added soon) It is now added.

UPD5: The contest is unrated due to problem copying. Details.

  • Vote: I like it
  • -1094
  • Vote: I do not like it

Comments Comments (238)
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15 months ago, # |
  Vote: I like it +337 Vote: I do not like it
As a tester
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    15 months ago, # ^ |
      Vote: I like it +65 Vote: I do not like it

    when asking for downvotes and asking for upvotes are unsurprisingly in the same SCC

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      15 months ago, # ^ |
        Vote: I like it -48 Vote: I do not like it

      As a participant, I will have participated.

      Spoiler for upvotes:
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    15 months ago, # ^ |
    ← Rev. 2   Vote: I like it +60 Vote: I do not like it

    posting mind blowing meme to get upvotes :

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15 months ago, # |
  Vote: I like it +60 Vote: I do not like it
As the leamder
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15 months ago, # |
  Vote: I like it +52 Vote: I do not like it
As an author
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15 months ago, # |
  Vote: I like it +30 Vote: I do not like it

Div 1+2 and 8 problems! Looking forward to it.

When will scoring distribution come out?

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15 months ago, # |
  Vote: I like it +9 Vote: I do not like it

As a tester, I just hope everyone's super-super-supercalifragilisticexpialidocious performance on the contest.

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15 months ago, # |
  Vote: I like it +30 Vote: I do not like it

As not a tester, I did not test this round.

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15 months ago, # |
  Vote: I like it -10 Vote: I do not like it

I want contribution, I also want rating, I can't have both of them, So I choose rating instead of contribution.

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15 months ago, # |
← Rev. 2   Vote: I like it +15 Vote: I do not like it

As a participant,I'm hoping for a great round!

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15 months ago, # |
  Vote: I like it -262 Vote: I do not like it
As an author
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    15 months ago, # ^ |
      Vote: I like it -8 Vote: I do not like it

    omg that thumbs up emoji looks fabulous

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    15 months ago, # ^ |
      Vote: I like it -21 Vote: I do not like it

    also YOU SHOULDNT BRAINSTORM THE SOLUTIONS WITH OTHERS DURING CONTEST ITS AGAINST THE RULES

    (yes, I couldn't believe my eyes after reading that line)

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      15 months ago, # ^ |
        Vote: I like it -22 Vote: I do not like it

      what's wrong with all this downvote, I'm just stating the facts. are they all cheaters or what

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        15 months ago, # ^ |
          Vote: I like it -251 Vote: I do not like it

        Come on dude! Isn't it obvious that, by "you guys", I meant everyone, brainstorm the solutions "individually"

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          15 months ago, # ^ |
            Vote: I like it +76 Vote: I do not like it

          Well, yes, except for the fact that brainstorming, in common sense and as stated on this wikipedia page, is a "group creativity technique". Using a group creativity technique where there isn't a group would be a pain...

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

    after the contest, I realized.

    You (plural) might have tried your best to make the problems as interesting as possible, but you (singular) didn't! Shame on you!

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    15 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    F*ck you problem stoler.

    Let's downvote him to make his contribution lower than zxyoi cuz it's the second time such situation happens

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

IIT KGP !orz

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

    What is IIT KGP?

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

      Indian Institute of Technology, Kharagpur. It's India's one of the best engineering institute.

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15 months ago, # |
  Vote: I like it +6 Vote: I do not like it
As a pupil contestant
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15 months ago, # |
  Vote: I like it +1 Vote: I do not like it

I hope I can do at least the first one

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15 months ago, # |
  Vote: I like it +91 Vote: I do not like it

ff23f4eac312f8598394447d6c543af0dc0a11f1

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15 months ago, # |
  Vote: I like it +50 Vote: I do not like it

Really happy to see IIT KGP finally setting contests in Codeforces. Kudos to the entire team of problemsetters.

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15 months ago, # |
← Rev. 2   Vote: I like it +10 Vote: I do not like it

I love div.1 + div.2 contests
Good luck for everyone!!
Hoping for a nice problems❤️

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

Is it rated for DIV 2 ?

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

Hoping to solve 4 problems, would be great

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15 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Hope to become as soon as possible Candidate Master.

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15 months ago, # |
  Vote: I like it +6 Vote: I do not like it
As a xxx

Here is: hided content

The above comment is edited as follows:

<div class="spoiler">
  <b class="spoiler-title">As a xxx</b>
  <div class="spoiler-content" style="display: none;">
    <p>Give me xxx</p>
  </div>
</div>

Here is: <span style="color: white;">hided content</span>
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15 months ago, # |
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good luck everyone and hope the problems this time is not mathforces qwq

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    15 months ago, # ^ |
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    some moderate amount of math is fine, last time it was kinda excessive but the problems were very interesting

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15 months ago, # |
  Vote: I like it -12 Vote: I do not like it

Contest time collides with IND VS AFG cricket match. Very unfortunate.

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15 months ago, # |
  Vote: I like it +1 Vote: I do not like it

All the Best EveryOne!

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15 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Good luck to all

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15 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Sometimes you win, sometimes you learn

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15 months ago, # |
  Vote: I like it +9 Vote: I do not like it

We just got trolled

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

How to solve D?

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    15 months ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    Color everything in blue. Now do a DFS from node 1. Since m is at most n+2, you will get at most 3 back edges. Store those back edges while traversing by dfs.

    Now you have obtained some tree rooted in node 1, with some back edges. Now if you have less than 3 back edges, simply color those back edges as red.

    If you have 3 back edges, and they form a triangle, in other words have form a-b, b-c, a-c, color a-b and b-c as red. Now look at a-c. Coloring this as red would be a waste, because a, b, c are already connected. So instead leave it as blue. Find out whether a or c has a greater depth in the DFS tree. Let’s say d = deepest(a, c). Color the edge d and parent[d] in red. This way we preserve the tree structure of the blue subgraph and decrement the number of connected components in the red subgraph.

    If you have 3 back edges that do not form such a triangle/cycle. Just color them in red.

    I highly suggest drawing a tree structure with back edges to understand it better.

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15 months ago, # |
  Vote: I like it +17 Vote: I do not like it

C was so good :D

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

Very goood problems A, B, C!

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

How to solve C? Any hints

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

    Look at number of index i, such that s[i]='(' and s[i+1]!=')'

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    15 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    I just use DSU.

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

      How is its time complexity?

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        15 months ago, # ^ |
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        I don't calculate but get only 31 ms.

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          15 months ago, # ^ |
          ← Rev. 2   Vote: I like it +22 Vote: I do not like it

          oh. that's fast. maybe the time complexity is $$$O(\text{fastenough})$$$.

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

      DSU here how? If you could explain your thought process it would be helpful

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

        I use DSU and at the last step count the different number or parent.If make count+1 at '(' and count-1 at ')' you will see that all i for 1 to 2*n where count =1 or count=0 are conected. I use this obserbation to solve. Sorry for my poor English.

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

        Link every opening bracket to it's corresponding closing bracket. Whenever there is a ")(" in the string link those two brackets to handle the "()()" case where 1 needs to connect to 4.

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

    the number of positions such that s[i] = s[i+1] = ')'

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15 months ago, # |
  Vote: I like it +233 Vote: I do not like it

F is shit.

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15 months ago, # |
  Vote: I like it +9 Vote: I do not like it

Problem C was nice.

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

    Wut it's easier than A, I'm mad because of B not saying it's hard but I'm so stupid and slow

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      15 months ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      me here, getting 150 on A :(((((((((((((((

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

      Good sir, can I know how you made that observation for C?
      I have never been so mad at a problem T-T

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

        I noticed that () () () for example is all connected then I assumed that all of are disconnected and everytime i see () then this one is connected to another sequence

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

          "I assumed that all of" you mean 'other' instead of 'of', right?
          Also, Thanks but I don't know why I couldn't think of this. Would it be lack of practice of 1400-1500 rated problems?

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

            It needs some luck i may haven't noticed it but if we are trained on graphs it won't be a problem

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

        I JUST use DSU and at the last step count the different number or parent.If make count+1 at '(' and count-1 at ')' you will see that all i for 1 to 2*n where count =1 or count=0 are conected. I use this obserbation to solve. Sorry for my poor English.

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

          Thank you! I guess having a bit more knowledge of dsu might have helped ;(

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

            It doesn't need any of that just a for loop would work

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

              Yea true, what I actually meant was, maybe knowing dsu well might have given me a different perspective because I couldn't make that observation.

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        15 months ago, # ^ |
        ← Rev. 2   Vote: I like it +3 Vote: I do not like it

        here's my crude drawing during contest for C, if it might help:

        notice how the component count increases for every two decreasing (or increasing, as others pointed out) steps in a row.

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

          Ohh nice, I will try this idea after few days and see if I can get the approach. Honestly, what I did was to notice the how deep the opening bracket's are going and counted them.
          For eg. ( ( ( ) ) ), depth was 2.

          And for this ( ( ) ) ( ( ) ), the depth for both first and second balanced sequence is 1. So, it becomes 2 and added 1 for the remaining component. And, it even passed for all given TCs, which made me believe theres something wrong in the code and not in observation.

          I get riddled in my own ideas ;-;

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        15 months ago, # ^ |
          Vote: I like it +1 Vote: I do not like it

        What i thought is when the string is ((())) complete then the answer is n but the time you take out a () outside, a component decreases. So calculate no of instant parenthesis. We already have 1 instant parenthesis in the starting case. So answer will be n-(instant-1).

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          15 months ago, # ^ |
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          Ohh wow, this actually makes sense to me. Wish I would have thought a bit in this direction.
          Alsoo, Congrats on specialist!!

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

    i messed up preety badly at c

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15 months ago, # |
  Vote: I like it +40 Vote: I do not like it

Problem F is boring and hard to implement :(

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    15 months ago, # ^ |
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    Any hints for D?

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

      The connected components are trees.

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        15 months ago, # ^ |
        ← Rev. 2   Vote: I like it +3 Vote: I do not like it

        Could you explain how to find spanning tree such that remaining 3 edges don't form a cycle( triangle )?

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          15 months ago, # ^ |
            Vote: I like it +30 Vote: I do not like it

          Random algorithm .

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          15 months ago, # ^ |
            Vote: I like it +34 Vote: I do not like it

          Determenistic solution: Do a DFS, for every vertex record parent and depth in DFS tree, find the 3 additional edges. If the edges do form a cycle, take any of the 3 edges (let's say (u-v), where depth[u] > depth[v]) and replace it with edge (u-parent[u]).

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          15 months ago, # ^ |
            Vote: I like it +32 Vote: I do not like it

          Find any tree, then, if the remaining edges form a cycle, build a new tree with one edge of this cycle in it. It will guarantee that the new tree is ok.

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

        Nope, for an example, in test case 2, you have one connected component and it is cycle.

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

    By the way , The $$$O(n\times \sqrt{nlog_2n})$$$ solution may be easier to implement.

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15 months ago, # |
  Vote: I like it +1 Vote: I do not like it

hints for D?

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    15 months ago, # ^ |
    ← Rev. 3   Vote: I like it +3 Vote: I do not like it

    I also failed it, my approach was this:

    1. use union find to color all red without loops

    2. at the end at most 3 blue edges will be left over

    3. if those three blue edges happen to form a loop, swap the last blue edge with one of the red edges -> there will be no blue loop any more

    4. my problem was I didn't know how to select the correct red edge to swap with, since for some red edges you will create a red-cycle

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

      You have to check if the edge you are changing from blue -> red should not be between an ancestor and child in the red tree

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    15 months ago, # ^ |
      Vote: I like it +4 Vote: I do not like it
    Hint 1
    Hint 2
    Hint 3
    Solution
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      15 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      How is this optimal?

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

        Each edge has to decrease number of either red components or blue components, and it can only decrese one of those by exactly 1. Thus, the optimal value is $$$2 * n - m$$$. In case of there not being a triangle we have exactly this value with the above method. In case of triangle also this value, but it's a little bit harder to see.

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15 months ago, # |
← Rev. 2   Vote: I like it +34 Vote: I do not like it

Defeated by the memory limit of problem F.

I wrote a segement tree without optimizing anything in the last minutes, using the memory of $$$O(n\log n)$$$. When I noticed the memory limits I didn't have time to make a change.

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

    Same here. 256MB is unnecessarily too tight! Lose the chance to gain rating.

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    15 months ago, # ^ |
      Vote: I like it +20 Vote: I do not like it

    My ST table didn't get MLE.

    However, I forgot to check whether there's a time which can reach n without passing any red lights and got WA on test 7. What a pity.

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

What is the expected time complexity for D?

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15 months ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

Can we solve D by randomizing? My idea is you only need to randomize m — n + 1 edges, and check if any of the two sets create a cycle. m-n+1 is at most 3 so it should be fast enough?

The proof of correctness is that, if the graph is a tree, obviously it makes no difference how the edges are allocated. If we have extra edges, then every edges will reduce the connected component by 1, so we make a tree for the blue, and the rest for red, as long as they are both trees.

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    15 months ago, # ^ |
      Vote: I like it +12 Vote: I do not like it

    Make MST. Pick the $$$m-(n-1)$$$ edges not in MST. Only when $$$m = n+2$$$ can these edges create a cycle. If they do, then take one edge and transfer it to other color and the edge to its parent to current color, provided that the swapped edge is not in its subtree.

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

What is a 'connected component' in Problem D? In the first example, it says blue edges form 2 components, but I only see 1.

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    15 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    $$$1,2,3,5$$$ was one component, $$$4$$$ was the other

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15 months ago, # |
  Vote: I like it +41 Vote: I do not like it

F is so annoying to implement (256MB memory limit makes it even worse!). But D and E are interesting.

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

    How to solve E? I thought its about calculating sums of $$$\frac{P^n_{2k}}{2^k k!}$$$ or smth for $$$1\leq k \leq \lfloor \frac{n}{2} \rfloor$$$

    (sorry about the big text, things were becoming unreadable)

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      15 months ago, # ^ |
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      Yeah, the way you do that subproblem is by thinking about it conceptually instead of combinatorially. You're trying to find the ways to pair some elements from n things. This can be written as a dp[n] = dp[n-1] + (n-1)*dp[n-2] (you either pair the first element with something, or u don't), which can be precomputed before you do the loop over number of 4-cycles.

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      15 months ago, # ^ |
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      Instead of permutations I thought about the problem with cycles (by building the graph of the permutation). To be almost perfect, cycles either have length 1 (pointing to itself) or length 2 (pointing to each other) or length 4, see picture. Rest was combinatorics. I will link my solution after the grading finished (was 7 minutes too late sadly...). I iterated over the amount of length-4 cycles.

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        15 months ago, # ^ |
        ← Rev. 2   Vote: I like it +8 Vote: I do not like it

        OHHHHHHH. I was thinking about cycles too, in this process I came up with that formula. totally missed the length-4 cycles though. E was a good problem after all!

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          15 months ago, # ^ |
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          171160691.

          First I calculated $$$P_i$$$ by this recursion: $$$P_0=1$$$, $$$P_1=1$$$, $$$P_i=P_{i-1}+(i-1)\cdot P_{i-2}$$$. This is the amount of partitions of $$$i$$$ elements into sets of size $$$1$$$ and $$$2$$$.

          Then we calculate the following:

          $$$\sum_{i=0}^{4i\le n} \binom{n-2i}{2i} (2i-1)!! \cdot 2^i P_{n-4i} $$$

          The sum goes over the amount of length-$$$4$$$ cycles. The binomial calculates how many different distributions for $$$2i$$$ neighbouring 2-tile-pairs there are. The Double Factorial is the amount of combinations to form $$$4$$$-tile-cycles by choosing $$$2$$$ each from the previously distributed $$$2$$$-tile-pairs. Each $$$4$$$-tile cycle then can have 2 different orientations, so we need to multiply by $$$2^i$$$. Then $$$P_{n-4i}$$$ is the amount of combinations to form $$$1$$$ or $$$2$$$ tile pairs from the remaining nodes.

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15 months ago, # |
← Rev. 7   Vote: I like it +5 Vote: I do not like it

why getting TLE at problem B? someone help me 

t = int(input())
while t>=1:
    n,m = map(int,input().split())
    if m<n:
        print("No")
 
    elif n%2 == 0 and m%2 !=0 :
        print("No")
 
    else:
        ans = []
        if n%2 !=0:
            ans = [1] * (n-1)
            a = m - (n-1)
            ans.append(a)
        else:
            ans = [1] * (n-2)
            a = (m - (n-2))//2
            ans.append(a)
            ans.append(a)
 
        print("Yes")
        for i in range(len(ans)):
            if i == len(ans)-1:
                print(ans[i])
            else: print(ans[i],end= " ")
    t-=1
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    15 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    @u1602016 You are not decrementing variable t, so it continues to loop forever :)

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

    Same problem. TLE with Python/PyPy3 (pretest 4), OK with C++ (system testing). Slow input/output in Python, maybe.

    Is it possible to get pretest 4 for experiments?

    Submission https://codeforces.com/contest/1726/submission/171110352 Room (with the other submissions with slightly smodified input/output) https://codeforces.com/contest/1726/room/464

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

      The way it works is that the input function in Python calls sys.stdout.flush. So everytime you call input you flush stdout, which can be really slow. The way to get around this is to use sys.stdin.readline instead of input. Worth noting is that input internally calls sys.stdin.readline.

      If you take a look at C++ submissions will probably see cin.tie(0), which is how you turn off this very same feature in C++.

      Knowing how to avoid flushing can be very useful in general. For example your (savsmail) C++ submission speeds up by a factor of 16 by avoiding to flush 171253842 .

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

        Wow, that explains a lot. Thank you!

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

    @u1602016 yes, the problem is with the slow I/O in python as it takes 5-10 times more to run compare to c/c++.

    For the cross-check part I have also converted your solution in c++, and tried to submit, and it works perfectly fine within time(561ms, note: your solution is correct) here

    Solution??? you could either use the fast I/O for python while doing CP (as @Varad2002 mentioned) or you could use c++ for CP (I would personally recommend this, as I have experienced the situation where my JAVA code throws TLE (even after using fast I/O) whereas same algorithm works within time limit in C++)

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

I couldn't enjoy this contest :(, might be very clever problems.

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15 months ago, # |
  Vote: I like it +9 Vote: I do not like it

hint for c?

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    15 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    )(

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    15 months ago, # ^ |
      Vote: I like it +12 Vote: I do not like it
    Spoiler 1
    Spoiler 2
    Solution
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      15 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Was this easy to catch this observation as i tried to solve trivally by making DSU but not able to do it

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

        Phew, well, depends on your understanding of easy. As I was solving the problem, the Spoilers/Solution which I posted reconstruct my thought process. And I'd say I have noticed them pretty quickly. I didn't even consider building the graph explicitly (and using DSU or similar) yet.

        I guess the first thing you want to do is find observations about the structure of the problem. Only if you can't find any structure, you take out the harder techniques. And then try to find structure on the harder techniques. And always user pen and paper!

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

Any hints for E?

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15 months ago, # |
  Vote: I like it +9 Vote: I do not like it

Really nice problems ! Keep it up authors ^_^

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

D: randomly color all edges. Break if both colors do not form cycles.

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    15 months ago, # ^ |
      Vote: I like it +20 Vote: I do not like it

    We tried our best to break such solutions, but it seems it was not enough :(

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      15 months ago, # ^ |
        Vote: I like it +13 Vote: I do not like it

      I would like to know how to break (or attempt to) these kind of solutions.

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

        My attempts were primarily focused on maximising the number of tests, with $$$n+2$$$ edges in every case, with graphs of ~1e3 (I think?) size. So mostly randomisation. I didn't have much hope of blocking everything because of the small number of extra edges compared to the number of possible spanning trees. If anyone has better ideas, I would be very interested to know...

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      15 months ago, # ^ |
        Vote: I like it +21 Vote: I do not like it

      Break? Isn't that correct?

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

        Some of our testers' solutions using the same idea didn't pass, so I assumed it is incorrect.

        Of course, if the number of iterations are sufficient it should pass with high probability. Admittedly, I did not do a thorough analysis of a randomised solution because the model solution does not use randomisation.

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          15 months ago, # ^ |
            Vote: I like it +17 Vote: I do not like it

          Well, this one looks way simpler than anything with some logic inside. I'm talking about the version with a guess that we can achieve maximum possible score ($$$2n - m$$$) and we repeat until we succeed.

          The intuition for me was that if it's anyhow hard/not straightforward to color the edges so that there are no cycles, these additional edges (yes, it's relative which ones are additional, but I'm talking about kind of a part of the graph in which something interesting happens) have to be close to each other, so the "hard" part of the graph is small, so there are not so many possibilities and we will find good one quickly.

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    15 months ago, # ^ |
    ← Rev. 2   Vote: I like it +1 Vote: I do not like it

    I did this: Colored red if nodes are in different disjoint sets, blue otherwise. However, for m=n+2, there is a possibility of a cycle of 3 formed.

    In that case, I made this observation: Suppose the 3-cycle is ABC then there is a node in A,B,C say A, such that AB and AC are connected using red. Then just take the last edge in the AB red path and make it blue, and make AB(blue) as red. It can be seen that this removes cycles in both red and blue.

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15 months ago, # |
  Vote: I like it +6 Vote: I do not like it

Really enjoyed the contest, especially because all the questions were not only mathematics based. Also, waiting to submit D, solved just after the time ended. Same thing happening for the second time :')

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

Code of Problem C <<< Solving Problem C

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15 months ago, # |
  Vote: I like it -13 Vote: I do not like it

wait what the heck

Spoiler
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15 months ago, # |
  Vote: I like it -14 Vote: I do not like it

I had to gain only 2 points to reach CM, and solved D on paper, started coding it more than 1 hr before the end, but in the end got TL 9 because i use maps which work slow. But my idea was fully correct. Now end up losing 7 points, and only 9 will be left before CM. this is fucking disqusting, it will be already 5th time when i have <10 points bafore CM!

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

Isn't $$$1.205710448301$$$ the answer for the first sample in H? I've checked that in geogebra and it seems to be so.

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    15 months ago, # ^ |
    ← Rev. 2   Vote: I like it +8 Vote: I do not like it

    Are you sure it's the first sample? I mean to say, you don't mean the second sample by any chance, right?

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15 months ago, # |
← Rev. 2   Vote: I like it +1 Vote: I do not like it

My solution to problem C passsed system tests, Even though I am pretty sure its so different from the intended solution

I solved with a sparse table and DSU, I am interested to know if the solution is correct and actually fast enough (I am 99% sure its fast enough but I don't know about its validity)

Can Someone Verify?
Submission : https://codeforces.com/contest/1726/submission/171128759

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    15 months ago, # ^ |
      Vote: I like it -8 Vote: I do not like it

    do note that a lot of the faster ones (usually in Div1) solved it with DSU

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

    I too solved it using seg tree and DSU.

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

Back to back bad performance from me, hope I can bounce back soon. People who are in a phase like me, "This shall pass too" :)

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

Check out this ridiculously easy solution for C: 171112911 (don't pay attention to that bin_power_of_two function — I forgot to delete it)

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

    The interesting thing is that I haven't found any of the red guys yet who would have approached the problem the same way.

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

      yes, because for them, the simpler (in the sense of observation) DSU implementation came first in mind.

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

        Could you explain the DSU solution?

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

          they just constructed the DSU like how they would do a normal graph and counted the components like how they would do a normal graph. only tricky part is what to merge and that's it

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      15 months ago, # ^ |
        Vote: I like it -8 Vote: I do not like it

      171072071

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

    literally ('('+input()+')').count('))') in python.

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    15 months ago, # ^ |
      Vote: I like it +1 Vote: I do not like it

    Even this works

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      15 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it 
          ll ans=1;
    forsn(i,1,2*n){
        if(s[i]=='(' && s[i-1]=='(') ans++;
    }
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15 months ago, # |
  Vote: I like it -10 Vote: I do not like it

These unfamiliar names kinda throw me off the track. Wish we could use simpler names

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

If the current problem C(bracket sequence) were problem A, would it have been as difficult as it is now?

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

My exactly same code got ac after system test

AC code : https://codeforces.com/contest/1726/submission/171155848 TLE code : https://codeforces.com/contest/1726/submission/171146706

code was same compiler was same.... Just for system test load my code got tle...Please @authors do something about this...MikeMirzayanov please do something

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    15 months ago, # ^ |
      Vote: I like it -10 Vote: I do not like it

    code was same, compiler was same, the difference is the seed

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

      what is seed?

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        15 months ago, # ^ |
        ← Rev. 2   Vote: I like it -10 Vote: I do not like it

        the seed is the value an RNG uses to start with. in your code this is chrono::steady_clock::now().time_since_epoch().count(), which varies by when the submission is run.

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

          I haven't used rng anywhere in my code...so it was not my fault at all.

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            15 months ago, # ^ |
              Vote: I like it -10 Vote: I do not like it

            oh, didn't notice it. turns out that your code barely just fit into the TL. I think you would've gotten hacked if this happened during competition, though.

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

    can you do something ? Newtech66

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    15 months ago, # ^ |
      Vote: I like it +15 Vote: I do not like it

    Do you really think 1996ms solution for 2s problem is correct?

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      15 months ago, # ^ |
        Vote: I like it -7 Vote: I do not like it

      How should I calculate this in contest time? Pretests got passed and even system tests too. But the judge load caused me TLE. Whats my fault here?

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15 months ago, # |
← Rev. 2   Vote: I like it +116 Vote: I do not like it

On A, I saw a solution which initializes the answer with $$$|a_n-a_1|$$$ and, while I could've simply made a test case which breaks that solution very obviously, I instead decided to make fun of the pretests by hacking it with a test case of 50 tests with $$$n$$$ being a random number between $$$1$$$ and $$$6$$$ and $$$a_i$$$ being random numbers between $$$1$$$ and $$$10$$$. While most obviously wrong solutions like the one I hacked FSTd due to hacks specifically made against such solutions, this test case of 50 random tests also unintentionally caused FST of almost 100 solutions that don't work if a test with $$$n=1$$$ has a test right before it such that $$$a_2$$$ of the test right before the test with $$$n=1$$$ is greater than $$$a_1$$$ of the $$$n=1$$$ test.

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    15 months ago, # ^ |
      Vote: I like it -13 Vote: I do not like it

    LOL that is awesome. this is a legend, as you are too

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15 months ago, # |
  Vote: I like it +23 Vote: I do not like it

Why is the D's constraints so big? I can't understand why sum of N is 1e6. It is hard for python... (I passed pretest but got TLE on system test)

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

Done C with DSU... how u guys come up with so easy solution...

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    15 months ago, # ^ |
      Vote: I like it -13 Vote: I do not like it

    observation, but observation isn't all so easy. observation comes along with experience as well

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

      I am not very comfortable with these graph problem as you can see from my rating that i havent encountered many graph problems till now . Do these graph problem usually have these kind observation asking wrt C

      And please guide my what i should do to reach specialist as while practicing i am comfortable upto *1500 problems but during contest i keep on getting stucked on problem C which are usually in the range of *1400 to *1600

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

        I am not very comfortable with these graph problem

        So am I! This was a different case though. I just drew some parentheses strings with pen and paper, like '(' = rising slope, ')' = falling slope. Then I connected what I needed to connect based on the statements. After that, it was a long phase of thinking, thinking hard. Usually time for thinking > time for coding, this is very normal. Accepting this is one crucial step to Specialist, imho.

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

Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

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

Got Accept with FST code for D.
Why is this so slow? It's just O(NlogN) and N <= 1e6
Can someone explain?

https://codeforces.com/contest/1726/submission/171146699
https://codeforces.com/contest/1726/submission/171158814

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

    Same happened to me and costed me losing CM. @admins should look at this.

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

    maybe try having a function fix(pair p) that returns a sorted pair(smaller element before larger element). idk if it fixes your solution but it did work for my friend

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

Btw, I FST for D, but when I try to resubmit the exact same code, it passes (barely). Is it possible to reverse the FST?

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

    Faced the same issue. Not sure if they will take it into consideration, but I hope they do.

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

Hello everyone, I'm a newbie participant. On problem A, although it runs perfectly on my computer, I got runtime error. How Can I fix this problem? I copy my code below. Thank you. (I use Microsoft Visual Studio BTW)

#include <iostream>

using namespace std;

int main() {

	int t;
	int n[1000];
	int s;
	int qr;
	int zr;
	int result;

	cin >> t;

	while (t--) {
		
		cin >> s;

		for (int i = 0; i < s; i++) {

			cin >> n[i];
		}

		for (int z = 0; z < s; z++) {
			
			for (int q = z; q < s; q++) {

				if (n[z] > n[q]) {

					zr = n[z];
					qr = n[q];
					n[z] = qr;
					n[q] = zr;

				}
				else {
					continue;
				}
			}
		}
		if (s == 1) {
			result = 0;
		}
		else {
			result = n[s - 1] - n[0];
		}

		cout << result << endl;
	}

	return 0;
}
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    15 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    you need to increase the size of array. notice n<=2000 but you have declared an array of size 1000 only

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15 months ago, # |
  Vote: I like it +4 Vote: I do not like it

After decades of the contest, Benq is on the top! Happy to see this (❁´◡`❁)

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

Just a shit

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15 months ago, # |
  Vote: I like it +1 Vote: I do not like it

Before I sleep,I saw my name turn green..What F???

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

lol everyone is downvoting this post and the editorial.

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15 months ago, # |
  Vote: I like it -15 Vote: I do not like it

The coding club of IIT KGP is gonna face a lot of disrepute now becoz of you guys

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

    Exactly , they may set 7 problems instead of 8 and time for 2 Hours would be better.

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15 months ago, # |
← Rev. 2   Vote: I like it +58 Vote: I do not like it

A sincere request to everyone!

Newtech66 and anubhavdhar have worked really hard for this round and they weren't aware of what happened in problem F. If you don't like the problems(except F), feel free to downvote the blog, but please refrain from doing so if this is the only reason! It would be very unfair and unjustice with them :(

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    15 months ago, # ^ |
      Vote: I like it -38 Vote: I do not like it

    This is bullshit. I support downvoting the blog post because of F, it's high time all the authors start asking their co-authors how they developed the problems, so situations like this don't repeat. Every author of the round should be held equally accountable for each problem, and the round in general.

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15 months ago, # |
  Vote: I like it -8 Vote: I do not like it

where is my ratings for this contest.

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

Why this contest is Unrated?

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

my rating where is gone?!! i just reached pupil pls any information?

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

This is has happened 2nd or 3rd time, that contest has been declared unrated due to picking problems from other platforms. Please make sure that problems are original beforehand. There goes a lot of effort in solving them and then finding that contest has been declared unrated just breaks the heart :( .

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