Codeforces Round #423: rated, based on VK Cup 2017 Finals
By KAN, 7 years ago, translation, In English

Hi!

On Tuesday, July 11, 2017, at 16:35 UTC Codeforces Round #423 will be held. Some of the problems will be from VK Cup 2017 Finals, some are prepared by Codeforces team to complete the problemset.

Note that there will be two rounds based on VK Cup Finals, so I want to ask finalists not to discuss problems until the end of the second round.

There will be six problems in each division, four of them shared. The problems of this round are proposed, prepared and tested by: vintage_Vlad_Makeev, demon1999, fcspartakm, Perforator, MikeMirzayanov, PavelKunyavskiy, qwerty787788, Belonogov, izban, tourist, vepifanov, AlexFetisov, winger, Errichto, Gassa, naagi, ashmelev, Endagorion, ifsmirnov, Arterm and me. Huge thanks to all who helped with the preparation!

There will be prizes from VK social network in this round! Namely, among participants solved five or more problems in second division, or three or more problems in first division, five are to be selected randomly. They will receive championship souvenirs. There is no country nor language restriction, everyone can win a prize. One don't have to have participated in VK Cup to receive the prize. Exact selection algorithm will be announced before the start of the round. There will be prizes in the second round using Finals' problems as well!

Good luck!

Congratulations to the winners!

Div. 1:

  1. W4yneb0t
  2. jqdai0815
  3. ko_osaga
  4. Radewoosh
  5. dotorya

Div. 2:

  1. ccz181078
  2. _rqy
  3. houruize
  4. _Is_It_Rated_
  5. liutianyi000

The analysis is here.

Corresponding problems from VK Cup 2017 - Finals:

The souvenirs winners are:

Eligible list place Contest Rank Handle
19 827 19 zeliboba
42 827 42 kevinsogo
114 827 115 triploblastic
125 827 126 lexuanan
138 827 140 gepardo

Congratulations!

  • Vote: I like it
  • +389
  • Vote: I do not like it

Comments Comments (277)
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7 years ago, # |
← Rev. 2   Vote: I like it +77 Vote: I do not like it

Thanks for your efforts

But how do you make sure that no one of the finalists has leaked the problems ?

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7 years ago, # |
← Rev. 7   Vote: I like it -96 Vote: I do not like it

Is it rated?

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    7 years ago, # ^ |
      Vote: I like it +51 Vote: I do not like it

    no it's not, they wrote the word "rated" as a joke :)

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    7 years ago, # ^ |
      Vote: I like it +325 Vote: I do not like it

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      7 years ago, # ^ |
        Vote: I like it +219 Vote: I do not like it

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      7 years ago, # ^ |
        Vote: I like it +22 Vote: I do not like it

      OMG, 300+ upvotes on a single comment. I want to thank my parents for bringing me up, my friends who have always supported me in everything I do and everybody who voted for this comment. Also, a huge "Thank you" goes to Mike Mirzayanov, without whom this would not be possible, for creating this awesome platform.

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7 years ago, # |
  Vote: I like it +66 Vote: I do not like it

Does souvenirs includes the cute cats?

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

Duration of the rounds? 3 hours, or more?

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7 years ago, # |
  Vote: I like it +63 Vote: I do not like it

15 reds

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7 years ago, # |
← Rev. 2   Vote: I like it +36 Vote: I do not like it

Wow! the number of problem setters exceeds the number of problems itself. xD

It must be a challenging round.

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it -8 Vote: I do not like it

    The problems of this round are proposed, prepared and tested by

    Not all of them are problem setters. So we can't be sure that the number of setters exceeds the number of problems.

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7 years ago, # |
← Rev. 3   Vote: I like it -35 Vote: I do not like it

The start time is too late

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7 years ago, # |
← Rev. 3   Vote: I like it -82 Vote: I do not like it

Okey, who wants to receive championship souvenirs, put dislike!

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Codeforces server seems awfully slow. Does anyone know what is the matter with vjudge1 , vjudge2 and so on. There is a huge number of submission from these ids.

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    7 years ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    Well, vjudge(.) accounts are used to submit solutions from VirtualJudge and there are not so many submissions to hack codeforces. However I worry that there can be problems with codeforces during the contest and also API is not working which I use participating in the contests(

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7 years ago, # |
  Vote: I like it +2 Vote: I do not like it

Good luck everyone :D

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7 years ago, # |
  Vote: I like it +8 Vote: I do not like it

I got so used to times specified in MSK timezone (which is also more convenient for me) that I thought this one is in MSK too and completely ruined my schedule, oops

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    7 years ago, # ^ |
      Vote: I like it +27 Vote: I do not like it

    In the English post it is written explicitly that time is in UTC, isn't it? On the contests page the times are noted with a timezone, right?

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Yeah I know it's completely my mistake, I was quickly checking it on my phone without paying much attention.

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7 years ago, # |
  Vote: I like it +27 Vote: I do not like it

For div 1 contests, will task be same as on official contest, or we will have one easier ?

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Thx for preparing the contest... GL all coders :) Let's Rock

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7 years ago, # |
← Rev. 3   Vote: I like it -42 Vote: I do not like it

.

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

delay :)))

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

10 minute delay? :(

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7 years ago, # |
  Vote: I like it +4 Vote: I do not like it

And again delayed...

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7 years ago, # |
  Vote: I like it +52 Vote: I do not like it

I think winner of div 2 will be ccz181078 :)

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7 years ago, # |
  Vote: I like it +20 Vote: I do not like it

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

delay------10 minute.

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7 years ago, # |
  Vote: I like it +21 Vote: I do not like it

I have a feeling that Div2 result will mostly be decided by the speed of solving the first 2 problems.

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7 years ago, # |
  Vote: I like it +59 Vote: I do not like it

You know what's worse than being in the friendzone for long? Being in the gray zone after a ton of contests. :(

Hoping to change that in this round. Need +40. Good luck to me! Good luck to everyone! :)

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7 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Where is the scoring distribution?

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7 years ago, # |
  Vote: I like it +52 Vote: I do not like it

To determine the winners we will use the following algorithm.

  1. List all participants from div. 1 who solved at least three problems from top to bottom of the standings, ties broken by the time of last AC.
  2. List all participants from div. 2 who solved at least five problems from top to bottom of the standings, ties broken by the time of last AC.
  3. Append the second list to the back of the first list.
  4. Use the following code to generate the indices of the winners in the resulting list. The seed is the score of first place in the next div. 1 contest. The length is the length of the list.
code
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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    "The seed is the score of first place in the next div. 1 contest."

    I can see people in the next contest posponing their submissions by few minutes and intentionally unsuccessfully hacking just to get a more favourable seed.

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    7 years ago, # ^ |
      Vote: I like it +20 Vote: I do not like it

    And? Who won? :D

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    7 years ago, # ^ |
    ← Rev. 3   Vote: I like it +20 Vote: I do not like it

    I wrote some scripts to make the process of winners determining easier, and they produce results slightly different then Xellos's. Here are my results (not official yet):

    List place Contest Rank Name
    19 827 19 zeliboba
    42 827 42 kevinsogo
    114 827 115 triploblastic
    125 827 126 lexuanan
    138 827 140 gepardo

    The seed is 5581, the length is 213. The output of the generator is 19 42 114 125 138. The scripts I used are below. Xellos, have you used any scripts? If so, can you see where the difference came from?

    python code
    c++ code
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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      The list places are the same for me. I didn't use scripts, just went through the scoreboards manually, marked off ineligibles and then added 1 to list places for each smaller ineligible place; maybe I added incorrectly for place 131.

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I believe that my results are correct, so now they're official.

      If anyone has comments about the code, please comment.

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7 years ago, # |
  Vote: I like it -24 Vote: I do not like it

what will be my contribution if i ask is it rated? and what the hell of lating?

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7 years ago, # |
  Vote: I like it +31 Vote: I do not like it

This contest has more hacks than the U.S. presidential election!

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    7 years ago, # ^ |
      Vote: I like it +20 Vote: I do not like it

    maybe because of the large number of russians here :p

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Is there anyone easier way to do E than FFT?

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

Easy div.2 A hacks

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    How can i hack other's code ?

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      You need to lock this problem firstly(problems page), then you can open other's code (room page, double left click / ctrl + click on their score on problem) , look through and try to hack (button will be at the end of code).

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        There are no constraints? e.g. rating only lock?

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          7 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          You can lock only task you have already solved(solution has passed pretests) and you cannot send any code on this problem after lock.

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7 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

4 1 2
1 1 1 2 Ez hacks :) P.S.: Sorry, room 125 ^_^

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

    Answer is 2 right?

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      yep

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        why 2 ? i thpught the answer is 0

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          7 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person. If there are still none of them, the restaurant denies service to this group.

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    7 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Hacked is better than system testing failed

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7 years ago, # |
  Vote: I like it +14 Vote: I do not like it

WA on pretest 11 on Div2 D. What could be this case?

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    7 3

    My code was giving wrong answer on this test case and I also got WA on pretest 11.

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7 years ago, # |
  Vote: I like it +3 Vote: I do not like it
Div2A Hack
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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    how is it 2? the first one will take an a table, the next 2 a b table and the last 2 the last b table. isn't this right?

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I assumed the same but..

      If a group consist of one person, it is seated at a vacant one-seater table. If there are none of them, it is seated at a vacant two-seater table. If there are none of them, it is seated at a two-seater table occupied by single person.

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

        oh god, i understand now lol. that restaurant has some stupid policies.

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

          Could it be said that the pretests were very weak?

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

            the pretests were fine. if i had wa i would have wasted too much time trying to debug, not realizing that i understood the statement wrongly.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I think very many solutions falls in this test, me too...

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7 years ago, # |
  Vote: I like it +8 Vote: I do not like it

Is the hack for C TLE ???

If so I'm screwed.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yeah same here, thanks to map<string, vector>

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Goddamn it the code was ready I was just about to resubmit it...

      Ah well...

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Hack for C div2 or A div1?

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7 years ago, # |
  Vote: I like it +30 Vote: I do not like it

people were hacking faster than eminem rapping

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7 years ago, # |
  Vote: I like it +8 Vote: I do not like it

TC 11 for DIV2 D any idea?

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    7 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Don't know if it's that, but check for 13 6, answer is 4.

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    7 years ago, # ^ |
    ← Rev. 2   Vote: I like it +17 Vote: I do not like it

    may be it was

    8 3

    what i think it was

    2        6
 \     /
  3   5
   \ /
    1
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    7
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    8 
    |
    4

    and not like

    1-2-3-4-5-6-7
  |
  8
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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Nope, I got WA 11 but I output 5 on this

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        That is why I wrote may be!! but after working on this my pretests were passed

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve E?

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7 years ago, # |
  Vote: I like it +42 Vote: I do not like it

After a really bad performance on A and B...I know if I continue to do this contest, as usual, I am definitely not going to get a good result, so I opened F. I did F for more than one hour and I didn't pass pretest 19 until the contest ended.

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    7 years ago, # ^ |
      Vote: I like it +31 Vote: I do not like it

    Why even bother with A and B? You can start with F straightaway :D Yes, this strategy might not give you the best result, but at least you will spend the contest solving an interesting problem.

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7 years ago, # |
  Vote: I like it +22 Vote: I do not like it

When you cannot debug a simple FFT solution for 1 hour :'(

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

4 1 2 1 1 1 2

Hack test in problem A :D :D Good Luck all :D

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

What is the hack for Div2C?

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve Div 2 D? And is Div2 E just a simple segment tree that remembers for every letter and every position % 10 number of occurences on every prefix?

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

    I solved Div2E nearly the same way you described and used bit instead of segment tree.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    From one node, the other nodes extend in k directions. like a bfs.

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Just like that? So simple? Any ideas how to prove that?

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    tried the same, but got mle. i suck at implementing seg trees.

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    7 years ago, # ^ |
    ← Rev. 3   Vote: I like it +5 Vote: I do not like it

    For Div2D.

    The tree wil be of star shape. So, put k vertices around centre. Keep doing this until you have less than k vertices. Then, put each of the remaining vertices in each of the branch. Maximum distance is

    • 2*floor(n-1/k) in case when n-1%k==0

    • 2*floor(n-1/k)+1 in case when n-1%k==1

    • 2*floor(n-1/k)+2 in case when n-1%k>1

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      How do you prove (even intuition behind that) that a star tree generates the lowest diameter?

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

        Followin is just an attempt to prove. Correct me if I am wrong.

        CASE: n > 2k

        We can prove this by induction.

        Let's say we know the best tree consisting of n - k vertices and k leaf nodes. So, our required tree will be formed by adding k vertices to the best tree consisting of n - k vertices and k leaf nodes , one at each of the k leaves.

        CASE: n <  = 2k (This is base case)

        We can clearly see that in this case it will be star. Any other shape of tree yeilds greater diameter.

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          7 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          The proof is pretty gappy. You don't prove that in the first case, there can't be a better solution.

          Also, when using induction in graphs, most times it's more efficient to go backwards, inspect a graph, take away a node/edge, and by knowing the statement was true for the previous graph, it can be shown, that it is for the current graph also.

          If you use the induction forward, there will be troubles to proove that you can achieve any graph this way.

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        7 years ago, # ^ |
          Vote: I like it +18 Vote: I do not like it

        There can't be a node with higher degree than k, because if a node had more edges, then the network would have more than k exit nodes.

        Let's assume the highest degree is smaller than k. Pick the highest degree node, and choose the two longest "branches" starting from it. At the end of these "branches" there has to be an exit node. If we would have a node instead of this one, with degree of k, then this two longest "branch" would be at most as long as it is now, because all the non-exit nodes, can be distributed into more parts.

        Also with this change, other lengths won't be longer, because it can be seen easily, that the longest path in this new network must go through the centre of the "star".

        In conclusion, the star network isn't worse than any other, hence it's one of the optimals.

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7 years ago, # |
  Vote: I like it +12 Vote: I do not like it

I found D easier than C in the contest.

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Hack for C div2 or A div1????

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7 years ago, # |
  Vote: I like it +14 Vote: I do not like it

Only 250points difference between C and D? Really?...

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7 years ago, # |
← Rev. 2   Vote: I like it +35 Vote: I do not like it

This was a CF round for me :D

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Also div2 A hack: 5 1 2 1 1 1 1 1

It worked, if somebody didn't count the number of split tables, just used a boolean.

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I tried to hack Div2 C for TLE. So I tried hack through generated input, then spammed some large string and same index. I care that sum of the size of the strings are equal to 10^6 and sum of k values are equal to 10^6 but I got invalid input with this error "Validator 'val.exe' returns exit code 3 [FAIL Expected EOLN (stdin)]". What is the possible reason?

If anyone succeded to hack via generated input, can they show their generation code?

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

    You forgot to output newline somewhere

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      So sorry to bother you again, can anyone find the mistake? Here is the generation code.

      using namespace std;
#define PB push_back
#define MP make_pair
typedef long long int ll;
int main(){
    std::ios::sync_with_stdio(false);
    cout<<1000<<endl;
    string t = "";
    for(int i = 0;i<1000;i++)
           t.PB('b');
    for(int i = 0;i<1000;i++){
        cout<<t<<" "<<1000<<" ";   
        for(int j = 1;j<=1000;j++)
            cout<<j<<" ";
        cout<<endl;
    }
    return 0;
}
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        7 years ago, # ^ |
          Vote: I like it +8 Vote: I do not like it

        You mustn't output " " at the end of the line.

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7 years ago, # |
  Vote: I like it -8 Vote: I do not like it

Div1 C memory limit too strict ???

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yup :( I just changed vector<pair<long, string> > to vector<pair<long, long> and string[MAX] and it passed. How to check difference in memory in this case?

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I changer my segtree size from maxn*4 to maxn*3 and it passed damn

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

What was that rogue test case in problem A Div 2 that many people missed?

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

in Div2 B if all the cells are black the solution will be -1 or 0 ?

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

div-2 A 5 1 3 1 1 1 2 2 i have hacked the code of user sugarRush his code gives 2 as a answer for above TC. but correct answer is 0 for above TC. some body please help me out ?? why

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

    ans is 2 bcoz first 3 group will occupy different tables as given in question than there will be only 1 table left for the 2 sized groups

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7 years ago, # |
  Vote: I like it +13 Vote: I do not like it

pretest 7 in E was an antitest for doing fft mod 2^27*3*5+1? Or did i make some stupid mistake?

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7 years ago, # |
  Vote: I like it -8 Vote: I do not like it

System test is started...

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7 years ago, # |
  Vote: I like it +30 Vote: I do not like it

Waiting for System Testing is like waiting for your HIV testing results...

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7 years ago, # |
  Vote: I like it +2 Vote: I do not like it

That feeling, when you couldn't solve task during contest and solved it for 6 minutes after contest

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7 years ago, # |
← Rev. 2   Vote: I like it +37 Vote: I do not like it

Div 2 A

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Yep, that's my hacking! My Room

    My hack tests:

    4 1 2
1 1 1 2

    ANS: 2

    5 1 3
1 1 1 1 2

    ANS: 2

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7 years ago, # |
  Vote: I like it +18 Vote: I do not like it

test 8 in Div2 C :))))

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

such shitty pretests for div2C

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

The contest is a massacre for any problem solver XD WOW

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

Memory Limit Exceeded on test 8 in Div2 C :(

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    7 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    If I understand the statement well, you can have a string of length 106 (e.g. aaaa...aaa) and then 106 positions for this string : 1, 2, 3, ..., 106. Maybe in this case you register 106 times a string of size 106 ?

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      yes. i was pushing the string to each of such indexes. Instead of pushing, could have just stored the max. length string for each index. (instead of creating a vector of string, sorting it and picking the last one. :P)

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7 years ago, # |
← Rev. 2   Vote: I like it -53 Vote: I do not like it

Very funny. Write some complicated statement in d2 C/d1 A. Do not add appropriate pretest and watch majority of the solutions fail. Did you guys bring enough popcorn?

Edit. Some people still don't know. The strings coincide, so the total amount of occurences can be huge, you should set each letter only once...

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    7 years ago, # ^ |
    ← Rev. 3   Vote: I like it +93 Vote: I do not like it

    Sorry, are you complaining about the fact that a solution that is obviously too slow gets TLE?

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      7 years ago, # ^ |
        Vote: I like it -33 Vote: I do not like it

      I'm complaining that it was not obviously too slow. It is the fact that you can't discuss as majority of submitted solutions failed.

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        7 years ago, # ^ |
          Vote: I like it +57 Vote: I do not like it

        Copying strings of length 105 from 106 different positions gives 1011 operations. What is not obvious about that?

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          7 years ago, # ^ |
          ← Rev. 2   Vote: I like it -40 Vote: I do not like it

          The fact that they can overlap. It was mentioned that they can coincide — I understood that they can have some common parts but not that the positions may overlap.

          The statement was not clear — perhaps examples showed that problem but it should not be required to read them to understand the task or at least it should be mentioned that "be careful, statement is not too good, read the examples". Besides it was not clarified what does it mean to be lexicographically minimal.

          Bottom line is — the statement was confusing which was proven by very high rate of unsuccessful submissions.

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            7 years ago, # ^ |
            ← Rev. 2   Vote: I like it +18 Vote: I do not like it

            You're just grasping straws. If you made the assumption that strings cannot overlap you cannot blame that on the statement not explicitly mentioning it.

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            7 years ago, # ^ |
            ← Rev. 4   Vote: I like it +31 Vote: I do not like it

            The statement doesn't say that they don't overlap. Thus, they may overlap. I don't see what can be unclear about it.

            P.S. The only thing a high rate of failed submission means that (surprise!) a lot of people submitted an incorrect solution. These solutions are blatantly wrong, it's not a corner case or something.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    i do, but still TLE... I'm curious.

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

      Probably java.

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Yes, and I used my own SortedMap implementation, which made the code about 2x slower than when using the standard TreeMap implementation. Still O(nlogn), pity that it was rejected.

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          7 years ago, # ^ |
            Vote: I like it +4 Vote: I do not like it

          I can confirm java is probably the problem. I translated my Java solution that failed systests into C++ and I got AC. It's annoying that problem-setters often seem to make bounds that are very unfavorable to java users, but understandable considering the main focus is on making sure suboptimal C++ solutions to not pass.

          Reference: 28433564 28453400

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            7 years ago, # ^ |
              Vote: I like it +4 Vote: I do not like it

            But there is a linear solution to this problem.

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

              I get that, but based on the bounds I expected am NlogN to pass in time. Maybe that's because I generally compare 10^8 operations ~ 1 second, which might be more accurate for C++ than Java.

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

                I don't get that there's a linear solution to this problem. But, that's why I'm not red and will have to wait for the tutorial...

                EDIT: Oh, yes, now I see, use an array instead of a (tree)map, that's linear indeed, thanks.

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

I don't get how my submission to div1A (28436078) can get a TLE... Can anyone help me understanding this? It seems to me that my program does something like 5·106 operations in any case...

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    7 years ago, # ^ |
    ← Rev. 9   Vote: I like it 0 Vote: I do not like it

    I was wrong. Deleted.

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The length of the answer can exceed 106. So you have an access violation that changed the loop counter or something like that.

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

      but I used array size 2e6+4, still I got tle on test case 8. Any reason for that?

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

      No it's not the problem, I have no access violation.

      Actually I got it : I wrote s = v[remember[i]] in a loop of size 106, and v[remember[i]] can be a string of length 106. So if I simply erase this and replace s by v[remember[i]] in the next lines, it should get AC... :'(

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    same doubt

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    7 years ago, # ^ |
      Vote: I like it +24 Vote: I do not like it

    You made the same mistake I did, copying the string each iteration instead of referencing it:

    s = v[remember[i]];

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

When you fail at A and C... Because of edge cases... I failed C because i used constant too low (2*10^6)

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

Div2C/Div1A how could this pass systest? http://codeforces.me/contest/828/submission/28450333

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    7 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    It takes about operations. The worst case is 1000 strings, each of length 1000, and each appearing at 1000 positions that are spaced 1000 apart.

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      guys i still struggling to get AC :( , if someone can help me

      I got WA on test case 5 and my solution works in O(|Needed String|)

      Here

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7 years ago, # |
  Vote: I like it +133 Vote: I do not like it

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Did O(m * log2(m)) pass in D?

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    7 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    Yes

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      HLD or smaller set to bigger set?

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        7 years ago, # ^ |
          Vote: I like it +25 Vote: I do not like it

        HLD , i started coding small to large with priority queues but that would be NlogN memory so i switched to HLD at the last moment.

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Actually small-to-big works as well; limits aren't too tight.

        28460132

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve E div2 ?

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

    I built segment tree for each length of e, every position in e, every letter. So it's about 220 segment trees. I didn't use custom allocator, so got TL. But solution with custom allocator gets AC.

    PS: can somebody explain me better solution?

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

      I also did something similar but using a space around 40*N and BIT trees.

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        ML is not problem. Every adding operation uses O(logn) memory, so it's O((n + q) * logn). My solution used 393 MB  <  512 MB.

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I didn't use a custom allocator either. But I made sure that when I was looking at length L, my Fenwick trees all had size n / L, maybe that makes a difference. Here is my solution: 28440263

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        I think because you used arrays you haven't troubles with memory allocation. And I'm not sure, but Fenwick works faster than segment tree

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7 years ago, # |
  Vote: I like it -64 Vote: I do not like it

Who wants it to be unrated? XD

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7 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

For Div 1 A/ Div 2 C, consider the following testcase:
4
bcc 1 3
c 1 5
bcbc 1 6
cc 1 4
The answer should be "bcbccbcbc" but the accepted solutions output is "aabccbcbc". Why so? The question states that there is at least k occurrences of each string!

EDIT: My mistake! Interpreted the question wrong!

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

    Question also states that you need to print lexicographically smallest string of all possible strings(answers). In that case u can assume that at least k means exactly k and put 'a' at all the remaining positions of string.

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7 years ago, # |
  Vote: I like it +39 Vote: I do not like it

Why is rating predictor not working ?

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

So why I'm getting TLE? :/ C SOLUTION

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone tell if the round was actually rated?

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Thank you for the contest, the problems were great. However, there is one tiny issue.

28439979

Look at this submit. It passes flawlessly pretests and then gets TLE 16. We can easily notice what happened — author is outputting answer via unsynchronised cout. The real question is — wasnt there any maxtest in pretests in such a valuable problem? KAN

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    7 years ago, # ^ |
      Vote: I like it +23 Vote: I do not like it

    In test 3 the output is 50000 lines. Should be enough to test I/O.

    In test 16 the output is 1 line. The problem is at some other place.

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

      I'm sorry then. Strange, though. The submit looks correct

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        7 years ago, # ^ |
          Vote: I like it +14 Vote: I do not like it

        Don't use ordered set

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          7 years ago, # ^ |
            Vote: I like it +12 Vote: I do not like it

          LOL! Gray retards put minuses again, they don't know that Fenwick tree should be used in such kind of problems as it is much faster.

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

            "Gray retards" ! Be polite man!

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              7 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              Look, it worked! What's happening with this site?

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      And what about div2 C ?

      TLE AC just for I/O

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

    The writer of this comment asks, who is the author?

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7 years ago, # |
← Rev. 3   Vote: I like it +18 Vote: I do not like it

A solution with policy-based DS fits exactly in the time limit in Div 1 C:

http://codeforces.me/contest/827/submission/28452503

It is such a sad fact that i didn't notice that i got WA during contest because i forgot to check the set size.

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7 years ago, # |
← Rev. 2   Vote: I like it +76 Vote: I do not like it

Time Limit 2 secs

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Problem C in div2 says "**It is guaranteed that the sum of lengths of strings ti doesn't exceed 10^6**".

I got RE in this submission: 28448052 Here array size was 1001000.

I got AC after increasing array size 2*10^6 in this submission: 28452964

Why?

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

can anyone tell me why i am getting TLE i dont think so that my code is taking more than 2 secs CODE->28448213

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    7 years ago, # ^ |
    ← Rev. 3   Vote: I like it -10 Vote: I do not like it

    Found the problem!

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      7 years ago, # ^ |
        Vote: I like it -18 Vote: I do not like it

      So what do we do now?? KAN please help us!!!!

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

        You are copying the string every time.. I changed your code Now it gives WA on 16, so there must be some other mistake 28454163

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          7 years ago, # ^ |
            Vote: I like it 0 Vote: I do not like it

          actually i got it earlier now i m just trying again and by the way thanks

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone explain their solution for Div 2C String Reconstruction ?

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

    Keep an array of indices, (we have an array a of length 2,000,000, if a[i] is marked with an index j, this means that there is an occurrence of string j at position i, if multiple strings occur at the same position, save the index of the longest string which occurs at that position). Then, we loop through all of the positions. For each position, if the position is marked and we are not currently printing a string, then we start printing the string marked by the current position. If a position is marked and we are currently printing a string, then we check to see whether the leftover string we are currently printing is longer than the string marked on the current position. If the string marked on the current position is longer than the leftover string we are currently printing, then we stop printing the leftover string and start printing the string marked on the current position. If a position is not marked and we are not in the middle of printing a string, then we just print 'a', since 'a' is the smallest letter.

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7 years ago, # |
  Vote: I like it +162 Vote: I do not like it

28448669

//lol;

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    7 years ago, # ^ |
      Vote: I like it +86 Vote: I do not like it

    Hey, don't steal my upvotes =)

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      7 years ago, # ^ |
        Vote: I like it +41 Vote: I do not like it

      I have to admit it, I'm proud of kpw29. #annealer

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      7 years ago, # ^ |
        Vote: I like it +20 Vote: I do not like it

      Ehh, I'm not so proud now (but still proud). Test were just very weak... First test that came into my mind: "VKVKVKVKVKVK..." with one random change was enough to make these submissions fail:

      28448669 — WA

      28448743 — TLE

      28447467 — TLE

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7 years ago, # |
  Vote: I like it +23 Vote: I do not like it

My Rating=2017 in year 2017 :), so to me seems like a notorious coincidence

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Why does this TLE for Div2C

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

What is the prize selection algorithm?

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

For Div2 C, can anyone tell me, why I am getting WA on test for my code: 28454008

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7 years ago, # |
  Vote: I like it +18 Vote: I do not like it

A bit of bit magic helps you to solve E div 2/ C div 1 in O((n / 32)2) time.

Code: 28454011

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7 years ago, # |
  Vote: I like it +56 Vote: I do not like it

Ugh, n = 1 on F... 28450889 vs 28454566. This is the worst way to get a problem wrong. I want a refund for my brain :-P

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

can somebody plz point out the mistake, I tried for hours now, here's the code 28455191

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Ah! String Shallow copy denied me a double digit rank.

string a = "This is a blunder. I knew it. But I forgot. Dont mess it up"

string b = a. //This takes not O(1) but O(n) time :(

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

Challenge in div2 C

Add this constraint

"the string can only be reconstructed using the input strings"

this is how i understood the problem during the contest and i thought that it was very hard problem

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I can't access neither my submissions page nor any other user submissions page since the round ended. Server returns page with "The page is temporarily blocked by administrator." message. Does anyone else experience this issue?

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can someone help me with the Div 2 part C 828C - String Reconstruction. I wrote a solution but get black spaces in between the final string for the answer. Can someone tell me what the mistake might be?

Here is my source code:

int main(){

//ifstream cin("testcase.txt");
ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

char s[1000000];
long long int n;
cin>>n;
long long int m=0;
while(n--){
    string str;
    cin>>str;
    long long int x=str.size();
    long long int t;
    cin>>t;
    while(t--){
       long long int a;
       cin>>a;
       if((a+x)>m)
         m=a+x;
       a--;
       for(long long int i=a,j=0;j<x,i<=a+x;++i,++j){
         s[i]=str[j];
       }
    }
}
//cout<<m<<endl;
for(long long int i=0;i<m;++i){
    cout<<s[i];
}
return 0;

}

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The given strings, don't fill the whole string. You have to put 'a' s on every position, which wasn't occupied by a character from the input.

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can Someone explain there solution of Div2C Please.

Thank You!

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    7 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Disjoint Set use this data structure and its O(αn) can solve this problem

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7 years ago, # |
  Vote: I like it -16 Vote: I do not like it

is this contest rated btw?

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Can anyone tell me why my submission is getting a WA on div1A/div2C ?

My Submission

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

    2

    aca 2 1 5

    aba 1 3

    Answer: acabaca

    Your answer: acaaaca

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Thanks for the help, was able to rectify that. Getting a TLE on TC 8 now. Will have to make the solution more efficient, I think. :)

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

        Yes. You have O(ANS·k) algorithm (where k is sum of ki), but exists O(ANS + k) algorithm that much shorter and easier

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      7 years ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      IgorSmirnov can you help me :/ ?

      I got WA on test case 5 while my algorithm works in linear time O(|Size of needed string|) that's good enough

      Here

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        7 years ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        Did you check for the test case provided above in the same comment thread?

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

        3

        aaa 1 1

        abbb 1 3

        bb 1 4

        Answer: aaabbb

        Your answer: aaabba

        After if in 42 line you have to write i = ii;

        But your solution seems to bee not linear too, because in lines 37 and 43 you copy the string, so I can give you a test when you will have to copy O(n2) symbols :-)

        BTW, your solution is very close to AC, you just have to think how not to copy all the string

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

          Ohh I thought that this operation works in O(1) but you made it clear to me :)

          but there's another mistake somewhere :''( , I got WA on test case 5 too

          Here

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

            But here you should do lst = ii in if, shouldn't you? I can't check it myself, because I'm not in front of my computer now

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              7 years ago, # ^ |
                Vote: I like it 0 Vote: I do not like it

              I think nope :/ because i compare two strings with each others so i used lst = i (the first position before the beginning of the last string) so i am getting its end with lst + size

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

                Yes, but in this if you take a new string to compare with others, but you don't update its start. Just try to add this line (lst = ii) in if and submit

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7 years ago, # |
  Vote: I like it 0 Vote: I do not like it

WA on test 62 421div2 B what is it ?

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7 years ago, # |
  Vote: I like it +29 Vote: I do not like it

When will the editorial be published?

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7 years ago, # |
  Vote: I like it +6 Vote: I do not like it

Where is the tutorial ?

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7 years ago, # |
  Vote: I like it +42 Vote: I do not like it

Will we know (after second round ofc.) which task from finals is which?

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    7 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Sure. I'll update the posts with this information.

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