Vovuh's blog

By Vovuh, history, 10 months ago, translation, In English,

Miss Div. 3 rounds? :)

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Hello!

Codeforces Round #540 (Div. 3) will start at Feb/19/2019 17:35 (Moscow time). You will be offered 6 or 7 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. Probably, participants from the first division will not be at all interested by this problems. And for 1600-1899 the problems will be too easy. However, all of you who wish to take part and have rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that the penalty for the wrong submission in this round (and the following Div. 3 rounds) is 10 minutes.

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

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

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

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of my work. Thanks to my good friends Mikhail PikMike Piklyaev, Maksim Ne0n25 Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round.

Good luck!

I also would like to say that participants who will submit wrong solutions on purpose and hack them afterwards (example) will not be shown in the hacking leaders table.

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UPD0: I also would like to thank zimpha and arpa for help with the round testing and finding some bugs!

UPD: Due to the problems being really interesting the round will last 2 hours 15 minutes.

UPD2:

Congratulations to the winners:

Rank Competitor Problems Solved Penalty
1 Ghajini 7 259
2 ACtest 7 271
3 akali 7 274
4 AuSquare 7 341
5 bonchinche 7 343

Congratulations to the best hackers:

Rank Competitor Hack Count
1 hytzongxuan 73:-9
2 MarcosK 59:-6
3 theroot 28:-1
4 yqdjl6 23:-6
5 2014CAIS01 24:-9
497 successful hacks and 574 unsuccessful hacks were made in total!

And finally people who were the first to solve each problem:

Problem Competitor Penalty
A 1021869 0:00
B 15Y 0:05
C oldpreisnerboy 0:11
D1 omeravci372742 0:17
D2 Ghajini 0:17
E __1900__ 0:12
F1 MoonLight_v0 0:10
F2 1021869 1:33

UPD3: Editorial is published!

 
 
 
 
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10 months ago, # |
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Div. 3 is back!

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10 months ago, # |
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Hoping to become expert with this contest.

GoodLuck and High Rating to everyone!

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10 months ago, # |
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I wish all rated(<1600) users in this DIV-3 will be Unrated in the next coming div-3 rounds!

:-)

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    10 months ago, # ^ |
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    the only way it's happening is if next Div-3 will be unrated because of some problems. Is that what you wish?

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      10 months ago, # ^ |
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      I'm pretty sure they wish those who are Div3 now soon to get over the 1600 threshold. They wish everyone a good performance, that is.

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    10 months ago, # ^ |
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    Hugdiya

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10 months ago, # |
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Can anyone explain to me please, how the hacking session works? I would like to try it for this round. Thanks

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    10 months ago, # ^ |
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    Iirc you get 12 hours after the contest to hack any solutions you want, but not for credit like the normal rounds. The only reward is to named as top 5 hackers (though the top spot is almost always halyavin!

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10 months ago, # |
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I love Vovuh's Codeforces Round!! I wish I would go to expert..

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10 months ago, # |
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I noticed that "(or 8)" was added next to the problem count in the copy-pasted part. On the last div3 round that part wasn't included. Does this mean that this round will have 8 problems?

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    10 months ago, # ^ |
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    Yes, I suppose :^) But one spoiler: two of them will be in two (easy-hard) versions.

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10 months ago, # |
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We need more div.3!!!

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    10 months ago, # ^ |
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    we also need div.4 !!!

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      10 months ago, # ^ |
      Rev. 2   Vote: I like it -17 Vote: I do not like it

      Sorry i have changed it due my wrong opinion.

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        10 months ago, # ^ |
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        This problem is too hard for Div.4, the sum overflows int.

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        10 months ago, # ^ |
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        do you mean, now we need less div 3?

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          10 months ago, # ^ |
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          yes so that will be better for the beginners

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            10 months ago, # ^ |
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            I don't know how having less contests is better for beginners. The more the better!

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              10 months ago, # ^ |
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              yaaa... more contests are not necessary. Because we can participate in virtual contest. But division 4 will be helpful for the participant whose have rating below the 1200. and it will also encourage the rating below the 1200 and beginners. Because it not possible for everyone to cross rating 1600.

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                10 months ago, # ^ |
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                if this comment gets more than 100 votes, I'll create a div4 contest in gym

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10 months ago, # |
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Is there going to be a problem with an Easy and a Hard edition ?

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10 months ago, # |
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I love cf

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good luck guys GG HF!

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Best of luck to all.

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10 months ago, # |
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I feel like I need a Div 4 too. :P

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10 months ago, # |
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That "(or 8)" smells like an easy-hard version

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An uncertain-problem number contest :D.

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I wish everyone high increase in ratings today. Happy Coding!

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Happy New year all...

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today we have a ratings party :))

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I hoped that there will be 7 or 8 problems,because a new problem can provide you a new skill.

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Please make Div 3 rounds as frequent as Div 2. It is really good from the competition point of view. Otherwise Div 2 rounds include upto 2100 rated competitors on the ranklist which push coders like me to the bottom of the ranklist. Its really hard to even get to the top 1000.

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All the way to Expert. :D

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Good Luck and Have Fun all!!

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10 months ago, # |
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Broken Div.3 again? Why not to use this problemset for Div.2, when its difficulty is even about Div. 1.5... Or division is adjusted only by problem A? Look at good examples of problemsets of Div.3 contests: Codeforces Round #479 (Div. 3), Codeforces Round #481 (Div. 3).

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    10 months ago, # ^ |
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    I know when you're this low rating you shouldn't really complain, and I am not. I am genuinely interested who should be able to solve these problems ? Is it aimed for people with rating around 1600? first ones were.

    Anyways, liked the round and thanks for doing this to help beginners.

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      10 months ago, # ^ |
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      This image was taken with "show unofficial" disabled, so it seems like many under 1600 rating were able to solve the problems (with the exception of F2).

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After reading the problem C I hated the problem and quit solving :)

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    10 months ago, # ^ |
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    Me too :v :v :v

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    10 months ago, # ^ |
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    C was very interesting, there are several different ways to implement it (I think so), and you are to choose (find) one which is easiest to implement.

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    10 months ago, # ^ |
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    The only thing which came to my head(after reading C) was to jump out of the window

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    10 months ago, # ^ |
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    I think quality of rounds must be more important than quantity of them, seen many implementation and brute-force recently...

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"Probably, participants from the first division will not be at all interested by this problems. And for 1600-1899 the problems will be too easy."

Sees reds failing to solve F2... O_O

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    10 months ago, # ^ |
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    Div3 Problem Setters these days —

    If (We have Div1F problem)
    __ Print "Let's announce a Div3 round."
    else
    __ Print "Let's Wait for a Div1F problem."

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      10 months ago, # ^ |
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      I don't think F2 is that hard. It is just a smart extension and careful implementation of F1 (provided I did this things correctly :p). Maybe a Div 2 F though.

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10 months ago, # |
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I think E is similar to Your text to link here...

I am sorry that it just have chinese description.

You can see the solution of this from Your text to link here...

And I think that it can swap C and E

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10 months ago, # |
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What's pretest 6 of problem C ?

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    10 months ago, # ^ |
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    me too ! i got WA at pretest 6 :((

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      10 months ago, # ^ |
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      It may well be a matrix with odd n and quantities of several numbers not divisible by 4, but divisible by 2. For instance, consider:

      input:
      3
      1 1 1 1 1 1 4 4 7
      
      output:
      YES
      1 4 1
      1 7 1
      1 4 1
      
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    10 months ago, # ^ |
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    maybe try: 3 1 1 1 1 2 2 3 3 4

    answer

    YES

    1 2 1 3 4 3 1 2 1

    my code accepted after handling this test case

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    10 months ago, # ^ |
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    try this

    5 1 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 8 8 9 9 7 7 6 6

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    10 months ago, # ^ |
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    My code works well for all the above cases.

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    10 months ago, # ^ |
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    Maybe a case when N is odd and not equal to 3. I rectified that case and got AC on C.

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how to solve D1 and F1

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    10 months ago, # ^ |
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    For F1:

    You need to count total number of red vertices and blue vertices in all the tree, then you can use simple dp to calculate for every vertex i the number of red children and blue children, let this value is pair<int,int> dp[i], now let's root the tree at node 1 and traverse using dfs, for each edge joining vertex "u" and vertex "v" you can remove it and check the number of red & blue vertices in each new tree in O(1) using (total number of red & blue vertices, number of red & blue vertices of the child), and count good edges

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    10 months ago, # ^ |
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    D1: binary search to check if we can get m pages in range [1, n] days

    optimal strategy to finish m pages in x days is to use largest value in 1st days, 2nd largest in 2nd days, ..., x largest in 1st days again (cyclic)

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    10 months ago, # ^ |
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    D2 : Binary search on number of days. To find maximum number of pages he can write in d days, Greedily pick largest a[i]'s and distribute them over d-days. This will work because in optimal solution, difference between number of elements in any two days is less than or equal to 1.

    F1 : First store the red and blue vertices in subtree rooted at v for all v, using dfs.

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    10 months ago, # ^ |
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    D1 or D2 : Let's apply binary search on the number of days required. For the current mid, let us greedily write pages restricting ourselves to at most "mid" number of days. We take the maximum value of a and write a[n] pages on day1, a[n-2] pages on day2 and so on. If at any point we exhaust the number of days, we continue by reading 1 less page from the current book. If at any point we are able to read m pages then mid is a candidate for the answer. submission

    F1 : let's root our tree at vertex 1 for simplicity. For every vertex, let's calculate the number of blue and red vertices in its subtree. We can do this using dfs. After this we can check for every vertex by removing an edge to one of its children. The number of red and blue vertices left in each of the new trees is blue[v], red[v], red[1] — red[v] and blue[1] — blue[v]. Here we are at vertex u and v is one of its children. We can calculate answer easily then. submission

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    10 months ago, # ^ |
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    For D1:

    I dont know if my solution will pass, but i used dp.

    First, i noticed that if we will drink k cups in some day, then we need to drink the lower first, keeping the higher amount of caffeine to the end, so i sorted the array of inputs in the beginning.

    Then, the dp state is dp[i][rem] is the number of days we need to write rem pages, using cups with indices from i to n, the recurrence is simple as we try to drink k number of cups in each day such that k is in this range [0,n-i+1].

    solve( idx , rem ) = min( solve( idx+1 , rem ) , solve( idx + k , max( 0 , rem-k ) ) + 1 )

    This solution is O(n*n*m) which is not bad. That's why i didn't manage to solve D2 as this solution will not pass the time or the memory limit.

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      10 months ago, # ^ |
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      for D1, 1. sort them in decreasing order. 2. apply binary search on ans. for F1, do a dfs to know ho many reds and blues for each subtree

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very good problems for DIV3 , enjoyed very much . Hope all my problems gets accepted after hacking phase.

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10 months ago, # |
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Author should stop setting problem C.

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what is testcase # 9 in D1?...T^T

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I think, I should change the first sentence in the announcement to "Miss Div. 1 rounds? :)"

But surely, this was expected that F2 is a pretty hard problem ¯_(ツ)_/¯

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The problem setter of C should be banned for some time for creating such bad problem

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I hate filling numbers in matrix...it's so time-costing and brain-f***ing to consider about symmetry and indices...my poor little brain...

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Is it possible to optimize the dp solution for D1 to solve D2?

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WTF Why B test 1 1 Answer 1 ?!?!??!?!?!??!

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    10 months ago, # ^ |
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    Because giving away 1 candy would mean even and odd sum are equal to 0 because no element remains. So the answer should be 1.

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    10 months ago, # ^ |
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    The amount of candies you eat in odd days is the same as the amount you eat in even days, which is 0 because you don't eat in any.

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    10 months ago, # ^ |
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    0 = 0

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Bonus:

Solve E with additional constraint:

first and last pairs are also considered adjacent for the 4th constraint (formally b1 ≠ bn and g1 ≠ gn)

Fun fact: seems like this problem can be solved with this limitation in all the cases when the original problem can be solved.

Who can prove it or find a counter-test (or bug) in my solution 50193010, that solves problem with this additional constraint?

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    10 months ago, # ^ |
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    i solved the problem with a simple pattern that also satisfy this constraint

    first use all pairs i(from 1 to k) and i + 1 mod k

    then all pairs i and i + 2 mod k

    after that pairs i and i + 3 mod k

    and so on,you can easily see that this patern satisfies all the constraints

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    10 months ago, # ^ |
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    I don't see how your solution solves this problem. It fails on the first testcase itself, I think (b1=b4).

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      10 months ago, # ^ |
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      For the test case 4 3 my solution prints

      YES
      1 2
      2 1
      1 3
      3 1
      

      I see no problem there.

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Can D2 be solved without binary search, maybe using priority queues?

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Really enjoyed the problemset. Best Div.3 round so far.

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How to solve F2?

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The contest would have been great if E and C were swapped.

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    10 months ago, # ^ |
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    What purpose swapping them will serve. Afaik both of them carries equal points??

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      10 months ago, # ^ |
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      Yes all carry equal points. Your point seems good, these things actually remind us to pick easy questions first (until you reach that level when C's implementation is a 8-10 min. cakewalk for you)

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    10 months ago, # ^ |
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    Problem E is rated 2000. If E and C were swapped,it would be rated no more than 1600.

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Can someone explain D2. I can't understand why we used a binary search in this.

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    10 months ago, # ^ |
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    Forget about the binary search for now, the idea is that for every number of days we know the optimal cups to drink, and their optimal order too, so we can get for every number of days the maximum number of pages we can write (we will talk about this optimal order later).

    So if we know this kind of information, we can simply check all numbers of days between 1 to n, and select the least one that allows us to write at least m pages, and that's will be our answer.

    Here's where binary search helps, as if we can write x pages in k days, then we can write x or more using more than k days.

    The optimal cups to drink for a given number of days k is:

    The largest cup in the first day, then the second largest one in the second day, and so on until we take k cups in the k-th day. Of course we can take more than one cup in a day, then we extend for the first day by drinking the k+1-th cup (don't forget to subtract the number of cups that are already drank as in the first day we actually will drink the k-1-th cup first then drink the largest cup, so we need to subtract it, same goes for all days), and for the second day we also extend it with the k+2-th cup, and so on.

    submission

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These two participants have same source code in problem A, B, C, D1, D2, F1

tim25871014 Fantasyice
pA 50167147 50183932
pB 50185024 50186012
pC 50179384 50186141
pD1 50192694 50193272
pD2 50194759 50195089
pF1 50191152 50191524

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Is there a way for which test case my code failed for B ques. It was hacked. I implemented this logic — store prefix sum of elements in odd and even indices in an array, and store the total sun. Then loop through given array once again , and check if sum to left of it the odd(/even) and right tO it the even(odd) prefix sun if equal to (totalSum-arr[I])/2. My submission is this: https://codeforces.com/contest/1118/submission/50179395[prob. B](https://codeforces.com/contest/1118/submission/50179395)

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    10 months ago, # ^ |
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    Hack Test:
    1
    6575

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      yeah found (in submission history). Thanks.

      My code is giving output as 0 for n=1. It should have been 1 instead.

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When does my brand-new rating reflects? I have to show off my [ LEGENDARY SHINING BLUE ] to my friends

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"I forgot to fix participant types (official <-> unofficial) if you registered before rating changes of the yesterday contest. It will be fixed after the contest. Thus, you will be a rated (official) participant if and only if your current rating is strictly less than 1600. Mike." **** Some people above 1600 had their ratings increased

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    10 months ago, # ^ |
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    Oh, sorry. I'll fix it soon. So don't be surprised with temporary disappeared ratings. I'll fix the issue and return ratings back.

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

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

    Check https://codeforces.com/contest/1118/ratings, in top 10 you can see 7 are giving the contest first time

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I received a message from codeforces system today regarding the similarity of my code with 2 other users. The reason for this may be that I was using ideone in public mode by mistake. Here are the links of my solution: https://ideone.com/s2sHpB https://ideone.com/Yax8pe https://ideone.com/9qbTAo

The only proof I can give you that I have not copied is that I coded the solution for problem for D1 and used the very same solution for D2 and got runtime error. The reason for that was the difference in array size I was using. In problem D1 I defined my array size=100 #define N 100 in my code. I submitted the very same code for problem D2 and got runtime error on a test case. I figured it out very soon due to large size constraints which the problem statement had and changed my array size by, #define N 200005

Had it been a copied code from somewhere I must have got accepted answer in first go or it would have taken me a significant amount of time to realize the error in the code.

The next proof I have is that you can see my coding style which is consistent across problems. I declare global variables with long long data type, use macros for defining array_size, #define N 200005

Request you to please give back my rating. It took me a lot of hard work and motivation to get a rank among top 1000 after such a long time.

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10 months ago, # |
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Can anyone tell why my solution for D2 fails for test 49?
In my machine its output is 1 as expected?
The problem is solved when is decrease the upper bound for binary search to n + 1.
¯_(ツ)_/¯

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

    Because in the test 49 n=1 , but in the fuction check ,

    for(ll i = 0; i < mid; i++) pgs += arr[i];

    the i is too large , and then the arr[i] is unexpected

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10 months ago, # |
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Does anyone know where we can find the editorial for this contest?

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

Why the updated rating has gone? Any particular reason?

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

I think this was the much tricky contest then last div3 contest.

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10 months ago, # |
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So, what is the method to think about the Problem A ? Couldn't get any idea.

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

Vovuh, Editorials please.

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10 months ago, # |
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the user ereased rating who has been skipped

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    10 months ago, # ^ |
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    the all my solution is skipped.....and i have rated ,down 128.

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10 months ago, # |
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i had a problem on submitting , after i submitted A i couldn't submit B after the contest i could submit B and now i can't submit C ? Has anyone faced this problem?

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

It looks like codeforces have a new way to tackle plagiarism. The one whose solutions are skipped have been given a score of 0, and their rating changes have been calculated by keeping that score in mind !!!!

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

How to solve F2? I don't have any idea about it. Waiting for Tutorial.

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10 months ago, # |
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This submission 50181596 got AC on D2 and WA on D1.

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10 months ago, # |
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can anyone plz provide me the link to editorial

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

So basically, no matter what, div1 coders will solve and win div 3 contests too. Great!