MateoCV's blog

Codeforces Round #774 (Div. 2)
By MateoCV, history, 3 years ago, In English

Hola Codeforces!

I am happy to invite you to Codeforces Round 774 (Div. 2), which will take place on 04.03.2022 18:35 (Московское время). Notice the unusual time! The round will be rated for participants with rating lower than 2100. You will have 2 hours to solve 6 problems. The problems were authored and prepared by me.

I would like to thank the following people who made the round possible:

See you in the standings!

UPD:

Scoring distribution: $$$750$$$ — $$$1000$$$ — $$$1250$$$ — $$$2000$$$ — $$$2500$$$ — $$$3000$$$

Thanks to ak2006, video editorials for some of the problems will be available on his channel after the round.

UPD: Editorial

UPD: Congratulations to the winners!

Official winners:

  1. lydia1
  2. mjhmjh1104
  3. Turret
  4. Vsinger_MoQingxian
  5. WAduck
  6. woolim
  7. Iam1789
  8. yagoo
  9. Koba.rde
  10. dtc03012

Unofficial winners:

  1. Vercingetorix
  2. Um_nik
  3. hank55663
  4. Umi
  5. dlalswp25
  6. disorientation
  7. rainboy
  • Vote: I like it
  • +643
  • Vote: I do not like it

Comments Comments (148)
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3 years ago, # |
  Vote: I like it +16 Vote: I do not like it

Good luck and high rating.

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3 years ago, # |
  Vote: I like it -48 Vote: I do not like it

Good luck to all Ukrainian contestants

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3 years ago, # |
← Rev. 2   Vote: I like it +4 Vote: I do not like it

Note The Unusual Timings (1 Hour Late)

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

gl ^ hf

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

best of luck for all♥️

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

Good luck and positive rating!!!

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

This will probably be my first contest in over a month...I'm awaiting -100 delta :P

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

    same... I'm awaiting losing my green after my first contest :P gl

    that wasn't true gonna get +20 :yayy:

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

      let's go! i'm having my series of defeat too

      i overslept so i didn't participate

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

Good luck everyone. Maybe this round will help improve people's mood at codeforces.

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    3 years ago, # ^ |
      Vote: I like it -20 Vote: I do not like it

    hello gagan plz guide me as i am stuck in newbie-pupil phase from a long long time and i don't see improvement also provide tips for dp if any. thnx in advance.

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

      The key to improvement is indeed just good-old practice. Now, everyone should choose a practice regime which suits them well. I just like you were shuffling pupil and newbie for 1 year. This year I started practicing more with a focus on solving questions rather than learning new algorithms. You only need a handful of algorithms like dfs-bfs, Dijkstra, dsu, segment tree to reach the expert but dp practice is very important. I practiced dp from problemset with filter-tag and started solving from 1000 rated questions and gradually increased rated questions. When I felt like I can do 1600 rated dp questions pretty confidently then I started randomly doing questions with gradually increasing difficulty from 1000 to 1600.
      Make sure you actually solve the questions and don't see the editorial until you solve it. If it seems impossible then try to read other people's code before checking out editorials. Do cp with your friends, discuss approaches and try to outrank your peers :p. This is what has worked for me till now, hope this helps you. Good luck!

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        3 years ago, # ^ |
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        I think DP is the most important thing. segment tree is not too important at pupil stage.

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          3 years ago, # ^ |
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          I was talking about the concepts which can land you to expert level.

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

            expert level doesn't need segment tree to be fair, binary search, greedy, simple DP, DFS/BFS and a little knowledge of graph.

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

Maybe we have the opportunity to just focus on this round itself!

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3 years ago, # |
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Look at the quality of the questions. My hands are too slow.Orz

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3 years ago, # |
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Before the usual time was 16:35, then 15:35, then 11:35 ... What is the actual usual time ?

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

ooo thanks

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

Are we going to fight against "Robot" in this round?

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

May C problem be solved by less than 2000 people. I want it to be an interesting one.

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

    Bruh This is every Cyan dream . That it has less than 2000 solves and we are among them . But sadly cheaters will never make it happen ;_;

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      3 years ago, # ^ |
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      I would say and pupil too — last 3 contests happens to me top 1000 after A,B (top 100 last round), and then somehow stuck on C, and -∆, -∆, -∆

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

First argentinian contest! Congratulations MateoCV

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

Contest after long time.Eager to participate

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

Good luck everyone! Have a good contest!

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

Argentinian round? Aguante el Diego, Talleres y FaMaF!

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

Good luck!

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

Unfriendly time for Chinese contestants.

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

Dear contest, How do you feel today? Yet another person commented on your announcement page rn (me). Since I said hi to you, please give +ve this time to me. yk the usual. thanks in adv

Regards, that random user asking for +ve in rating

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

Very cool scoring distribution 1000 — 750 = 1250 — 1000 , 3000 — 2500 = 2500 — 2000

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

there is no as a tester comment.

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3 years ago, # |
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SpeeeeedForcesssss

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

    why? the number of people who solved C isn't high or too low

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

antonforces

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

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3 years ago, # |
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I haven't coded recursive code in 2 years lol good to remember it with problem C

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    3 years ago, # ^ |
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    I did it without recursion

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      3 years ago, # ^ |
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      How?Plz share your approach..

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        3 years ago, # ^ |
        ← Rev. 3   Vote: I like it +1 Vote: I do not like it

        Store the factorials numbers <=1e12 in an array (I think the count of them is 13 or 14)

        Just iterate on all subsets of them (by iterating from 0 to 214) and then subtract the factorials number from n (let's call the resulting number as temp). After each subset, count the number of set bits in temp. This will be the number of powers of 2 required. Do this for all and keep track of min so far. Submission https://mirror.codeforces.com/contest/1646/submission/148374185

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

        Solved it using bitmask dp on factorials <= 1e12 and pop count

        Submission: https://mirror.codeforces.com/contest/1646/submission/148339954

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

          I don't think there is any DP in your solution :p.

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

            sorry used bitmasks to generate all possible subsets of size 15 :)

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

            Why did the setters confuse with -1 where ans is always possible?

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              3 years ago, # ^ |
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              Just to check that we are smart enough to think that . As its quite obvious that every no can be represented in binary form. So there no chance for answer to be -1.. But some ppl i saw do handle that case ;_;

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      3 years ago, # ^ |
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      I think it's still the same complexity recursion can be done in a different approach of course

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    3 years ago, # ^ |
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    I did with a mask-compressed DP...

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

It took me 2WA's and 50 minutes to figure out that for long long integers __builtin_popcount doesn't work and we need to use __builtin_popcountll ... :-/

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

    I realized it just before click submit button...

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

      I am just happy that at least I figured it out before the contest ended :)

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

    You should keep such things as #define (macro), easier to remember

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it +1 Vote: I do not like it

    I suggest using std::popcount(). The only issue with this function in terms of competitive programming is that it requires x to be of unsigned type, so we have to cast explicitly if x is long long. 

    int popcnt = popcount(uintmax_t(x));

    Of course, the simple macro solves this problem: #define popcnt(x) popcount(uintmax_t(x)).

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3 years ago, # |
  Vote: I like it +17 Vote: I do not like it

5th test case of problem D...

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3 years ago, # |
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it was so fun to write 15 nested for loop in problem C(p.s. I know recursive solution would also work)

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

    that means you don't know this

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      3 years ago, # ^ |
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      I mentioned fun, actually you don't know what I know so please read comments carefully next time

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3 years ago, # |
← Rev. 4   Vote: I like it +67 Vote: I do not like it

Imagine wasting 20 mins debugging E because you "simplified" $$$\frac{lcm(x_1, x_2, \ldots, x_k)}{x_1}$$$ to $$$lcm(x_2, \ldots, x_k)$$$ without thinking properly T_T.

Also I can't exactly describe it, but CDE feel like they are somehow standard yet fairly interesting at the same time. Maybe its because they have extremely simple and natural formulations which don't feel like they've been forced to match an idea?

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

A nice contest for me!! I loved the problem C. Would love to share my experience while solving problem C,

Initially after reading the problem , I immediately thought of solving it using subset generation through bitmasks , because i thought the size of factors + powers of 2 would be around 30 , but turned out it was ~52. So i drifted a bit & thought of solving it using DP(using coin change approach) but that is also not feasible.

Solution :

Spoiler
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    3 years ago, # ^ |
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    you only need to bitmask factors, then greedy choose powers of 2 for the rest.

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      3 years ago, # ^ |
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      Yup, thats what i did. I don't think you need to greedy choose powers of 2 for the rest , because the remaining number's binary representation containing the set bits already is the least number of powers of 2 required to represent it.

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3 years ago, # |
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Logic for C?

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    3 years ago, # ^ |
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    Note that n can allways be constructed with number of its bits, since these are powers of 2.

    So construct all possible sums of factorials (that are at most 2^14 since there are only 14 factorials in range), subtract each one from n and count the bits of the difference.

    Smallest number of set bits in difference plus bits in factorial sum is ans.

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3 years ago, # |
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How to do F?

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3 years ago, # |
  Vote: I like it +5 Vote: I do not like it

What d hell is test case 5 of D -_-

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

    Either a large graph or something that kills the incorrect bipartite idea I guess.

    Try the following graph if you don't get why bipartite is incorrect:

    Input Graph
    Answer
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    3 years ago, # ^ |
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    There are cases where it is optimal to not take two neighboring nodes. Splitting the tree to a bipartite graph and then taking one of the two parts is not always optimal. The idea is similar to the well-known DP problem "Maximum sum over array such that no two elements are adjacent" but implemented on a tree.

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

    2,3-dimethylbutane

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3 years ago, # |
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My Idea for D :

  • Take all leaves to be good vertices
  • Now all parents of the leaves (call it buds) are assigned with weight = 1
  • Now the graph is split into separated sections with "buds" as boundaries for each section
  • For each section, I solve them independently by coloring it "bipartitely". In each section, I find which gave me the best arrangement whether the color "black" / "white" on this section are good vertexes and bad vertices for its counter-color.

I got WA on pretest 6. Not sure if the idea is wrong or my implementation My submission : 148381092 Any thoughts?

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    3 years ago, # ^ |
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    I tried the same thing (but failed pretest 5) :(

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    3 years ago, # ^ |
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    it is true that a good vertex(blue) must be surrounded by bad vertices(black) but it is not necessary that bad vertices be adjacent to good vertices. so if u two color the tree, u wouldn't take into consideration cases where two adjacent nodes are bad

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    3 years ago, # ^ |
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    Pretty sure it's finding the max independent set.

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      3 years ago, # ^ |
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      I did it in sort of dp style. Rooted the tree at 1.

      dp[i][0] = <C,S> means ith node is not good and fill its children however you want such that max good vertices below my subtree is C with sum of edges as S.

      dp[i][1] = <C,S> means ith node is good and fill its children however you want such that max good vertices below my subtree is C with sum of edges as S.

      For transitions dp[i][1] depends on dp[c][0] where c is children of i since you are good, children can't be good.

      dp[i][0] picks dp[c][0] or dp[c][1] according to which is best.

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        3 years ago, # ^ |
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        did it work ? i tried the same thing but kept getting WA in pretest 5 or 6

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          3 years ago, # ^ |
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          You also need to keep track of the cost in case dp[i][0]==dp[i][1].

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

            Yeah the reason for storing S in the dp is to handle the case of dp[c][0] == dp[c][1]

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    3 years ago, # ^ |
    ← Rev. 3   Vote: I like it +1 Vote: I do not like it

    How does the problem change after removing all the leaves?

    If there exists some tree for which bipartite doesn't work, I can just create a new tree by attaching a single node to each leaf and your soln won't work for it.

    And try the following graph if you don't get why bipartite is incorrect:

    Input Graph
    Answer
    A Correct Solution
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      3 years ago, # ^ |
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      i defined dp[i][0] = max good vertices if vertex i is good and dp[i][1] for bad. and dp[i][0] = 1 + all dp[ne][1] in neighbours and dp[i][1] = max(dp[ne][0] and dp[ne][1]) for all neighbours. Is is correct ? i kept getting WA in either pretest 5 or 6

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        3 years ago, # ^ |
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        I guess in dp[i][1] case if dp[ne][0] and dp[ne][1] are equal then how are you ensuring which one to pick to get minimum sum.

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          3 years ago, # ^ |
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          Yup, this is likely the issue.

          One way to solve this is to store the dp value as a pair $$$<\text{max number of nodes}, -(\text{min sum of degree})>$$$, then taking max works as expected.

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      3 years ago, # ^ |
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      Oh I see. I get it. Thanks!

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      3 years ago, # ^ |
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      Wait, in your explanation should it be $$$dp_{x, 0} = max(dp_{v,0} , dp_{v,1})$$$ ?

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3 years ago, # |
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Could someone share their approach for B ?

I tried solving it by sorting the array (minimize the Blue sum) and using prefix sum, but failed. I think this is because my approach colored all the numbers.

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    3 years ago, # ^ |
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    Use sum of two smallest numbers as small sum, and biggest number as big sum.

    If both sums are ok ans=YES else add next smallest number to small sum, next biggest number to big sum. Repeat.

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

    Well, I also sorted it first, and then kept on calculating Blue Sum from the start and Red Sum from the end of the array. Also, to maintain the count(Blue) > count(Red), I initialised the starting index to be 1(0-based indexing) while initialising Blue Sum to be equal to the first array element.

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3 years ago, # |
← Rev. 3   Vote: I like it +4 Vote: I do not like it

submitted C in the last minute

took more than 1 hour to realize brute-forcing all subsets of factorial values is feasible

but brute-forcing all subsets of 2-power values is infeasible which gave me a wrong intuition that the factorial counterpart is also infeasible

I hope in the future I can be more familiar with the thresholds of feasibility.

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    3 years ago, # ^ |
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    Why we Can't use one factorial two times?

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      3 years ago, # ^ |
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      because we need distinct numbers

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        3 years ago, # ^ |
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        Got it now. I Read the question again. Sucks when missed important point.

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      3 years ago, # ^ |
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      That would make it way harder so be glad about it

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      3 years ago, # ^ |
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      the problem description said distinct.

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    3 years ago, # ^ |
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    Even if it was feasible to brute force all powers of 2, how would you find the needed amount of factorials?

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      3 years ago, # ^ |
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      I don't know how to find the needed factorials except for brute-force, so I should have taken that approach initially. Unfortunately the infeasibility of brute-forcing 2-powers gave me the wrong intuition that brute-forcing factorials is also infeasible.

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      3 years ago, # ^ |
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      No Idea. That is why I was not able to solve C. Was expecting there is some other trick to solve the problem which I was not able to think.

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      3 years ago, # ^ |
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      There are not much factorials in range, it is only 14 numbers. So you can create only 2^14 sums from these numbers. Just try all of them.

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    3 years ago, # ^ |
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    Went through the same trajectory, except I took even more time to realise this and couldn't submit my code in time :( It also somehow escaped by mind that combinations of powers of 2 literally define every single number in exactly one possible way.

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

Thanks for the contest, I don't know why I did so dumb on A and B, But after all I really liked the problems.

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3 years ago, # |
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In c, finding all the numbers powerful in a set then take a number which are just less than n and subtracting them from n is not feasible why?

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

    For 184 the correct answer is 120 + 64 but your approach would yield 128 + 32 (edited from 56) + 24.

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      3 years ago, # ^ |
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      How did you get 56 here....

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        3 years ago, # ^ |
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        184 — 128 = 56. The point is that you can't pick the max number <= n.

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          3 years ago, # ^ |
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          But 56 won't be in set, so it would look like 184 → 128 + 32 + 24 (wrong approach)

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

            Ah sorry, I meant 128 + 32. Thanks :)

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3 years ago, # |
  Vote: I like it +5 Vote: I do not like it

what is pretest 6 of D? ಥ_ಥ

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

    If you were doing BFS solution from leaves, It would fail. Optimal solution will be reached by dynamic programming.

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      21 month(s) ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Can you please explain why bfs from leafs doesn't work ?? Any test-case ???

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3 years ago, # |
← Rev. 2   Vote: I like it +14 Vote: I do not like it

I think D is just Maximum Independent set with least total degree which can be solved by dp. I realised this, but could not reconstruct the solution from the dp before the contest ended.

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3 years ago, # |
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How to solve E? I was looking for numbers with the same prime factors

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

    Take 2, 4, 8, ... as example. In the row starting with 2, you get 2^1, 2^2, ... In the row starting with 4, you get 2^2, 2^4, ..., so what in the cells are these powers of 2: 1 ~ M, 2 * 1 ~ 2 * M, 3 * 1 ~ 3 * M, ..., depending on how many powers of 2 are less than or equal to N. Then, what you need to do is calculate how many distinct number is in this list. Sum up all the length of lists of a number that is not power of another one. Don't forget 1.

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

Any counterexample on D solving with 2-coloring?

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    3 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it 
    6
1 3
2 3
3 4
4 5
4 6
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    3 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    6

    1 2

    1 3

    1 4

    4 5

    4 6

    The output should be:

    4 6

    1 1 1 1 1 1

    The idea is that although there can be no 2 adjacent good nodes, this does not mean that there can be no 2 adjacent bad nodes. Unfortunately I realized that a short time just before the contest ended.

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    3 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it
    Input Graph
    Answer
    Visualization
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3 years ago, # |
← Rev. 2   Vote: I like it 0 Vote: I do not like it

Hey,
In the first problem, why this works cout << s/(n*n) << '\n';
but using cout << (int)(s/pow(n,2)) << '\n'; gives wrong answer in test 2?

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    3 years ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    because pow() returns double which may cause precision issue.

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

    It turns out s/(n*n) is bounded within INT_MAX because bounds of s depend on n. Problem was caused due to precision issue of pow() as Aging1986 pointed out.

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3 years ago, # |
  Vote: I like it +9 Vote: I do not like it

tried dp in D but it doesn't work... so retried bipartite and failed again.

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3 years ago, # |
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Is it possible to output -1 in c ??

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

    no, any number can be represented as sums of powers of two (its binary representation).

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3 years ago, # |
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Getting MLE on test 39 on problem D just because I used longs and a relatively high memory constant is the most unfair thing ever...

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3 years ago, # |
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Auto comment: topic has been updated by MateoCV (previous revision, new revision, compare).

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3 years ago, # |
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Auto comment: topic has been updated by MateoCV (previous revision, new revision, compare).

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

Nice contest!

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

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

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

Would this idea work for D?

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

    It won't work

    Edit:

    Actually I don't know.

    Edit2:

    I've coded it — it doesnt find the minimal value correctly.

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