By pigstd, 2 months ago,

Hello, Codeforces!

Members of team EZEC are glad to invite you to participate in Pinely Round 1 (Div. 1 + Div. 2), which will start on Nov/20/2022 17:35 (Moscow time). You will be given 7 problems, one of which has a subtask, and 2 hours and 30 minutes to solve them. The round will be rated for everyone. It is greatly recommended to read all the problems.

There is at least one interactive problem, so please see the guide of interactive problems if you are unfamiliar with it.

We would also like to thank:

Fun facts:

• Anton reviewed 18 problems in total.
• This is the 13th round of EZEC. Here are some previous rounds: Round 11 on Luogu, Round 12 on nowcoder.
• The team has an anime character EZEC-chan (the girl in the image), hope you like it :3
• orzdevinwang found a better solution of F the day before the contest.

This round is made possible with the support of Pinely!

Pinely is a privately owned & funded algorithmic trading firm, our main focus is set on high frequency and ultra low latency trading.

We are a team of mathematicians, programmers, engineers and computer scientists driven by the immense passion for knowledge. We constantly face various challenges such as developing strategies for trading, optimizing trading systems to achieve the lowest latency reactions to various market events, saving and processing large volumes of historical data. The work we do requires a high ability to create effective C++ code, algorithmic thinking and mathematical intuition, which attracts winners, awardees and medalists of various competitions in the respective fields such as ICPC, IMC, HITB PRO CTF and Google HashCode etc.

Find out more about us on our website pinely.com or from our employees registered here on CF.

Apply

Top 30 contestants and 10 random contestants placed 31-100 will receive a branded Pinely hoodie :)

The statements were made as briefly and clearly as we could. GLHF!

Scoring distribution: 500 — 1000 — 1250 — 1750 — 1750 — (2500+2000) — 3500

UPD2: Congratulations to the winners:

• +590

 » 2 months ago, # |   +46 Auto comment: topic has been updated by pigstd (previous revision, new revision, compare).
•  » » 2 months ago, # ^ |   +8 Where is fyy Memorial Valley School
•  » » 2 months ago, # ^ |   +109 .
•  » » » 2 months ago, # ^ | ← Rev. 3 →   -89 In fact there's something more about the problem D: As a tester, I give out a solution with BGF and ODE for it. So in fact, this problem still can be strengthened. :)But to make the gap less large, the problem provider didn't do that. :)However, the gap between D and E might be even larger than such between C and D. :(
•  » » » » 2 months ago, # ^ | ← Rev. 2 →   -66 oops
•  » » » » 2 months ago, # ^ |   +8 i am agree with him
•  » » » » 2 months ago, # ^ |   -8 In fact there's something more about the problem D: As a tester, I give out a solution with BGF and ODE for it. So in fact, this problem still can be strengthened. :)But to make the gap less large, the problem provider didn't do that. :)However, the gap between D and E might be even larger than such between C and D. :(
 » 2 months ago, # |   +144 All hail our emperor anton
•  » » 2 months ago, # ^ |   +110 All hail our emperor anton
•  » » » 2 months ago, # ^ |   +104 All hail our emperor anton
•  » » » » 2 months ago, # ^ |   +112 All hail our emperor anton
•  » » » » » 2 months ago, # ^ |   +132 All hail our emperor anton
•  » » » » » » 2 months ago, # ^ |   +87 All hail our emperor anton
•  » » » » » » » 2 months ago, # ^ |   +88 All hail our emperor anton
•  » » » » » » » » 2 months ago, # ^ |   +68 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   +89 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   +44 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   0 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   +11 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   0 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   -192 haha I broke your silly chain
•  » » » » » » » » » 2 months ago, # ^ |   0 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   -41 All hail our eperor anton
•  » » » » » » » » » 2 months ago, # ^ |   -16 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   -12 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   -9 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   0 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   0 All hail your emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   0 All hail our emperor anton
•  » » » » » » » » » 2 months ago, # ^ |   -15 haha I broke your sacral spine
•  » » » » » » » » » 2 months ago, # ^ |   -8 are you broke :))))
•  » » » » 2 months ago, # ^ |   0 All hail our emperor anton
•  » » 2 months ago, # ^ |   0 All hail our emperor anton
•  » » 2 months ago, # ^ |   -9 broke another chain
 » 2 months ago, # |   +103 orzdevinwang orz
•  » » 2 months ago, # ^ |   +84 orzdevinwang orz
•  » » » 2 months ago, # ^ |   +79 orzdevinwang orz
•  » » » » 2 months ago, # ^ |   +48 orzdevinwang orz
•  » » » » » 2 months ago, # ^ |   +63 orzdevinwang orz
•  » » » » » » 2 months ago, # ^ |   +45 orzdevinwang orz
•  » » » » » » » 2 months ago, # ^ |   +29 orzdevinwang orz
•  » » » » » » » » 2 months ago, # ^ | ← Rev. 2 →   +24 orzdevinwang orz
•  » » » » » » » » » 2 months ago, # ^ |   -16 orzdevinwang orz
•  » » » » » » » » » 2 months ago, # ^ |   -132 haha broke your silly chain
•  » » » » » » » » » 2 months ago, # ^ |   +11 recursion base case reached
•  » » » » » » » » » 2 months ago, # ^ |   0 orzdevinwang orz
•  » » » » » » » » » 2 months ago, # ^ |   -10 orzdevinwang orz
•  » » » » » » » » » 2 months ago, # ^ | ← Rev. 2 →   0 orzdevinwang orz
•  » » » » » » » » » 2 months ago, # ^ | ← Rev. 2 →   0 orzdevinwang orz
•  » » » » » » » » » 2 months ago, # ^ |   -8 orzdevinwang orz
•  » » » » » » » » » 2 months ago, # ^ |   -8 no
•  » » » » » » 2 months ago, # ^ |   +17 orzdevinwang orz
•  » » » » » » » 2 months ago, # ^ |   +29 pocafup orz
•  » » » » » » » » 2 months ago, # ^ |   0 orzdevinwang orz
•  » » » » » » » » » 2 months ago, # ^ |   -8 orzdevinwang orz
•  » » » » » » » 2 months ago, # ^ |   0 orzdevinwang orz
•  » » » » » 2 months ago, # ^ |   -8 orzdevinwang orz
•  » » » » » » 2 months ago, # ^ |   -8 orz orz orzdevinwang orz
•  » » 2 months ago, # ^ |   0 orz
 » 2 months ago, # |   +400 As a tester, i'd say guessAnyone who upvote me can get high ranks in this contest
•  » » 2 months ago, # ^ |   +102 If my rating dropped,I will downvote you!!!
•  » » » 2 months ago, # ^ |   +18 If it's tried, "You cannot vote twice. You have already voted for this comment before."
 » 2 months ago, # |   +13 Anime round is coming :|
 » 2 months ago, # |   +110 As a tester, I'm cute (｡･ω･｡)ﾉ♡
 » 2 months ago, # | ← Rev. 2 →   +13 pigstd orz
•  » » 2 months ago, # ^ |   0 pigstd orz
•  » » » 2 months ago, # ^ |   +3 pigstd orz
•  » » » » 2 months ago, # ^ |   +2 pigstd orz
•  » » » » » 2 months ago, # ^ |   +7 pigstd orz
•  » » » » » » 2 months ago, # ^ |   -22 pigstd orz
•  » » » » » » » 2 months ago, # ^ |   +8 pigstd orz
•  » » » » » » » » 2 months ago, # ^ |   0 pigstd orz
•  » » » » » » » » » 2 months ago, # ^ |   -19 pigstd orz
•  » » » » » » » » » 2 months ago, # ^ |   +5 pigstd orz
 » 2 months ago, # |   +62 As a tester, I strongly recommend this round for straightforward statements & high-quality and interesting problems. GLHF!
•  » » 2 months ago, # ^ |   0 I agree by being a contestant on luogu and their problems have high quality in just about every round. Hope to get higher ratings this round!
•  » » » 2 months ago, # ^ |   0 by the way sto Ecrade_ orz
 » 2 months ago, # |   +18 Who is she?
•  » » 2 months ago, # ^ |   +15 my mommy
•  » » » 2 months ago, # ^ |   +6 Can I get inside?
•  » » » » 2 months ago, # ^ |   -18 strange)
•  » » » » » 2 months ago, # ^ |   -18 strange)
•  » » » » » » 2 months ago, # ^ |   -18 strange)
•  » » » » » » » 2 months ago, # ^ |   -18 What do you mean "inside" strange)
•  » » » » » » » » 2 months ago, # ^ |   -18 you can make something stranger than a stranger when it becomes strange
•  » » » » » » » 2 months ago, # ^ |   0 strange)
•  » » » 2 months ago, # ^ |   0 No she is mine :(
 » 2 months ago, # |   0 WAIFU!!!! We really need an anime-mascot-cute-girl for codeforces <3
•  » » 2 months ago, # ^ |   0 I can't wait to see that!
 » 2 months ago, # |   +14 Just noticed Technocup 2022 Elimination Round 1 is going to be held tomorrow as well (which in the past years has a live mirror on CF). Are we going to have 2 rated CF rounds in the same day ?
•  » » 2 months ago, # ^ |   +7 This year Technocup will be held not on Codeforces, but on the Allcups platform made by VK so unfortunately it seems to me there will be no mirror.
 » 2 months ago, # |   +41 Wow, EZEC Round! EZEC is one of my favourite Chinese problemsetting groups. Looking forward to solving the problems with high quality.
 » 2 months ago, # |   0 Is it hard?
•  » » 2 months ago, # ^ | ← Rev. 2 →   0 Maybe hard I think.
•  » » » 2 months ago, # ^ |   +52 In this round, we tried our best to avoid weird, difficult and confusing problems. Let's discuss them after the contest!
•  » » » » 2 months ago, # ^ |   -18 are you sure about that?
 » 2 months ago, # |   +3 What a cute blog ._.
 » 2 months ago, # |   +7 She is so cute!
 » 2 months ago, # |   +5 Looks like a good contest, good luck everyone!
 » 2 months ago, # |   0 500 — 1000 — 1250 Fabulous!! ^_^
 » 2 months ago, # |   0 Wow! 60 people are coordinating in this round.
 » 2 months ago, # |   0 May the anime girl bring me a high +ve delta in this contest.
 » 2 months ago, # |   0 Good luck!
 » 2 months ago, # |   +5 After seeing the photo I really have a good feeling about this round
 » 2 months ago, # |   0 Patalok!
 » 2 months ago, # | ← Rev. 3 →   -11 This ROUND seems interesting one (^o^) ,eagerly waiting for this round ★^^★
 » 2 months ago, # |   +22 happy world cup day
 » 2 months ago, # |   -40 Stop putting otaku images in codeforces!
•  » » 2 months ago, # ^ |   +34 I apologize if it bothers you. The picture is a representation of our team spirit. She is not from an anime. We designed the picture and spend a lot of time and money to make her alive. I appreciate your honesty and hope you like her
 » 2 months ago, # |   +17 is pigstd a girl? pigstd
•  » » 2 months ago, # ^ |   +35 If you take a look at her profile, you will know that she is Chisato Nishikigi, obviously a girl. She even has a picture of herself uploaded as proof.
•  » » 2 months ago, # ^ |   +40 Yes!I'm her classmate,she is so cute! :d
•  » » » 2 months ago, # ^ |   +6 cute~~ girl pigstd！！！
•  » » 2 months ago, # ^ |   0 Dude, this is not important, the important thing is that pigstd is really cute.
•  » » 2 months ago, # ^ |   -16 It depends on your imagination :).
 » 2 months ago, # |   +5 Hopefully I can hit 500+ this round :)
 » 2 months ago, # |   -13 Rated?
•  » » 2 months ago, # ^ |   +8 Rated for everyone
 » 2 months ago, # |   +255 The problems are very nice, recommend participating.
 » 2 months ago, # |   +20 I hope there will be no anime in the problem description.
•  » » 2 months ago, # ^ |   +13 The statements were made as briefly and clearly as we could.Don't worry :)
•  » » 2 months ago, # ^ |   0 I think this is the hope of alot of coders.
 » 2 months ago, # |   +5 dXqwq orz
 » 2 months ago, # |   0 orz
 » 2 months ago, # |   +5 I hope I can be red this time!
 » 2 months ago, # |   -9 Is this contest rated for everyone?
•  » » 2 months ago, # ^ |   -7 1
 » 2 months ago, # |   -16 is this rated?
•  » » 2 months ago, # ^ |   0 Yes, this is rated and it is the an enjoyable contest.
 » 2 months ago, # |   +11 Great progress on the announcement
 » 2 months ago, # | ← Rev. 2 →   0 Expecting Interesting Problems
 » 2 months ago, # |   +1 CF plus World Cup night！
•  » » 2 months ago, # ^ |   +5 Is China in the world cup?
•  » » » 2 months ago, # ^ | ← Rev. 2 →   +32 yes! in the same group with Italy (
•  » » » » 2 months ago, # ^ |   0 Centainly Not. Honestly, neither China nor Italy has made it into the World Cup this year.
•  » » » » » 2 months ago, # ^ |   -21 As Chinese all know,China football team will never get the oppotunity to join the world cup,unless they train hard enough.In China,this has become a joke.
•  » » 2 months ago, # ^ |   0 They can't because they are small.
 » 2 months ago, # |   0 CF round + FIFA world cup day... OMG :)
 » 2 months ago, # |   0 All hail our emperor Anton :)
 » 2 months ago, # |   0 how did you draw the anime girl? did you get it commissioned?
 » 2 months ago, # |   0 Red Round and The World Cup.... just I need tourist beside me and a tub of popcorn... Rank 1 and amazing World Cup guaranteed ^,^
 » 2 months ago, # |   0 Good luck everyone :)
 » 2 months ago, # |   0 Ah, why it couldn't have been moved a bit earlier so it doesn't clash with WF opening? :/
 » 2 months ago, # |   +8 How young is orzdevinwang?
•  » » 2 months ago, # ^ |   +28 Grade 10
•  » » » 2 months ago, # ^ |   +16 :depression:
 » 2 months ago, # |   0 good luck everyone !
 » 2 months ago, # |   +8 Weebs Get a life
 » 2 months ago, # |   +11 EZEC-chan is so cute！
 » 2 months ago, # |   +28 Off topic: where does EZEC-chan's tail come out from?
 » 2 months ago, # |   +13 Is pigstd really a girl?
•  » » 2 months ago, # ^ |   +9 Of course a cute girl~
•  » » 2 months ago, # ^ |   +21 Don't be cheated by his anime girl profile. sure.
•  » » 2 months ago, # ^ |   +3 What is your definition of girl?
•  » » » 2 months ago, # ^ | ← Rev. 2 →   -42 Definition This
 » 2 months ago, # |   0 as a newbie downvote me
 » 2 months ago, # |   0 All hail our emperor anton.
 » 2 months ago, # |   0
 » 2 months ago, # |   +9 how is this div1+div 2? where is the div 2 in this lol
•  » » 2 months ago, # ^ |   +3 average newbie
•  » » » 2 months ago, # ^ |   +6 ironic
•  » » » » 2 months ago, # ^ |   +3 I would love to get castrated by both of you.
•  » » » » » 2 months ago, # ^ |   +3 mm yes
•  » » » » » 2 months ago, # ^ |   +3 so inspirational
 » 2 months ago, # |   +6 It is difficult for me,I wrote problem A and went to bed. After reading B and C, I have no idea. I want someone to teach me, after the game
•  » » 2 months ago, # ^ |   0 Didn't solve even A :) Have the same feelings after this competition.
•  » » 2 months ago, # ^ |   +3 me too... -120+ yay
 » 2 months ago, # |   +2 wasted starting 10 minutes figuring out wtf is common prefix and sufifx
•  » » 2 months ago, # ^ |   +17 How are you a specialist?
•  » » » 2 months ago, # ^ |   +18 By solving question in contest and ignoring people like you.
•  » » » » 2 months ago, # ^ |   0 smortie
•  » » 2 months ago, # ^ |   0 TRUE AF
•  » » 2 months ago, # ^ |   0 Me too but i wasted like 25 min something
 » 2 months ago, # |   +25 Oh, the round seems to be more difficult than the writers expected, but as difficult as I expected, because I'm just a low-level taster.btw, the writers are watching the World Cup.
•  » » 2 months ago, # ^ |   +8 Why not try the next div4 Codeforces Round #835 (Div. 4) to gain your confidence and rating?
•  » » » 2 months ago, # ^ |   0 Because it's on Monday and not on weekend.
 » 2 months ago, # |   0 Just every single time I think that I can't get results worse than previous time, I somehow manage to do this anyway. At least the next competition is div 4 round...
 » 2 months ago, # |   +12 Spend 1h on E because I forgot to check whether the vertex is a cut point and 1h on an incorrect solution of F. :(
•  » » 2 months ago, # ^ |   0 is there a nice way to cover this case? i ended up directly checking the validity of my candidate answers in $O(n^2)$ with bitset.
•  » » » 2 months ago, # ^ |   0 what language are you guys talking in?
•  » » » 2 months ago, # ^ |   +8 I did the following:Find arbitrary spanning tree, choose any leaf in it. It is definitely not cutpoint. In case it has degree less than $s - 1$, (where $s$ is the size of current connectivity component), you can switch it and you're done. Otherwise, switch any other node, for example, adjacent of the leaf in the spanning tree (as I did it).
•  » » » » 2 months ago, # ^ | ← Rev. 2 →   +10 hmm I'm not sure that any node works in that case, since there can be other nodes in the spanning tree that are also degree $s-1$. for example, the case 1 6 011100 100100 100100 111000 000001 000010 which looks like  1 / | \ 2 | 3 \ | / 4 5-6 hacks your solution 181782667. If the spanning tree is a star graph about 1, then choosing node 4 as your leaf will cause you issues, as it and its parent are both connected to everything.
•  » » » 2 months ago, # ^ |   +15 Any smallest degree vertex in the component works (unless it's a clique which is a different case)
•  » » » » 2 months ago, # ^ | ← Rev. 3 →   0 Doesn't this break for this example:1-2, 1-3, 2-4, 2-5, 4-5, 3-6, 3-7, 6-71 has the smallest degree of 2, but it is a cut point.Edit: Oh, I guess it works for the overall problem, though.
•  » » » » » 2 months ago, # ^ |   +8 Yes, doesn't matter if it's a cut point, it will always give an answer set of size 1.
 » 2 months ago, # |   0 Lets hope I dont get TLE in C
 » 2 months ago, # |   +6 How to solve D?
•  » » 2 months ago, # ^ |   0 put $a$ and $b$ on top of each other, and lets build a string $c$ of length $n$ on the alphabet $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$ (i can't get the column vectors to play nice with the inline latex, but the first one goes to $a$ and the second goes to $b$). we will count by starting off with the empty string, and sequentially inserting characters into valid positions.carries are started by a $(1,1)$, are continued leftwards by either $(0,1)$ or $(1,0)$, and are terminated by $(0,0)$ or another $(1,1)$.suppose we will place $x$ chains of carries, where $1 \leq x \leq k$. we will place $x$ of $(1,1)$ and $k - x$ of $(0,1)$ or $(1,0)$ that will continue these chains. we have $2$ characters to choose for each of the continuations, and we must place these $k-x$ items into one of $x$ groups, so there are $2^{k-x}\binom{k - 1}{x - 1}$ to make this assignment. now, for each of these $x$ chains, either it will be terminated by a $(0,0)$ or by the $(1,1)$ that starts the next chain. let's manually terminate $y \leq \min(x, n-k)$ chains with $(0,0)$. there are $\binom{x}{y}$ ways to place these.we have $n-k-y$ non-carry positions left to place. each of them can go in one of $y+1$ groups: either to the left of one of the $(0,0)$'s, or on the very right before any of the chains have begun. there are $3$ legal characters for these: $(0,0)$, $(0,1)$, and $(1,0)$. thus, we have $3^{n-k-y} \binom{n-k}{y}$ to fill in these last positions.our answer comes out to $\displaystyle \sum_{x=1}^{k} 2^{k-x}\binom{k - 1}{x - 1} \sum_{y=0}^{\min(x,n-k)} \binom{x}{y} 3^{n-k-y} \binom{n-k}{y}$from here, i brute forced my way though to the end using a convolution, so there is likely a more clever solution. nevertheless, i manipulated the equation with \displaystyle \begin{align} & \sum_{x=1}^{k} 2^{k-x}\binom{k - 1}{x - 1} \sum_{y=0}^{\min(x,n-k)} \binom{x}{y} 3^{n-k-y} \binom{n-k}{y} \\ =& \sum_{x=1}^{k} 2^{k-x} \frac{(k-1)!}{(x-1)!(k-x)!} \sum_{y=0}^{\min(x,n-k)} \frac{x!}{y!(x-y)!} 3^{n-k-y} \frac{(n-k)!}{y!(n-k-y)!} \\ =& 2^k 3^{n-k} (k-1)! (n-k)! \sum_{x=1}^{k} \frac{x}{2^x(k-x)!} \sum_{y=0}^{\min(x,n-k)} \frac{1}{3^y y!^2(n-k-y)!} \cdot \frac{1}{(x-y)!} \\ =& 2^k 3^{n-k} (k-1)! (n-k)! \sum_{x=1}^{k} \frac{x}{2^x(k-x)!} \sum_{i+j=x,i \leq n-k, j \leq k} \frac{1}{3^i i!^2 (n-k-i)!} \cdot \frac{1}{j!} \end{align}note that the inner summation is a convolution of two sequences $p$ and $q$, where \displaystyle \begin{align} p_i &= \frac{1}{3^i i!^2 (n-k-i)!}, i \leq n-k \\ q_i &= \frac{1}{i!}, i \leq k \end{align}precompute the inner summation with your convolution tool of choice in $O(n \log n)$, which allows us to compute the outer summation in $O(k)$.
 » 2 months ago, # |   0 how to do D without breaking out the arbitrary mod FFT/NTT?
•  » » 2 months ago, # ^ |   +24 I got to the following formula: $\sum_{i=1}^{min(k,n/2)} \binom{n-k}{i} \binom{k-1}{k-i} 3^{n-(2 i)}+ \sum_{i=0}^{min(k-1,(n-1)/2)} \binom{n-k}{i} \binom{k-1}{k-1-i} 3^{n-1-(2 i)},$where $i$ is the number of blocks of consecutive carries (11 followed by 11,10 or 01 any number of times, ended by 00). The binomial coefficients count the placement of these blocks within the n bits. The second sum accounts for the case when the last block is not ended by 00.
•  » » 2 months ago, # ^ |   0 Root ideaIf you want a continuous sequence of k carries then consider two binary strings a and b of length k+1. The first bits of two strings must be 1 and the last 2 bits must be 0 and for the rest of the positions at least one of the bits must be 1.Note: first means the least significant bit.Based on this, we can solve this using a bunch of combinatorics.
•  » » » 2 months ago, # ^ |   0 I could think of this during the contest, than my idea came out to be that we want to divide a block of size n + 1 into f different blocks of size >= 1, f = n + 1 — k as the root idea is a block of size k + 1 dimineshes it into k, its true for all k >= 0, so we need size of k + 1 >= 0I think somewhere in the middle i started to think too much and it all became a mess but can you explain how to move ahead after this, or what is wrong in my approach
•  » » » » 2 months ago, # ^ | ← Rev. 2 →   0 Let's consider i blocks with sum lengths of all blocks being k, then there are two cases Last block at the end. The last block is not at the end
 » 2 months ago, # |   -21 I demand the castration of the authors of this round!!!
•  » » 2 months ago, # ^ |   0 Based
 » 2 months ago, # | ← Rev. 2 →   0 GREAT ROUND, How to solve Problem C? and can anybody prove why answer always exists?
•  » » 2 months ago, # ^ |   +8 It doesn’t. The checker guarantees that the answer exists for all testcases.
•  » » 2 months ago, # ^ | ← Rev. 3 →   0 The answer does NOT always exist. The problem specifically implies this. However, the problem guarantees that an answer must exist for each test case, so you don't have to worry about these no-answer scenarios. A simple counterexample is when every row has at least one 1. Since there is no all-zero row, this means every set is a proper subset of another set, which must lead to a cycle of proper subset chains, so each set in this cycle is a proper subset of itself, which is impossible.Anyway, my answer was to enforce a rule that the number $i$ will only appear exactly in set $i$ and its proper subsets. I'll refer to $i$ as the ID of set $i$. This fulfills the three conditions: Sets are non-empty (must contain its ID) and values are from $1$ to $n$ (all IDs are from $1$ to $n$, and no other values are considered). All sets are distinct, because if we consider any pair of sets, they cannot both be proper subsets of each other (this leads to a contradiction as I mentioned above), so the one that isn't a proper subset will lack the ID of the other. Binary Matrix representing proper subsets is proven by how we construct the sets: for row $i$ of the matrix, we add the ID $i$ to each column $j$ for which $b_{i, j} = 1$. This fulfills the third condition, while also following the ID rule. My submission: 181766895
 » 2 months ago, # |   0 In problem C:what should be the answer for this test case: 1 4 0001 0010 0001 0000 
•  » » 2 months ago, # ^ |   0 my code gives this hope its correct:->1 1 1 2 2 2 3 3 1 3 4
•  » » 2 months ago, # ^ |   0 My code returns this: 1 1 1 2 2 2 3 4 1 2 3 4 
•  » » 2 months ago, # ^ | ← Rev. 3 →   0 My code gives this: 1 1 1 1 2 1 2 3 1 2 3 Is it correct?Edit: it is wrong as a2 is same as a1.
 » 2 months ago, # |   -17 B and C were pretty "sucky sucky" problems.
 » 2 months ago, # |   +21 Hopefully I lose enough rating to do tomorrow's Div 4 contest
 » 2 months ago, # |   0 For Problem E, if there is a component which contains a cycle, how do we determine the minimum non-zero number of operations that can be performed on it such that the component remains connected?I realized that the answer is $|C_u|$ if $C_u$ is complete, and 1 if $C_u$ is acyclic but not complete, but I couldn't figure out the solution for general graphs. Can anyone provide a hint on how to solve this general case, or at least let me know whether there is an easier alternative approach for solving E?
•  » » 2 months ago, # ^ | ← Rev. 3 →   0 I have thought of this solution: Let's think the graph is not connected. Let's think there is no single vertex, else we could invert its edges. If there is a connected component that is not complete, let's invert the edges of one its vertex that is not a joint and is not connected to all vertices of the component. Then it will be connected to at least one of the other vertices and the other vertices themselves will be connected together. Note that if there is a vertex of the component that is connected to every other vertex, the other vertices are all not joints, and if there is none, the leaves of the DFS tree still are all not joints. If all the components are complete, there are either two components or more. One possible answer is to invert all the vertices of any component. There is another way to do it if there are >= 3 components. We can see that if we invert a vertex in each of some two components, the graph will be connected. (Note that after we invert some vertices, the not inverted vertices of each component will be connected together and the inverted vertices of each component will also be connected together and also with the not inverted vertices of other components.)
•  » » » 2 months ago, # ^ |   0 I understand that if there is a degree-1 vertex in a component with $\geq 3$ edges, then this vertex alone is enough. But there is a broad set of graphs for which no vertex has degree-1, but the graph is also not complete. Inverting all vertices in this component might not be optimal (which pretest 4 thankfully demonstrates), so how do we determine the optimal number? Is there some clever way of choosing which vertices to invert that leads to an optimal solution?
 » 2 months ago, # |   +5 Spent more than half of contest trying to figure out amount of ways to choose t blocks of k elements from n positions and didn't seem to succeed(problem D). How to solve it?
•  » » 2 months ago, # ^ |   +5 First N choose t for the places, then for the sizes, we have t integers summing up to K, each must be greater than 1, it's a case of bars and stars.
•  » » » 2 months ago, # ^ |   +3 Thanks! But shouldn't we take care about cases where for example n=5 last block is at position 3 and it's size more than two(0-indexed)?
•  » » » » 2 months ago, # ^ |   +5 The way I imagined the places, was rather than based on a fixed position, based on insertion between blocks of n-k integers. As in, if we have n-k integers and we need to place t blocks of a total of k between them, we have n-k+1 possible places a block.
•  » » » » » 2 months ago, # ^ |   +3 Thanks! Got it now.
 » 2 months ago, # |   +30 In problem G, the time limit seems not to be enough to make the allowed number of queries interactively.
 » 2 months ago, # |   0 I'm wondering if there will be an editorial for this.
•  » » 2 months ago, # ^ |   0 Yay. There is an editorial
 » 2 months ago, # |   -6 dalbaiob
 » 2 months ago, # | ← Rev. 2 →   +5 Can problem $D$ be solved using $DP$ and matrix exponentiation trick to optimize the $DP$ to not end up in a complexity of $O(N\cdot K)$?If so, how can we use matrix exponentiation in this case?
•  » » 2 months ago, # ^ |   -34 In fact we can solve the bonus of D with DP, but need some hard knowledge. :(You will see it in the tutorial soon.
 » 2 months ago, # |   0 how to solve E? because I don't get what's wrong with my code
•  » » 2 months ago, # ^ | ← Rev. 2 →   +1 I don't know your approach, but here is a counterexample: graph with two components A-B-C and D-E-F, both paths. The answer should be 1, because if you apply the operation on any vertex other than B or E, the graph becomes connected. The test-case format is: 6 010000 101000 010000 000010 000101 000010 More generally, if there is a connected component with at least 3 vertices and at least one vertex has degree-1 (which is always true for a tree btw), then choosing this vertex would always be sufficient. It will connect to all other components, and although it loses an edge with its neighbor, it will gain edges to the other neighbors of this neighbor, so connectivity within its own component is preserved.(I didn't solve E btw, but this particular subcase I was able to establish)
•  » » » 2 months ago, # ^ |   0 Thank tou for the counterexample!
 » 2 months ago, # |   +4 Penalties for wrong submissions are killing me. And it's not just in this contest, but in almost every contest I participate in. Last 6 contests: 15 wrong submissions of which 9 are in last 2 contests...Any advice on how to have mental energy to spend extra few minutes checking if the solution is actually correct?
 » 2 months ago, # |   0 how to solve D?
 » 2 months ago, # |   +35 A really huge gap between C and D.
•  » » 2 months ago, # ^ | ← Rev. 2 →   +1 Oh I so agree !!
 » 2 months ago, # |   +1 How to solve E ?
 » 2 months ago, # |   0 Is C is to hard or I'm just a rookie ?
•  » » 2 months ago, # ^ |   0 if (User.rating <= 1300) { cout << "ROOKIE\n"; } else cout << "HARD\n";
 » 2 months ago, # |   -8 FSTforces.
 » 2 months ago, # |   +6 If anyone is curious, my take on F1 and F2Consider the subarray of length 3 case. Since the median cant be in the middle, this rules out cases where for a 3-tuple (a, b, c), either a > b > c holds true, or a < b < c holds true. So this lets us know that there is a zig-zagging throughout the whole array, it always will go up and down. Some elements will always be less then their adjacent neighbor elements and some will always be higher, we will denote those low numbers and high numbers respectively. Now, notice that in a subarray of length 5, WLOG suppose the middle number is a low number. In order for it not to be the median, either the first, the last, or both the first and last numbers must be greater than it. If this is not true, than it is the median. For example, (3 5 1 7 2) is valid, so is (5 9 4 7 2), but not (4 9 7 8 5). Let us look only at this pattern that happens between the first, middle, and last numbers in a 5-tuple, suppose the 5-tuple is (a, b, c, d, e), we shall call them (a, c, e) respectively. As we said before, either a, e > c, a > c > e, or a < c < e. This tells us, that if a < c, then it must be true that c < e, and likewise if e < c, then it must be true that c < a. From this we can deduce that when the 'low' numbers start increasing, they will stay increasing, and likewise for decreasing. What we get is a 'V' shape in the 'low' numbers, they will start relatively high, reach some minimum, and only increase from there. WLOG this can be extended to the 'high' numbers. Since we have a 'V' shape at the bottom, and an upside down 'V' shape at the top for the high numbers, we now know that all the high numbers are larger than all the low numbers (which is something that was not obvious to begin with). This knowledge allows us to determine that the first n/2 numbers are low, and the last n/2 numbers are high (some edge cases). So you can treat them separately. This allows us to take the odd and even indices, separate them, and squeeze them back together to get two equivalent problems:We have now reduced the problem to: "given a partially filled permutation, find the number of valid permutations that have only one minimum/maximum (depends on whether we are handling the 'high' or 'low' numbers case". It would take a visual representation to show you how to do this efficiently, but it is certainly doable within O(n log(n)) time, possibly faster.
•  » » 2 months ago, # ^ | ← Rev. 4 →   +5 But for those wondering, the next part of the solution is along the lines of: thisyou can come up with an interval where you know for sure the lowest x numbers are (including the minimum) for some X which you can find by looking at the pre-filled numbers in the permutation, and following the direction of increasing/decreasing pairs. (or x = n if there are no pre-filled), then, by doing the same method, you can look at the interval between all the pre-filled numbers in the permutation, working your way up from the minimum, and then, based on that determine that the numbers a-b, for some a and b, will all be placed in two intervals of certain size. Continue doing this until you reach the edges of the array, and you are done. Once you get those assignments of "there are x numbers that I must put in decreasing order among y positions, and increasing order among x-y positions", such as: putting 3-7 in [-, -, -], [-, -] which you can do like this: [3, 5, 7], [6, 4]. The way you get the number of possibilities is by saying "I will place numbers starting from the highest going to the lowest, either in the left or the right array". So in my example, you would have 01010 (left, right, left, right, left). In other words, you would have (x choose y) ways of setting that up. Once you have all your various X's and Y's, your answer would just be the product of (X_i choose Y_i) for all i. Repeat for the 'high' numbers, multiply the two together (this will also ensure that if one of them has no possible solution, you get zero), and you are done! oh, add mod as well :]
 » 2 months ago, # |   -43 shitty fucking contest im crying so hard!!!
•  » » 2 months ago, # ^ |   +2 you are not alone
 » 2 months ago, # |   0 Anyone help me.....What is the output for this test case in problem B.14 1 2 3 2 1 2 1 2 1 2 1 2 1 2If it give answer n ? how it is work ?
•  » » 2 months ago, # ^ | ← Rev. 2 →   0 Answer will be n (Size of the vector) which is 15 here! For cases like a,b,a,b...the answer is (n/2)+1 and otherwise it'll be n because we can choose the number next to the different number and delete it without causing two equal numbers becoming adjacent to each other (Since initially no two equal numbers are adjacent)
•  » » » 2 months ago, # ^ |   0 first one (14) is the number of element. But why?? maxxi can chose any element and delete it. if I delete first element 2(i=2) and 2(i=n) will be adjacent. again chose 3(i=3) again 2(i=2) and 2(i=4) will be adjacent. So, Why we get answer 14 for this test case??
•  » » 2 months ago, # ^ | ← Rev. 2 →   0 remove 14 then 2 next to 3 then 1 next to 3 and so on until everything is removed
•  » » » 2 months ago, # ^ |   0 thanks.
 » 2 months ago, # |   +136 ALL of the following simultaneously being true for problem D is pretty infuriating: 1s time limit for no good reason Test 30 not in pretests The fact that n*log(n) on n=1e6 TLEs in Java The fact that this TLE on system tests is not something anyone would ever think to hack :(
 » 2 months ago, # |   0 A speed-round for contestants who didn't solve D
 » 2 months ago, # |   0 D is a problem of distributing 1s into blocks, which is easy.how do you do the case for distributing 0s? The first 0 of each block has to be both 0, and other 0s has 3 ways to get it
 » 2 months ago, # |   +6 stupid e
 » 2 months ago, # | ← Rev. 2 →   0 Why is my solution getting WA for problem E?My Logic: Find all connected components. If the number of connected components = 1 then return 0 For each component, I will check if it is strongly connected. If it is not strongly connected then the answer is 1, because we can choose a node in this component then it will still remain connected after 1 operation and also it will get connected to all other components. Now we are left with all strongly connected components. If the number of strongly connected components is> 2 then the answer is 2 because after one operation we get to the same situation as the second point. Else (no. of comps ==2) then anser is minimum size of two components. Please correct me if my approach is not correct.
•  » » 2 months ago, # ^ | ← Rev. 4 →   +10 First of all, the graphs are undirected, so every connected component is "strongly connected". I don't understand how you arrived at Steps 4 or 5, but neither are correct.Counterexample for Steps 4 and 5: consider two or more trees (a component with no cycles within it) of say, size 50 each. Then the optimal answer is 1. Just pick any leaf (degree-1 vertex); the operation will connect it to all other trees, and although it loses the edge to its parent, it gains an edge to its grandparent, so it is still connected to its parent and all other nodes in its tree.
•  » » » 2 months ago, # ^ |   0 Sorry I meant "complete components" instead of Strongly connected.I just realized step 4 is wrong.But step 5 is still correct because in the counter-example you have given a tree that is not a complete component. So the answer is 1 from step 2.I still claim that my approach is correct ignoring step 4.
•  » » 2 months ago, # ^ | ← Rev. 4 →   0 You missed a case in step 2. Consider the following example: 1 6 011100 100000 100000 100010 000100 000000 Your solution will operate on node 1, but it will not make the graph connected.
•  » » 2 months ago, # ^ |   0 Your solution is (almost) correct, look at test 4 to find error in logic. I one-line-fixed your solution: 181807261
•  » » » 2 months ago, # ^ |   0 From guitarhero_0221 's comment test case, I realized that we can't directly remove any node from the non-complete-component. We should first check if removing a node from this component still preserve the connections between other nodes in the same component.But, why does reversing the component fix the solution?
•  » » » » 2 months ago, # ^ |   0 Unlike first vertex, last vertex in dfs order will never break condition "check if removing a node from this component still preserve the connections between other nodes in the same component". This vertex will be one of the leaves or vertex on some cycle.I fell into this trap too. And could not understand this until I saw test above. Tourist solves this "issue" by taking last vertex of bfs order if all vertices are not connected to each other
•  » » » » » 2 months ago, # ^ | ← Rev. 2 →   0 Wow, thanks a lot. I understood it. Also I was thinking instead of reversing( or getting the last vertex of bfs) we could sort the nodes in each component based on the degree of that node. Will this work?Upd : Yea Its working, Thank you Igorjan94 :)
•  » » » » » » 2 months ago, # ^ |   0 Yes. This way we will get leaves first (or if there is no leaves we don't care at all about order and can use any vertex)
 » 2 months ago, # |   +114
•  » » 2 months ago, # ^ | ← Rev. 2 →   +7 One down, one to goUpdated: Accepted!
•  » » 2 months ago, # ^ | ← Rev. 2 →   +59
 » 2 months ago, # | ← Rev. 3 →   +3 Seems like this solution of mine for B is similar to editorial. May I know a testcase where it'd be WA?submissionThank you in advance!
•  » » 2 months ago, # ^ |   +8 When n <= 3 you don't read the rest of the input
•  » » » 2 months ago, # ^ |   0 Oh my... thank you so much! :')
 » 2 months ago, # | ← Rev. 2 →   0 why this is wrong for the first test case ?! 3 1 2 3 1 1 2 1 2 4 1 2 3 4 1 1 2 1 2 3 1 2 3
•  » » 2 months ago, # ^ |   0 set 3 [2 1 2] can't be a proper subset of set 1[3 1 2 3]
 » 2 months ago, # |   +67 Congratulations to dorijanlendvaj for reaching LGM with a 2nd place finish!
•  » » 2 months ago, # ^ |   +10 dorijanlendvaj ANIMALL!!!
 » 2 months ago, # | ← Rev. 3 →   -41 Can somebody please check where my toposort solution of C is wrong #include using namespace std; #define mod 1000000007 #define f(i,k) for(int (i)=0;(i)<(k);i++) void findTopoSort(int node, vector < int > & vis, stack < int > & st, vector < int > adj[]) { vis[node] = 1; for (auto it: adj[node]) { if (!vis[it]) { findTopoSort(it, vis, st, adj); } } st.push(node); } vector < int > topoSort(int N, vector < int > adj[]) { stack < int > st; vector < int > vis(N, 0); for (int i = 0; i < N; i++) { if (vis[i] == 0) { findTopoSort(i, vis, st, adj); } } vector < int > topo; while (!st.empty()) { topo.push_back(st.top()); st.pop(); } return topo; } int main(){ int t;cin>>t; while(t--) { int n;cin>>n; vectora(n); f(i,n) { cin>>a[i]; } vectoradj[n+1]; f(i,n) { f(j,n) { if(a[i][j]=='1')adj[i].push_back(j); } } vectorlvl = topoSort(n,adj); mapmp; f(i,lvl.size()) { mp[lvl[i]+1]=i+1; } for(auto it:mp) { cout<
•  » » 2 months ago, # ^ |   +1 put your full code in ideone , we don't know what the "topoSort" does
•  » » » 2 months ago, # ^ |   0
•  » » » » 2 months ago, # ^ |   0 The solution is wrong because of this part of the problem statement: "In other words, bi,j is 1 if Ai is a proper subset of Aj and 0 otherwise.".Also if you get WA on pretest 1 (the samples from the problem statement), then you can look at the failure details even during the contest and the checker says "wrong answer Not meet the requirements of b_{3,1} (test case 1)". I made exactly the same mistake in my initial submission: 181794076
 » 2 months ago, # |   0 in problem B i'am not able to think of any approach due to circular array term getting confused each time a small hint to the problem will be a great help
 » 2 months ago, # |   0 as contester go on and do more rounds
 » 2 months ago, # | ← Rev. 2 →   0 Really liked problem E. Thank for authors of contest!
 » 2 months ago, # |   0 Why is my solution getting WA for problem E? sorry for my bad english and thank you in advance https://codeforces.com/contest/1761/submission/181809468
•  » » 2 months ago, # ^ |   0 Take a look at Ticket 16475 from CF Stress for a counter example.
 » 2 months ago, # |   0 for problem B . my approach was i would check for every index if( index && index+2 are equal ) i'd increment a duplicate variable by 1 . for n-2, n-1 th index i'd check with 0 , 1 index. if duplicate == n . ans == n/2+1else ans = n.why wouldn't this work
 » 2 months ago, # |   0 500 delta and 250 comments! So, 500/250 = 2; we want 2nd round in near future!
 » 2 months ago, # |   +11 My solutions (video):A. Two Permutations SolutionIf $n - a - b \le 1$, then you can see that $p=q$, because either there is one element not in the common prefix or suffix, making the position of that element forced, or the common prefix and suffix intervals are touching/overlapping, so check if $n=a=b$. Otherwise, $n - a - b \ge 2$, and you can take $q = p_1, ..., p_a, p_{n-b}, ..., p_{a+1}, p_{n-b+1}, ... p_n$, so the answer is "Yes."B. Elimination of a Ring SolutionIf there is $1$ distinct element, then $n=1$ so the answer is $1$. If there are $2$ distinct elements, they alternate between two elements, and two elements are deleted each operation no matter what, except for the last two operations, where one element is deleted each time, so the maximum number of operations is $n/2+1$ ($n$ will be even).If there are $\ge3$ distinct elements, the main idea is to make it so that there is a unique element in the ring. Then you can just delete elements next to the unique element one at a time. So assume the frequency of each element is at least two. There will be two equal elements on the ring with distance at least $3$ apart, going clockwise or counter-clockwise. Take one of those elements $x$. If its two neighbors are different, delete $x$. Otherwise, delete one of its neighbors, then you are guaranteed able to delete $x$. Repeat the process until $x$ is unique, and then we can do as stated before. The answer is $n$.C. Set Construction SolutionIf $A_i$ is a subset of $A_j$ make a directed edge from $j$ to $i$. Then, run a depth-first search algorithm on all the nodes $u$ with no parents. Let $x$ be the smallest number not used yet, (this is guaranteed to be $\le n$, since there are $n$ nodes). Let $N$ be the set of all the neighbors of $u$. Then you can set $A_u=(\bigcup_{v \in N} A_v)\cup x$. $A_u$ is guaranteed to not be a subset of anything we don't want because of $x$.D. Carry Bit SolutionConsider all blocks of carry bits and non-carry bits. Loop over $i$, the number of carry bit blocks.If a carry bit comes right before a non-carry bit, there is only one way. If a carry bit comes right before another carry bit, there are three ways.If a non-carry bit comes right before a carry bit, there is only one way. If a non-carry bit comes right before another non-carry bit, there are three ways.So you can go through each of the four cases, whether it starts/ends with a carry bit or non-carry bit, and calculate the number of ways in that case, using the stars and bars formula. Special cases need to be handled for $k=0$ and $i=1$.\begin{array}{|c|c|c|c|c|} \hline \text{First block} & \text{Last block} & \text{Ways to put carry blocks} & \text{Ways to put non-carry blocks} & \text{Ways to set bits} \cr \hline \text{Carry} & \text{Carry} & \binom{k-1}{i-1} & \binom{(n-k)-1}{(i-1)-1} & 3^{n-i-(i-1)}\cr \hline \text{Carry} & \text{Non-carry} & \binom{k-1}{i-1} & \binom{(n-k)-1}{i-1} & 3^{n-i-(i-1)} \cr \hline \text{Non-carry} & \text{Carry} & \binom{k-1}{i-1} & \binom{(n-k)-1}{i-1} & 3^{n-i-i} \cr \hline \text{Non-carry} & \text{Non-carry} & \binom{k-1}{i-1} & \binom{(n-k)-1}{(i+1)-1} & 3^{n-i-i} \cr \hline \end{array}
•  » » 2 months ago, # ^ |   0 Sir,please check MY CONTEST CODE OF C of PROBLEM C , i am applying the same APPROACH as you but got wrong on test 2 UNABLE TO FIND OUT WHAT IS WRONG IN MY CODE SINCE YESTERDAY
•  » » » 2 months ago, # ^ |   0 Take a look at Ticket 16476 from CF Stress for a counter example.
 » 2 months ago, # |   -84 Trash problem E, if use cin or scanf("%1d",&x) will get TLE.Meaningless.
•  » » 2 months ago, # ^ |   +4 Try C++20. It's faster.btw after the contest, plenty of my classmates were complaining about this meaningless TLE in problem E. Extremely thankful for this. Skill issue? maybe :(
 » 2 months ago, # |   0 Nice round, i like b, d, e!although im still annoyed at myself for accidentally printing n^3 on the k=0 case instead of 3^n LOL its worse when the one sample with k=0 had n=3 XD
 » 2 months ago, # |   0 Finally became pupil. Please suggest how can I be a specialist? What things should I learn. I am not doing well with DP.Can I become a specialist without using DP?
 » 2 months ago, # |   0 The editorial link is made private. I can't open it.
 » 2 months ago, # |   +2 ()
 » 2 months ago, # | ← Rev. 2 →   0 Hey Everyone , This is My Code for C ,but i got wrong on test 2 but don't know why?my approach is :) - i am applying DFS in PROBLEM C assume like directed tree- i is subset of j then we can say that j is ancestor of i - we got a adjacency list from given matrixthen we set of parent = set of parent + set of its all sons please go through my code , please give any counter cases where the code is falling , is there anything i missed?
•  » » 2 months ago, # ^ |   0 you write lot of code bro
•  » » 2 months ago, # ^ |   +1 Take a look at Ticket 16477 from CF Stress for a counter example.
 » 2 months ago, # | ← Rev. 2 →   0 I deeply love this contest. But the data in problem B is a little bit weak.My program passed the final test but it uses the wrong initialization.
 » 2 months ago, # |   0 First comment as specialist ! :)Round was really good for me than usual
 » 2 months ago, # |   +5 TLE with std::cin for std::string in problem E is sad. :( I usually use scanf/printf for everything except strings, but here is an example where it may screw you, so you either should use char* with scanf or ios_base::sync_with_stdio(0) and your lovely cin/cout...
 » 2 months ago, # |   -18 It would be nice to finally fix language selector issue and remove excessive use of flags.Sources with supporting arguments: fix language selector blog post flags are not languages, www.flagsarenotlanguages.com Codeforces Language Picker -- chrome extension to see how fixed codeforces language picker would look like.Please support the initiative and stop reinforcing poor UX practices.
 » 2 months ago, # |   +8 I really enjoyed this one, I'm a big fan of dX!! Although I didn't play this game, I like E and D very much. Personally, I think the difficulty of E is much less than that of D. It may also be my math problem.
•  » » 2 months ago, # ^ |   +3 Interesting how both your codes coincides with the same person (Nighthe).
•  » » » 2 months ago, # ^ | ← Rev. 2 →   0 The first code is not at all matching
•  » » 2 months ago, # ^ |   0 i too got this for problem c. similar code can be coincident
 » 2 months ago, # | ← Rev. 3 →   0 jayanth777/181777290 (mine)KaranBardhan/181760840 was flagged for cheating . i dont know the other guy. it was coincidence . how to prove this
»
6 weeks ago, # |
0

Congratulations to hoodie winners! In a few weeks you will be contacted via private messages with instructions to receive your prize.

As usual, we used the following two scripts for generating random winners, seed is the score of the winner.

get_tshirts.py
randgen.cpp
List place Contest Rank Name
1 1761 1 tourist
2 1761 2 dorijanlendvaj
3 1761 3 mnbvmar
4 1761 4 fivedemands
5 1761 5 Benq
6 1761 6 jiangly
7 1761 7 maroonrk
8 1761 8 inaFSTream
9 1761 9 hitonanode
10 1761 10 gisp_zjz
11 1761 11 ksun48
12 1761 12 BurnedChicken
13 1761 13 noimi
14 1761 14 TheQueenOfHatred
15 1761 15 ugly2333
16 1761 16 tranquility
17 1761 17 fallleaves07
18 1761 18 Maksim1744
19 1761 19 ecnerwala
20 1761 20 wangziji
21 1761 21 tabr
22 1761 22 WYZFL
23 1761 23 353cerega
24 1761 24 nantf
25 1761 25 Vercingetorix
26 1761 26 Egor
27 1761 26 Nyaan
28 1761 28 SSRS_
29 1761 29 Magpie_
30 1761 30 suyiheng
55 1761 55 Little99
57 1761 57 Tlatoani
63 1761 63 fengzhengwei
81 1761 80 kmjp
84 1761 84 oleh1421
85 1761 85 Jimanbanashi
88 1761 88 acceptedzhs
95 1761 95 TrivialMan
97 1761 97 Ayanami_desu
99 1761 99 Everule
•  » » 5 weeks ago, # ^ |   0 Congratulations :)
•  » » 5 weeks ago, # ^ |   +43 Suprise!