By pikmike, history, 13 months ago, translation,

Hello Codeforces!

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

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

The problems were invented and prepared by Roman Roms Glazov, Adilbek adedalic Dalabaev, Vladimir Vovuh Petrov, Ivan BledDest Androsov, Maksim Ne0n25 Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

Our friends at Harbour.Space also have a message for you:

Hello Codeforces,

We want to know what you think about Harbour.Space’s involvement in the Codeforces community, and how you believe we can improve.

So we created this short, 5 question survey to hear your thoughts about how we can provide you with more stuff that you’re interested in, so that we can improve your experience on Codeforces.

We really value your feedback, so it would mean a lot if you could take a minute and fill out the survey. Thanks in advance!

GO TO SURVEY→

Congratulations to the winners:

Rank Competitor Problems Solved Penalty
1 kefaa2 7 192
2 jiangly 6 152
3 betrue12 6 187
4 I_love_Tanya_Romanova 6 196
5 pekempey 6 198

Congratulations to the best hackers:

Rank Competitor Hack Count
2 apoorv024 87:-4
3 Kolo_Ray7 73:-31
4 lollihunter 52:-7
544 successful hacks and 568 unsuccessful hacks were made in total!

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

Problem Competitor Penalty
A Chereshnya_ 0:01
B Chereshnya_ 0:02
C YahiaAshraf74 0:04
D sysjuruo 0:08
E kefaa2 0:31
F Benq 0:35
G Benq 1:02

UPD2: Editorial is out

• -168

 » 13 months ago, # | ← Rev. 3 →   -32 Are the educational rounds a bit on the tougher side ?
•  » » 13 months ago, # ^ |   +16 Hahaha you are so funny.
•  » » 13 months ago, # ^ | ← Rev. 2 →   +20
 » 13 months ago, # | ← Rev. 2 →   +12 PikMike your Educational Contest are great <3 !!
•  » » 13 months ago, # ^ |   0 You must be new here...
•  » » » 13 months ago, # ^ |   +49 PikMike gives us so many good rounds, we are very grateful <333333
•  » » 13 months ago, # ^ |   0 Great way to farm karma eh?
•  » » 13 months ago, # ^ | ← Rev. 2 →   +38 I agree with you :)
 » 13 months ago, # |   -23 Finally a contest with a 12 hour hacking period Hope to turn purple this contest
 » 13 months ago, # |   +75 I hope all of you who is >= master can register for the round unofficially. Shouldn't take part in with new account. Not nice! Thanks!
•  » » 13 months ago, # ^ |   +7 You are absolutely right!
•  » » 13 months ago, # ^ |   +1 is div2 harder to raise rating for CM than div1 or div1+2 because of new accounts?
•  » » 13 months ago, # ^ |   -8
 » 13 months ago, # |   +19 thanks codeforces for preparing several contests with low time interval. hope to have a great contest !
•  » » 13 months ago, # ^ |   +5
•  » » » 13 months ago, # ^ | ← Rev. 2 →   +5 or is this just fantasy?
•  » » » » 13 months ago, # ^ |   0 Caught in a landslide, no escape from reality
•  » » » » » 13 months ago, # ^ |   0 Open your eyes, look up to the skies and see
 » 13 months ago, # |   +7 Thanks for the contest Pikmike
 » 13 months ago, # |   +2 I don't want to become a specialist again.
•  » » 13 months ago, # ^ |   0 same to you...
•  » » » 13 months ago, # ^ |   0 All the best to you!
 » 13 months ago, # |   +1 Good luck everyone!!!! I wish everyone to rise!!!
 » 13 months ago, # |   +8 I hope the pretest is strong enough because it's the first time I have solved 4 problems in a live contest. Thanks God!
•  » » 13 months ago, # ^ |   0 but round is unrated now :(
•  » » » 13 months ago, # ^ |   0 who said so?did they announce anything about that?
•  » » » » 13 months ago, # ^ |   0 see the anouncement under problems
•  » » » » » 13 months ago, # ^ |   0 It is rated. My rating has changed.
•  » » » » » » 13 months ago, # ^ |   +3 i think unrated for who solved B wrong
 » 13 months ago, # |   +5 you should've used quotations ( 'X' , '.' ) in problem E, It would be simpler to read
 » 13 months ago, # |   +12 So many unbalanced rounds these days!
 » 13 months ago, # |   +19 How to solve E?
•  » » 13 months ago, # ^ |   +3 Hint: If there exists segment of length x, where b<=x
•  » » » 13 months ago, # ^ |   0 So for a, we need to to prioritize decreasing length by a+b-1 always?
•  » » » » 13 months ago, # ^ |   +3 Not really. You can find a countertest for that. You can check the comments below, there is a spoiler to E.
•  » » » » » 13 months ago, # ^ |   0 ok got it!
 » 13 months ago, # |   +46 Yes, the problem G was correct...
 » 13 months ago, # |   +7 how to solve D with DP? i keep thinking about back tracking...
•  » » 13 months ago, # ^ |   +13 it's easy to see that each element should increase at most 2 times
•  » » » 13 months ago, # ^ |   0 I thought same but then I made this case.$a=3 \enspace 3 \enspace 4 \enspace 4$$b=5 \enspace 1 \enspace 5 \enspace 7This case made me feel there might not be limitation in height increase. •  » » » » 13 months ago, # ^ | ← Rev. 4 → 0 I guess the answer is 6 (a = 3 \enspace 4 \enspace 5 \enspace 4), and none of them increase more than twice. •  » » » » » 13 months ago, # ^ | 0 Yes I realized it after running over AC code. I thought answer would be 8. •  » » 13 months ago, # ^ | 0 SpoilerYou never add more than 3 to a single position •  » » » 13 months ago, # ^ | +1 2 or 3? got AC with the same insight but for 2. •  » » » » 13 months ago, # ^ | +7 2 is correct aparently. I used 3 just to be sure. Some submissions even use a few more •  » » » » » 13 months ago, # ^ | 0 Here is the argument for 2. Suppose you've done everything you want with all but the i-th board. Now note that there are at most 2 boards next to the i-th board, so you never need to increase the height of the i-th board more than twice in order to make the heights unique. •  » » 13 months ago, # ^ | ← Rev. 2 → 0 but u need to use FAST IO, it was my mistake  » 13 months ago, # | 0 Really confused in E...  » 13 months ago, # | +11 How to solve E? Don't even know how to do it in O(n^2). •  » » 13 months ago, # ^ | ← Rev. 2 → +136 Case 1: When there's a segment with length x where b\leq x< a, Bob always wins.Case 2: When there's at least two segments with length at least 2b, Bob can always divide one fitting into case 1 and thus win.Case 3: When there's no segment with length at least 2b, the winner depends on the parity of number of segment with length \geq a.Case 4: When there's exactly one segment with length at least 2b, enumerate over all possible moves for Alice with this segment, then it reduces to case 3. •  » » » 13 months ago, # ^ | +14 simply wow •  » » » 13 months ago, # ^ | +1 How you came up with this solution? •  » » » » 13 months ago, # ^ | +17 B wins if in his move there is some segment with length at least 2b, because he can make a segment of length b, therefore he will have at least 1 more possible move than player A. •  » » » 13 months ago, # ^ | +5 How did you come up with intuition of segment with length atleast 2b •  » » 13 months ago, # ^ | +3 Check wheter there exists segment of length x, where b<=x=2*b and now we only take care of segments >=a. If( cnt>=2) Then B always win because no matter what A does he can split segment of length x, where b<=x •  » » 13 months ago, # ^ | +3 My observation: in B move if there is some segment of size >=2*b he can win. •  » » 13 months ago, # ^ | 0 At first consider only blocks with no less than b letters. If there is a block with at least b but less than a letters then second player wins. If it is a turn of a second player and there is a block with at least 2 * b letters then second player can create a block of length b and win. So, check at first if there is a block of length between b and a — 1, if there is, then second player wins. Then count amount of blocks of length at least 2 * b. If it is 0, then in every other block each player can make only one turn, so total number of turns is amount of blocks. If it is at least two, then second player wins. If it is one, bruteforce the move of the first player, and check if he can win(after his move all the blocks should be the length less than 2 * b so number of moves will be the number of blocks of length at least b).  » 13 months ago, # | +5 How to solve E?I thought I'd compute for every X from 1 to size(s) winAlice[X] and winBob[X].winAlice[X] will be true <==> Alice can win a segment of size X if played optimally.Then I'll be left with the segments of the give string(contiguous '.') and for each one(say it's length is X) I'll have 4 possible cases, depending the pair (winAlice[X], winBob[X]).Anyone else thought like this? •  » » 13 months ago, # ^ | 0 Same thoughts, but can't figure out how to obtain the answer from those 4 values.  » 13 months ago, # | ← Rev. 2 → 0 why my submission problem C is wrong? I tried greedy method.. submission •  » » 13 months ago, # ^ | 0 Same as me, I don't know why I got WA on test 2 also •  » » 13 months ago, # ^ | 0 There is a strictly numerical solution to C •  » » 13 months ago, # ^ | ← Rev. 2 → 0 1 3 3 0 answer is 2. 1 team: c c m 2 team: c m m •  » » » 13 months ago, # ^ | ← Rev. 2 → 0 I found counter example 1 18 13 0 answer is 10 •  » » 13 months ago, # ^ | +2 The answer was simple min(a,b,(a+b+c)/3)) •  » » » 13 months ago, # ^ | 0 What is the intution behind this...? I mean to ask what guarantees that by doing this ; you always end up with that many teams satisfying all the constraints in the given question ( I am concerned obviouslsy only with the (a+b+c)/3 part ; i understand why other two terms 'a' and 'b' are in the min expression )...Can you please present a formal proof for the claim... thanks...! •  » » » » 13 months ago, # ^ | 0 (a+b+c)/3 is a constraint because each team must consist of three members. •  » » » » » 13 months ago, # ^ | ← Rev. 2 → 0 ok...Doubt : How do you know that this ALSO returns you the number of teams with 2 coders and one mathematician ( Or 2 mathematicians and 1 coder ) •  » » » » » » 13 months ago, # ^ | +3 Because a+b+c is the total number of people, including the coders and mathematicians.  » 13 months ago, # | 0 Who knows problem B test 2?? •  » » 13 months ago, # ^ | ← Rev. 2 → 0 never mind... sorry •  » » » 13 months ago, # ^ | +5 how can this be a testcase of B? •  » » » » 13 months ago, # ^ | 0 i’m sorry •  » » » » » 13 months ago, # ^ | 0 It is OK Thank YOU •  » » » » » 13 months ago, # ^ | 0 :) Btw i got wa on case 2 for this case :( •  » » 13 months ago, # ^ | 0 Input: 100 Apparently, the checker was buggy, so many wrong solutions passed during the contest and now they re-judged the solutions.  » 13 months ago, # | +12 I am stupid enough because my brain cannot get the ways to solve the problem which everyone except me have been solved. :( •  » » 13 months ago, # ^ | 0 dude... •  » » 13 months ago, # ^ | 0 I feel the same :(  » 13 months ago, # | 0 Can anyone tell where my test case in Perfect team go wrong here is my short code, It will be a great help! #include using namespace std; int main(){ int t; cin>>t; while(t--){ long long a,b,c; cin>>a>>b>>c; long long ans,sum; ans=min(a,b); sum=a+b+c; if(ans==0) cout<0){ if(3*ans<=(sum)){ cout< •  » » 13 months ago, # ^ | 0 Test case 1 1 0 No output in your codeI think this will be while(ans >= 0)  » 13 months ago, # | 0 Can anyone tell test 2 for C? Really desperate to know •  » » 13 months ago, # ^ | 0 1 18 13 0 answer: 10 •  » » 13 months ago, # ^ | 0 try 10 10 0  » 13 months ago, # | ← Rev. 2 → 0 How to solve D ? I tried to solve it with dp but I couldn’t. •  » » 13 months ago, # ^ | +5 You can find an optimal solution where no border is increased more than 2 units, so it can be solved by dp using this observation.Proof: If adjusting border i to be finally of length a_i will make it equal to some adjacent border, and adjusting it to be of length a_i+1 will make it equal to the other adjacent border, so for sure adjusting it to be a_i+2 will make it different from both adjacent borders, and there is no point in increasing it more than this. •  » » 13 months ago, # ^ | 0 Each elements in a should increase at most 2.Let dp(i, j), j \in [0;2] is the minimal cost to make the fence from 1 to i (inclusive) great.Remember keep tracking the last element in previous state.Let prev[j], j \in [0;2] is the last element of the great fence after we increase a_i with j units.Base case:- dp(1, 0) = 0- dp(1, 1) = b[1]- dp(1,2) = 2*b[1]and- prev[0] = a_1- prev[1] = a_1 + 1- prev[2] = a_1 + 2 Recurrence: dp(i, j) = j*b[i] + min\{dp(i-1, k) | k \in [0;2]\}Check my submission for more detail.  » 13 months ago, # | ← Rev. 2 → +58 Can you check the checker of B? This submission seems incorrect (https://codeforces.com/contest/1221/submission/60861923) •  » » 13 months ago, # ^ | ← Rev. 4 → 0 deleted  » 13 months ago, # | 0 Is it possible to solve E using grundy numbers ? If yes How ? •  » » 13 months ago, # ^ | +8 No, because game isn't impartial.  » 13 months ago, # | ← Rev. 5 → 0 Is G answer = 2^n - 2^c - 2*s+ 2 ? (c = # of components and s = # of independent sets). •  » » 13 months ago, # ^ | ← Rev. 5 → +31 No, you also need to take into account the number of bipartitions.Let f(C) be the number of arrangements such that none of the numbers in set C appears. Then our answer is 2^n - f({0}) - f({1}) - f({2}) + f({0,1}) + f({0,2}) + f({1,2}) - f({0,1,2})$$f({0})$ and $f({2})$ are the number of independent sets.$f({1})$ is $2^{(\text{number of components})}$$f({0,2})$ is the number of bipartitions.$f({0,1})$ and $f({1,2})$ is $2^{ \text{number of isolated vertices}}$finally $f({0,1,2})$ is $2^n$ if the graph has no edges and $0$ otherwise.
•  » » » 13 months ago, # ^ |   0 $f(0,1,2)$ should be $2^n$ in the no-edge case.
•  » » » » 13 months ago, # ^ |   0 Oh yeah, thanks.
 » 13 months ago, # |   -38 Is it possible for me to become unrated, as I was affected by G's wrong sample outputs. The announcement was too late and by that point I have already been too focused on debugging. If the announcement was done earlier I would have been able to complete the task.
 » 13 months ago, # | ← Rev. 2 →   -8 Problem A is interesting.In addition to using greedy algorithms (*), priority_queue data structures (**) can also be used. Solution (*): 60890951 Solution (**): 60890114
•  » » 13 months ago, # ^ | ← Rev. 2 →   +3 Actually, because it's guaranteed that all elements of the multiset are powers of two, you can solve it simply and directly just like this.
•  » » » 13 months ago, # ^ |   0 Can you give a proof of this?
•  » » » » 13 months ago, # ^ | ← Rev. 2 →   +3 I can't give strict proof, but imagine the worst situation, after some mergence there is 1 + 2 + 4 + ... + 1024 = 2047. You can get 2048 if any number appear once more, even if it's 1. Or you can get 2048 when get a number once more but lose those smaller numbers. All in all, it seems the answer just depend on the sum.
•  » » » » 13 months ago, # ^ |   0 It can be proven inductively on 2^i.
 » 13 months ago, # | ← Rev. 3 →   +22 Here's my thinking to problem $G$:Clearly we should use inclusion-exclusion principle, then we need to count the number configurations without some certain sets of edges.without $0$/ without $2$: Use meet-in-the-middle, together with SOS DP. The complexity should be $O(2^{n/2}m)$. Heavy code.without $1$: answer is $2^c$, where $c$ is the number of connected componentswithout $0$ and $2$: the answer is $2^c$ if the graph is bipartite, $0$ if not, where $c$ is the number of connected componentswithout $0$ and $1$/ $1$ and $2$: answer is $2^s$, where $s$ is the number of singletonswithout $0$ and $1$ and $2$: answer is $2^n$ if there's no edges, otherwise $0$.However, this requires too much code... Is it intended? Or I think this problem can be solved in $O(fib(n))$ using some recursive searching approach after inclusion-exclusion?
•  » » 13 months ago, # ^ |   0 The only long code is SOS DP though, so I don't really see much of a problem.
•  » » 13 months ago, # ^ |   0 I think $s$ should be $c$ right?
•  » » » 13 months ago, # ^ |   0 I think not, in such case if the conneted component isn't a singleton, then there's only one possible way to color it, otherwise $2$.
•  » » » » 13 months ago, # ^ |   0 You can flip the $0$s with the $1$s and it is still valid.
•  » » » » » 13 months ago, # ^ |   0 That's the case when there's no edge $1$. I'm talking about the case when there's no edge $0$ and $1$/$1$ and $2$.
•  » » » » » » 13 months ago, # ^ | ← Rev. 2 →   0 Consider the graph $K_2$, there are two ways to paint the vertices so that there is no $0$ or $2$. One way is $(0,1)$ and the other is $(1,0)$
•  » » » » » » » 13 months ago, # ^ |   0 Oh, you mean that case. Corrected. Thanks.
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 How do you come up with this? Where does your idea come from? Could you please show your thinking flow? I just don't know where I should start with even though given enough time. Thanks in advance!
•  » » » 13 months ago, # ^ |   0 The condition says 'at least one edge with $0$, at least one edge with $1$ and at least one edge with $2$', which is a union of three sets, implying that one should apply inclusion-exclusion principle.
•  » » » » 13 months ago, # ^ |   0 six six six
 » 13 months ago, # |   +53 My B got accepted, and it doesn't even pass the sample. 60857895
 » 13 months ago, # | ← Rev. 2 →   +27 I would have solved 6 problems if I had just submitted G, I had a working solution which didn't match the output on the 3rd testcase, and now I'm just pissed off. You should have made an announcement that the 3rd sample had a mistake and to refresh the page.
 » 13 months ago, # |   0 Is there anyone who also used INT_MAX instead of 1e18 in D and got wrong :(
 » 13 months ago, # |   0 Please explain to me logic for problem E? Thanks so much <3
 » 13 months ago, # |   +22 Is there anything wrong with the checker of B? Many wrong solutions seem to pass the samples.
 » 13 months ago, # |   +39 Problem B should be removed from problemset.
 » 13 months ago, # |   +23 Checker in problem B is uncorrect. Solution passes tests even if answer wasn't true.
 » 13 months ago, # |   +7
•  » » 13 months ago, # ^ | ← Rev. 2 →   +9 The checker was wrong apparently. They rejudged all the Bs i think. It shows WA for both your submissions now.
•  » » 13 months ago, # ^ |   +1 wait for some time, that one will get WA too
•  » » 13 months ago, # ^ |   0 Well, they should have announced while changing the test cases and rejudged our submissions during the contests while we still had time. This is somewhat an unexpected behavior from Codeforces :(
 » 13 months ago, # |   +40 Does anyone notice that there are $>50$ participants who got AC in the last minute of the contest? It's amazing!!! Deadline is the strongest productivity.
•  » » 13 months ago, # ^ | ← Rev. 3 →   -8 Wonderful =))
 » 13 months ago, # |   +13 the checker of problem B is wrong ,i got hack by the pretest,please unrated.
•  » » 13 months ago, # ^ |   -10 Got WA on 2nd pretest after the contest, problem B should be removed from the problem set or this contest should be made unrated.
 » 13 months ago, # |   -11 please remove problem B and make it rated !
•  » » 13 months ago, # ^ |   -8 I think so. B is unpleasant
•  » » » 13 months ago, # ^ |   -8 Me too. :(
•  » » 13 months ago, # ^ |   +38 Then it will be unfair for people who actually solved it.
•  » » 13 months ago, # ^ |   0 second that
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 Unlikely to happen, what about participants who spent a significant amount of time on B?I only see two realistic possibilities : Contest unrated for all. Unrated for those who got changed to WA afterwards due to flawed checker. Also generally a problem is only removed if the answer is completely wrong. In this case it appears the checker has been fixed, so it'll probably remain in the contest/problemset. Anyway if a problem is removed, it nearly always entails the round becoming unrated.
 » 13 months ago, # |   +19 is it rated?
 » 13 months ago, # | ← Rev. 2 →   0 Thanks to the author for posting problem B to this contest 60892327
 » 13 months ago, # |   +15 Getting wrong answer on pretest after the contest ends, nice checker
 » 13 months ago, # |   -12 you must do this contest unrate
 » 13 months ago, # |   +17 Wrong answer on pretest 1 after the contest? WTF!
 » 13 months ago, # |   -6 My solution was accepted during the contest. But astonishingly after, I saw my submission was not accepted and was also not hacked. That means the author's code was not correct. Either this round should be unrated for me or this problem should be removed from problem set.
 » 13 months ago, # |   +8 It seems there was something wrong with the checker for problem B...
 » 13 months ago, # |   +8 Checker for problem B was wrong so please make it unrated!
 » 13 months ago, # |   0 Problem B should be removed from the problemset.
•  » » 13 months ago, # ^ |   +1 Problem B is fine. Problem is with checker.
•  » » » 13 months ago, # ^ | ← Rev. 2 →   0 Then how I can understand if the problem with checker if already my solution passed during contest time and now I saw WA on test 1. How funny.
•  » » » » 13 months ago, # ^ |   +1 So the contest should be unrated. Problem should not be removed.
 » 13 months ago, # | ← Rev. 3 →   +24 hack checker is faster than the pretest checker to tell me my solution in problem B wrong.
 » 13 months ago, # |   +2 please make it unrated !
 » 13 months ago, # |   -8 There is a bug, my second question was wrong on 2nd pretest, yet that was accepted, and they say wrong answer..... Now my rating is being highly deducted.... Huge bug This contest must be unrated....(atleast for me)
 » 13 months ago, # |   0 PikMike or anyone else, can please let me know what's the problem with B??It was accepted earlier during the contest, but now it is showing WA. I think the system testing has not been done yet, and the solution has not even been hacked.Link to the Solution
 » 13 months ago, # |   +2 My problem B was accepted in contest time and now it shows wrong answer on pretest 1. it's unfair for us(who face same issue)!
 » 13 months ago, # |   0 Wrong answer on pretest 2 after contest ? WTF!!
 » 13 months ago, # |   0 Problem B basically had really weak pretests. I do not find it very convincing to make the round unrated
•  » » 13 months ago, # ^ | ← Rev. 2 →   +5 actually, it has no pretests during contest
 » 13 months ago, # |   0 Educational Codeforces Round 73 [Rated for Div. 2] problem no.2 my solution got accepted initially but it was not able to pass even the testcase given in the question. That created confusion, If the at that time the contest would have said that it is wrong on testcase 1. I might have corrected it. This is wrong codeforces.
 » 13 months ago, # | ← Rev. 2 →   0 There were some problems with checker in problem B during contest. It showed that your solution is right, even if it is not. In my opinion it is not fair and round should be unrated, because during contest i thought that my B was right
•  » » 13 months ago, # ^ |   -8 So every round where there are hacks after contest should be unrated? These are preliminary results for a reason.
•  » » » 13 months ago, # ^ |   0 There was problem with checker during contest. Why do you say about hacks? If i knew that my B was incorrect, i would definitely submit it during contest and get AC
 » 13 months ago, # |   0 Nice Checker on B... Best one I've seen at Codeforces. Make it unrated. tlqkf
 » 13 months ago, # |   +5 No point to make the contest unrated.Assume it to be hacked instead of a WA.
•  » » 13 months ago, # ^ |   0 This is Educational Round instead of a common round so there is no something can be called pretest. I have never seen so many people be hacked in a Educational Round.
•  » » » 13 months ago, # ^ |   +1 Go check last education round's B
•  » » » » 13 months ago, # ^ |   0 :eyes: I'm sorry for this wrong.
•  » » 13 months ago, # ^ |   0 And even if there can be pretests in Edu Rounds, it's not even weak, it just literally can't be called a test.
•  » » » 13 months ago, # ^ |   0 Imagine you print out something you don't know what that is, and you get "Pretest Passed".
 » 13 months ago, # |   0 The checker of B is literally a joke. And the solution to this is just rejudging in several minutes AFTER the contest, I'm literally surprised.
 » 13 months ago, # |   +6 I failed B after rejudging. Still think it should be rated.
 » 13 months ago, # | ← Rev. 2 →   +11 I think than round should be unrated because there are problems with checker in problem B. And it influenced on results.
•  » » 13 months ago, # ^ | ← Rev. 3 →   0 yes i agree with you
 » 13 months ago, # | ← Rev. 3 →   +39 I understand people whose results have been impacted by the rejudge of B, but I suggest that you look at this from a different angle: 700-odd people have WA instead of AC after the contest. There have been many times when a single problem had north of 1000 hacks and the round remained rated. (e.g. some problem about inversion had 1200 AC's during the round, which fell to under 200 afterwards.) Therefore, the solution to make the round unrated for everyone would be a bit overkill (IMO), because it seems that the standings table will have changed less after the hacking phase than in some rated contests (pretests for all problems, except B, are very strong). I know that I'm going to be downvoted to oblivion, but it's just an alternative point of view to consider.
 » 13 months ago, # |   +8 B was trivial, I have -delta but still think it should be rated
 » 13 months ago, # |   +8 I assume that the checker was maybe just checking whether the output is N*N? lol
 » 13 months ago, # |   -23 We are really sorry for the issue with problem B checker. The contest will be definitely unrated for the participants affected by it.
•  » » 13 months ago, # ^ |   +7 Only for them?
•  » » 13 months ago, # ^ |   -6 Not for all?
•  » » 13 months ago, # ^ |   -10 Should be unrated for everyone! This does not make sense at all
•  » » » 13 months ago, # ^ |   0 arpanadhingra10 Exactly Bro.
•  » » » 13 months ago, # ^ |   +11 There are people who solved it with correct method. Think about others. The decision taken by Codeforces is the best, I think. : )
•  » » 13 months ago, # ^ |   -8 Please unrated for all participants who have at least one submisson on problem B.
•  » » » 13 months ago, # ^ |   0 Dude! People's have spent good amount of time solving B. It can be unrated only for those who have affected by it, make's sense.
•  » » » 13 months ago, # ^ |   0 I don't think there was a submission on B which don't get accepted unless there is presentation error. :3
•  » » 13 months ago, # ^ |   +4 For all or not? ;D
•  » » » 13 months ago, # ^ |   0 I voted by mistake Sorry ;(
•  » » 13 months ago, # ^ |   0 How do you determine people affected by problem B ?
 » 13 months ago, # |   0 In Problem B: My solution was wrong and it give wrong answer for pretests also,still it passed the all Pretest. After the Contest Over I realized that my solution was wrong. Why This happened???https://www.imageupload.net/image/d7ft
•  » » 13 months ago, # ^ |   +3 Read the comment section before asking please
 » 13 months ago, # |   +12
 » 13 months ago, # |   -20 Just putting forth a point. All participants affected by the checker of B are all not a specific group of people, either indirectly or directly. Kindly consider this point for the fate of outcome of this contest.
•  » » 13 months ago, # ^ |   +6 what do you mean they are not a specific group of people? You just specified who they are?
•  » » 13 months ago, # ^ |   +14 Rizu Bond
•  » » 13 months ago, # ^ |   0 Probably riz_1_ has a point, but is getting ignored as the officials do not want to make this unrated, and hence the downvotes for the comment by other users are for making this comment to be ignored. BledDest should address the this point also.P.S. Don't downvote me please for putting across a general point
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 Finally, someone who understands. Probably I was misinterpreted the first time.
 » 13 months ago, # | ← Rev. 4 →   -9 Wtf, is this contest a joke? Making it unrated for only those who got affected by it isn't the solution. I did not give the contest from my official handle ( DunnoY ) to waste 2 hrs.
 » 13 months ago, # |   +22 I hope Codeforces could give the affected participants the right of choosing if he/she will be rated when the similar sitiation happens again.
•  » » 13 months ago, # ^ |   +16 I want it to be rated for me even though my rating is decreasing.
•  » » » 13 months ago, # ^ |   +6 Div3 is coming :p
•  » » » » 13 months ago, # ^ |   +6 Why not! Totally fair!
•  » » » 13 months ago, # ^ |   -12 Remember,you won't get extra points for being good.
 » 13 months ago, # | ← Rev. 2 →   +84
 » 13 months ago, # |   +66 What is the difference between having really weak pretest and having a bad checker (if corrected before the main tests). I think making unrated for participants affected by problem B isn't the right decision, but it at least try to find a fair solution. Reading the comments I see a lot of people asking for unrated contest for everyone. I know, you are upset if you lose points, but there is no reason to make it unrated for everyone. The contest was interesting with lot's of good problems. An upvote from me, really not understanding the downvotes. Problemsetters are human too, they make mistakes. They made a lot of work to make this contest, but they get lots of downvotes because of 1 mistake.
•  » » 13 months ago, # ^ |   +9 There was also a mistake in problem G, don't forget that.
•  » » » 13 months ago, # ^ |   +16 Problem G has affected only 2 participants at the moment of clarification. Moreover, the round wasn't rated for them.
•  » » 13 months ago, # ^ |   +2 I don't think that the fact that sometimes there are weak pretests justifies a bad checker. Unjustified ammounts of hate are bad, but mistakes shouldn't necessarily be applauded.
•  » » 13 months ago, # ^ |   0 Problem B and G had problems. I agree with you about the Contest overall, the tasks are really good, but, for me, it is clearly an unrated round.
 » 13 months ago, # |   -18 I hope, BledDest and his friends will understand their fault and make this contest unrated for EVERYONE, because imho either contest should be rated for everyone or in seldom times, like this, it should be unrated. It's weird that they are trying to find the third way :/
•  » » 13 months ago, # ^ |   +5 Yeah, it's not like the ranking doesn't change when you remove a few hundred people.
•  » » 13 months ago, # ^ |   +43 Why it should be unrated for people who have not been affected? Why do you want to discount their efforts? Please, explain.
•  » » » 13 months ago, # ^ |   -41 in this way your rating doesn't show your skill(in my opinion), it is just useless number not more
•  » » » » 13 months ago, # ^ |   +35 How did you come up with a thought that rating won't show the true skill of a participant, if the round is unrated for a specific group of people? Can you elaborate on that?
•  » » » » » 13 months ago, # ^ |   +2 maybe because "rating delta" changes when you remove some participants?
•  » » » » » » 13 months ago, # ^ |   +23 Well, yes, it does. But it's not that tremendous to make the rating system useless. You can notice, that the predicted rating delta of most of the participants has changed by about 10, not more. Everybod knows, that such error in rating is just nothing
•  » » » » » » » 13 months ago, # ^ |   +69 if you don't care about rating, every round is unrated
•  » » » » » 13 months ago, # ^ |   +5 You are interested in the fact that the round was rated for you.
•  » » » » » » 13 months ago, # ^ |   +10 In my comment I didn't tell anyone that I want the round to stay rated. I was just trying to understand, what he said, so, your pretension doesn't make any sense.
•  » » » » 13 months ago, # ^ |   +1 Wait, what?? You solved a problem correctly and got the right verict, how doesn't this show your actual skill? You took more than 1 hour to solve D, because of which you are getting a rating change of -30. How would that have changed if the checker for B was right. I mean you anyways had done it right.
•  » » » » » 13 months ago, # ^ |   -6 "How would that have changed if the checker for B was right" i will open the secret, my little Indian bunny, delta depends on the performance of all contestants, and my delta had changed when they removed some contestants.
•  » » » » » » 13 months ago, # ^ |   +5 Your point is right, but you took the effort to open someone's profile to see if he is Indian to make you sound more cooler? Why?
•  » » » » » » » 13 months ago, # ^ |   +6 dude, that's a joke, i'm not trying to offend anyone
•  » » » » » » 13 months ago, # ^ |   0 Ok. I know that russians are great mathematicians. But are you saying that before every contest you calculate your probability of winning with every other contestant, your expected rank and then decide your strategy and if a few people were removed, your strategy would have changed. Besides, the only change which happened in the verdict was from AC to WA. So your rank would have only been worse had those participants not been removed.Btw I just read that cats eat bunnies. So I am sorry if I offended you.
•  » » » 13 months ago, # ^ |   0 But what about about the efforts of User who solved ABCD, and B is gone.
•  » » » 13 months ago, # ^ | ← Rev. 2 →   +30 As long as sum of all rating changes is negative (recalculate rating over the set of people that passed, without taking any info into account from unrated users) it should be OK. Otherwise you get inflation and it devalues work.UPD: actually I thought about it again based on below comments and my past experiences and I conclude it should be all unrated.
•  » » » 13 months ago, # ^ |   +4 If you make contest unrated for affected users only, you discount their efforts. Such distribution does not show real skills, some of affected people could solve task correctly in short time after getting WA, so really they are stronger than people who overtook them because of wrong checker in B.
•  » » » » 13 months ago, # ^ |   0 second that
•  » » » 13 months ago, # ^ |   +3 Is hard to meassure if someone was "affected", because the ones that "got affected" could have scored better if there were no issues on problem B, and therefore affect to the contestants that "got not affected".
•  » » » » 13 months ago, # ^ |   0 yes. that's what i want to say.
•  » » 13 months ago, # ^ |   +3 Ivan, let's be honest, you just don’t want to lose 30 rating points
 » 13 months ago, # | ← Rev. 2 →   -40 Round should be unrated for everyone.
•  » » 13 months ago, # ^ |   +14 Please, read the rules: hacks don't give points on Educational Rounds.
•  » » » 13 months ago, # ^ |   -43 Still unrating for everyone is a better choice. Everyone in the comment section is asking or this.
•  » » » » 13 months ago, # ^ |   +24 No. Some dozens are asking. More than 5 thousand participate in the contest.
•  » » » » » 13 months ago, # ^ | ← Rev. 2 →   -38 But this is not a fair contest anymore and the best thing is to unrate it. And I think by some dozen you mean more than half the people who are commenting on this issue.
•  » » » » » 13 months ago, # ^ | ← Rev. 2 →   +9 Lol, what kind of metric is this. Technocup 2019 only 291 comments were made and there were 7500 participants. Should that have been rated?
•  » » » » 13 months ago, # ^ |   0
•  » » » » 13 months ago, # ^ |   +9 If you start taking people's opinion for deciding if a round is to be kept unrated, every round will become unrated.
•  » » » » » 13 months ago, # ^ |   -23 Yes, but every round does not have checker problem like this round.
 » 13 months ago, # |   0 Auto comment: topic has been updated by PikMike (previous revision, new revision, compare).
 » 13 months ago, # |   0 Unrated.
•  » » 13 months ago, # ^ |   0 starboy_jb unrated for everyone?
•  » » » 13 months ago, # ^ |   -10 No, I am just requesting for that.I am an affected user.
 » 13 months ago, # |   +74
•  » » 13 months ago, # ^ |   +81
•  » » » 13 months ago, # ^ |   0 XD
 » 13 months ago, # |   +11 Weaker pretests and many hacks doesn't justify make a round unrated.
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 Why is everyone forgetting about issues with problem G? ... I spent some time thinking about the Meet in the Middle and about the code ...
•  » » » 13 months ago, # ^ |   +12 Because too little people were affected. And for those affected, the solutions were rejudged and problem G was the last.
•  » » » 13 months ago, # ^ |   0 Big lol !! You shouldn't have thought about meet in middle , in the middle of getting your C accepted xD . You want this round to be unrated becuase you get a negative delta that's what most believe we are all humans . Don't throw up things like , " I was affected by the issues with problem G because I spent time thinking about it".
•  » » » » 13 months ago, # ^ |   0 hahaha, like if I care about my rating. Have u seen my rating history?
•  » » » » » 13 months ago, # ^ |   +5 ok lord leonardo_paes you should now be going ahead with your meet in the middle solution for G the issues of which affected you instead of trying to reason if this round should be rated or not because hahaha like if you care about your rating .
•  » » » » 13 months ago, # ^ |   0 Nice position at Round 585, btw
•  » » » » » 13 months ago, # ^ |   +5 heard about bad days , kid ?
•  » » » » » » 13 months ago, # ^ |   0 Yes, I am in one today :/ Sorry for the comment
•  » » » » » » » 13 months ago, # ^ |   0 I know mate. I have been in more bad days than the good ones on this platform .
•  » » 13 months ago, # ^ |   0 There is literally nothing can be called pretests.
 » 13 months ago, # |   +12 MikeMirzayanov What happen for people failed in problem B and their rating will be positive also?
•  » » 13 months ago, # ^ |   0 unrated i believe :(
•  » » 13 months ago, # ^ |   0 For the people who are affected by checker,it will be unrated for them. For the rest of people, it's rated. It looks fair to me!! Problems were nice and for one mistake, it seems unfair to problem setters and unaffected users to make round unrated.
 » 13 months ago, # |   -19 spend 2 hours for nothing. thanks for a stupid setters.
 » 13 months ago, # | ← Rev. 2 →   -13 :)
 » 13 months ago, # |   0 Can some one tell the proof that putting 'B' and 'W' alternatively is best answer in Problem B ?
•  » » 13 months ago, # ^ |   0 Let's consider that cell (0;0) is white.Is is true that cell with coordinates (i; j) is white if and only if ((i+j)%2 == 0).From cell (i; j) you can reach the following points:(i+1; j+2),(i-1; j+2),(i+1; j-2),(i-1; j-2),(i+2; j+1),(i-2; j+1),(i+2; j-1),(i-2; j-1). Here you can observe that if you get from (i;j) to (new_i; new_j) the parity of (new_i+new_j) is always switching, so the color will switch as well.
•  » » 13 months ago, # ^ | ← Rev. 4 →   0 On a checkered chessboard, a knight only attacks squares of the opposite color of the square it is currently on. Formally, the parity of the square, $(i + j) \mod 2$, always differs between a knight and the squares it attacks. Thus placing white knights on one parity and black knights on the other will always ensure that each knight is attacking as many opposing knights as they could, as there's no chance of "wasting" attacks on own knights.
 » 13 months ago, # |   +20 There is no reason i could think that round should be unrated for all user.Mistake was unfortunate but MikeMirzayanov and PikMike found a good solution to handle this situation. Thanx
 » 13 months ago, # | ← Rev. 2 →   +2 Keep it rated, at least for people not affected by B.
 » 13 months ago, # |   +41
 » 13 months ago, # | ← Rev. 4 →   0 I can see/sense that many solutions for problem D are being hacked using the extreme inputs like queries count = 3*10^5, n = 1, and a = b = 1000000000. The solutions are getting TLE'd just because they haven't used fast I/O. I don't think this is the expectation of any Educational round problem to focus on using efficient I/O and get hacked if you don't.EDIT : I myself have now hacked around 16 solutions using the same test mentioned above, purely on the basis of inefficient I/OEDIT-2 : Have hacked 85 solutions now. Can go on and on but too tired to continue. Let the downvotes come while I sleep like a baby !
•  » » 13 months ago, # ^ |   +11 It should be a given that you should always use fast IO
•  » » » 13 months ago, # ^ |   +2 Then such a test case should be a part of the pre-tests and not be used as a hack.
•  » » » 13 months ago, # ^ |   0 This is a sad day for me but yea I definitely learned that lesson
•  » » 13 months ago, # ^ |   0 The same thing happens to me. Feel so sad.
 » 13 months ago, # |   +50
•  » » 13 months ago, # ^ |   0 you are so funny, bro
 » 13 months ago, # |   0 Why only 256mb in F? Are sparse segment-tree suppose to MLE?
•  » » 13 months ago, # ^ |   0 ML for F is too strict..
 » 13 months ago, # |   0 The first time solving problem D. 60907231
•  » » 13 months ago, # ^ |   -8 idol gio?i qua' a. =))
•  » » » 13 months ago, # ^ |   0 :v
 » 13 months ago, # |   +8 When will the editorial be published?
 » 13 months ago, # |   0 Why do we need to print inf as coordinates for F when the score is 0 ?
 » 13 months ago, # |   0 Hello admin, I have just thought of a test that some participants will get wrong, but they are still accepted in lesson A, how to fix the test
•  » » 13 months ago, # ^ |   0 PikMike please reply me !
•  » » 13 months ago, # ^ |   0 can you give me your test case??
•  » » » 13 months ago, # ^ |   0 1 3 1000 1000 48
•  » » » » 13 months ago, # ^ | ← Rev. 2 →   0 The answer is NO, My friend issued YES but he is still accepted in problem A
•  » » » » » 13 months ago, # ^ |   0 "Every integer in this multiset is a power of two." Read the statement carefully.
•  » » » » » » 13 months ago, # ^ |   0 oh, sorry
•  » » » » 13 months ago, # ^ | ← Rev. 2 →   0 you can public AC solutions link^.^
•  » » » » » 13 months ago, # ^ | ← Rev. 2 →   0 ;)
•  » » » » 13 months ago, # ^ |   +1 This statement is written in the question:- "It is guaranteed that all elements of the multiset are powers of two." 1000 and 48 are not powers of 2
•  » » » » » 13 months ago, # ^ |   0 Sorry, I did not read carefully
•  » » » 13 months ago, # ^ | ← Rev. 2 →   +1 Sorry, I did not read carefully
 » 13 months ago, # |   0 https://codeforces.com/contest/1221/submission/60867022 Can anyone mathematically explain why this is failing for C?
•  » » 13 months ago, # ^ |   0 You can try 6 6 0, your answer is 3, the best answer is 4 2 1 2 1 2 1 is your statrgy, The great statrgy is 2 1 2 1 1 2 1 2
•  » » 13 months ago, # ^ | ← Rev. 2 →   +1 Consider the scene after your n has become 0... and lets dive straight into the test where your code fails...Test : 18 13 0 Your intent : you are now wanting to subtract 2*buf from the greater number ( 18 ) and buf from the smaller number...Now you wonder how many times this subtraction you should be doing... And you answer this as : as many times so that both v[0] and v[1] remain positive... and till when this holds...? You answer this as min ( v[1]/2 , v[0] )...Mistake : You forgot that in due course of subtraction... somewhere in between... the bigger at the moment might turn up getting smaller...i.e. at the moment ; 18 > 13 ( lets call them a and b ) a > b ( Currently )But after some subtractions ; since a is reduced by a greater amount at each step ( 2*buf ) than b ( buf ) ; so inequality might transform from a > b to a < b...so you should pause at that moment ; and swap a and b again ; and then only continue the remaining steps...So your deductions look like : 18 13 -> 16 12 -> 14 11 -> 12 10 -> 10 9 -> 8 8 -> 6 7 ( and now pause and swap )7 6 -> ...So you should start thinking about the third scene also while calculating buf... and it should be min ( {a/2 , b , a — b} ) ;
 » 13 months ago, # |   0 Why did I join this game, the rate has not changed? my submission is "*name", what does * stand for?
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 '*' stands for U participated virtually. Maybe U registered after the start time of the game.
•  » » 13 months ago, # ^ | ← Rev. 2 →   0 If you initially got AC on problem B, but the verdict was changed to WA later (which I think was the case), then it's likely that you were affected by a bug in the checker, and so this contest is made unrated for you.You can read more about it here.
 » 13 months ago, # |   +1 How to solve Problem F, I did not see a comment about it, Thanks!
 » 13 months ago, # |   +4 Editorials?
 » 13 months ago, # |   +15 Please publish the editorials. Thank You.
 » 13 months ago, # | ← Rev. 3 →   0 [Deleted]
 » 13 months ago, # |   0 This contest is gone from my profile. There is no such contest and submissions for any problem of this contest in my profile. How to know the verdict of my solution for this round?
•  » » 13 months ago, # ^ |   0 If you initially got AC on problem B, but the verdict was changed to WA later (which I think was the case), then it's likely that you were affected by a bug in the checker, and so this contest is made unrated for you, which means that it's hidden from your profile.You can see the problems by clicking the checkbox "Show unofficial" at the top of the problems page in your profile.You can read more about the issue here.
•  » » » 13 months ago, # ^ |   0 Thanks!!!
 » 13 months ago, # |   0 Auto comment: topic has been updated by PikMike (previous revision, new revision, compare).
 » 13 months ago, # |   +10 the first time to get mentioned in round, very awesome ^_^
 » 13 months ago, # | ← Rev. 2 →   0