By awoo, history, 4 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 Adilbek adedalic Dalabaev, Vladimir vovuh Petrov, Ivan BledDest Androsov, Maksim Neon Mescheryakov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

UPD: Editorial is out

• +196

 » 4 months ago, # |   +13 Me to chance pupil coder again,,,,,
•  » » 4 months ago, # ^ |   -7 Hope your rating can get over 1200 after this contest! By the way, your profile photo looks interesting, so I used it as my profile photo. :)
•  » » » 4 months ago, # ^ |   +3 Thanks you for your wising,,,, By the way i need perform like your profile graph and reach 1600+ rating,,, how can i do..../?
•  » » » » 4 months ago, # ^ |   0 Maybe you can try to solve some problems that have a higher rating than you currently have in the ProblemSet？For example, try to solve problems with a rating about 1400？
•  » » » » » 4 months ago, # ^ |   +38 At first I thought you were talking to yourself . . .
•  » » » » » » 4 months ago, # ^ |   0 me too
 » 4 months ago, # |   +17 Good luck everyone !
 » 4 months ago, # |   0 Very sad that I can't participate
•  » » » 4 months ago, # ^ |   0 lol
•  » » » 4 months ago, # ^ | ← Rev. 3 →   0 hahahaha, the cleverest
•  » » » 4 months ago, # ^ |   0 lol
 » 4 months ago, # |   0 Good luck everyone!!!
 » 4 months ago, # |   0 Good luck everyone!!
 » 4 months ago, # |   +2 Chance to get to pupil....
•  » » 4 months ago, # ^ |   +1 Chance to leave pupil
 » 4 months ago, # |   +7 Sadly, overlaps with Reply Challenge.Probably, if there's lot of participants wishing take part in both, could be moved one hour earlier?
 » 4 months ago, # |   -8 Hope everyone to get higher rating in this round!
 » 4 months ago, # |   0 good luck guys
 » 4 months ago, # |   +1 Good luck everyone!
 » 4 months ago, # |   0 I like Educational round.Thanks to the organizers.
 » 4 months ago, # |   +6 contest is a kind of festival to me!
 » 4 months ago, # |   -27 Kindly note unusually usual time.
 » 4 months ago, # |   -11 Any Interactive problem?
 » 4 months ago, # |   +17 My first comment. plz I really hope to be the expert
•  » » 4 months ago, # ^ |   +1 I should be practice more
 » 4 months ago, # |   0 Could some one create a group to collect all educational contests like Div. 3 ?
•  » » 4 months ago, # ^ |   0 Use this website here you can filter contest based on contest types and many more things you can do https://clashofcodes.herokuapp.com/codeforces
•  » » » 4 months ago, # ^ |   0 Why ask to signup first? How is signing up going to benefit the user?
•  » » » » 4 months ago, # ^ |   0 I guess they made it mandatory recently as i made an account previously so i don't know about it but it is a good website can make accout there google auth.
 » 4 months ago, # |   +2 Considering my downward trend of performance in the recent Educationals, I can't help but Spoiler
 » 4 months ago, # |   0 I have a question how do we know the difference between final standings after system tests and when hacking phase is finished? from what I have seen they don't write anything to distinguish that
 » 4 months ago, # | ← Rev. 2 →   -16 C was tough one
•  » » 4 months ago, # ^ |   0 Nice joke..
 » 4 months ago, # |   +2 DistanceForces
•  » » 4 months ago, # ^ |   +20 "absolute"ly.
 » 4 months ago, # |   +1 $|$ codeforces $|$.
 » 4 months ago, # | ← Rev. 2 →   0 SpeedForces :(
 » 4 months ago, # |   -6 Why such a huge difference between problem B and C, kinda unfair for pupil / specialists
•  » » 4 months ago, # ^ | ← Rev. 7 →   +4 Actually it was not so unfair since you required just one observation for C. HintFor optimum result, it is necessary to connect the corners of both networks a and b to some computer of opposite row. ReasonIf the leftmost or rightmost connected computer of any row is broken further left or right computers respectively wont be part of combined network.
•  » » » 4 months ago, # ^ |   0 I am sorry but ... shouldn't connecting any 2 pairs do it? I mean having 2 connections between 2 distinct pairs, this way if one fails the any other is reachable.I spent about 30 minutes asking myself why the hell does the first test case need to connect 4 pairs.
•  » » » » 4 months ago, # ^ | ← Rev. 2 →   +3 No it is not necessary. ReasonIf the leftmost or rightmost connected computer of any row is broken further left or right computers respectively wont be part of combined network. Hence corners must be connected to other row.
•  » » » » » 4 months ago, # ^ | ← Rev. 2 →   0 Shouldn't they be reachable from the other functioning corner? Or you're saying connections are directed (one way)?
•  » » » » » » 4 months ago, # ^ |   +3 No they wont be. Why?Look, consider any row whose corner is not connected. Lets say it is row B and its left corner is not connected. Now, if the leftmost connected system of row B is broken than any computer left of that broken computer would not be a part of combined network anymore. Hope it helps :)
•  » » » » » » » 4 months ago, # ^ |   0 Thank you so much for your detailed responses. I get it now, senpai XD.
•  » » » 4 months ago, # ^ | ← Rev. 2 →   +2 Actually I figured out all of this, but I don't know what I am missing, maybe I'll find out tomorrow, pretty stressed out from today's contest
•  » » » » 4 months ago, # ^ |   +1 did you figure it out?
•  » » » » » 4 months ago, # ^ |   0 yeah for the X, |X and X| type of cases, I was doing a loop from 1 to n — 1, but 0 to n — 1 works, I am not sure why, because I cant see the test cases. Maybe some corner cases were getting missed
•  » » » » » » 4 months ago, # ^ |   +1 The idea is that if both a0b0 and a0bn-1 have the same distance, you might choose the wrong one i.e. if you pick a0b0, there might exist a smaller cost that requires you to pick a0bn-1
•  » » » » » » » 4 months ago, # ^ |   +1 Ahhh, you are making sense, damnit
•  » » » 4 months ago, # ^ |   +3 I think most of the people got the approach but got messed up in corner cases, including me. :(
•  » » » » 4 months ago, # ^ |   0 Yes, same for me :(
•  » » 4 months ago, # ^ |   0 welcome
 » 4 months ago, # | ← Rev. 3 →   +2 I think i know my mistake in C but not enough time damn I wasted so much time thinking about stupid possibilities that doesn't help in anythingwe could do it in three wires right ? like this 1 0 0 2 0 3 1 0 0 3 0 2i tried only 2 wires or 4 but not 3.
•  » » 4 months ago, # ^ |   0 you just need 2 wireslink 1 1 and 3 2make it a loop
•  » » » 4 months ago, # ^ |   0 That's not the values it's how it's possible to link it the ones with the same number are linked sorry it wasn't clear
•  » » » 4 months ago, # ^ |   0 no he need 3 wires. 1_1 2_2 3_3 ans=0;
 » 4 months ago, # |   +11 Experts and CMs after solving A-D: ImageBtw Im writing this comment 23 minutes before the end of the contest. I have no idea what to do with all this time on my hand:))
•  » » 4 months ago, # ^ |   +9 And some Ms, too
•  » » 4 months ago, # ^ |   0 I don't know what you are talking about...
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 ...
•  » » 4 months ago, # ^ |   +3 shake hands...
 » 4 months ago, # |   +1 Isn't problem c only about first and last index of both arrays and finding optimum values in their counterpart arrays?
•  » » 4 months ago, # ^ |   0 Yes. But you need to take into account what happens, if you connect 2 of those 4 nodes to each other.
•  » » » 4 months ago, # ^ |   0 I considered it, seems i need to see the test case to understand :(
 » 4 months ago, # |   +16 hints for problem E pls
•  » » 4 months ago, # ^ | ← Rev. 4 →   0 Dlete
•  » » 4 months ago, # ^ |   +5 Hint 1The bipartite graph consists of disjoint cycles. Each cycle contains an equal number of nodes from either part. Hint 2Say that we know the sum of $MM(G(l,r,L,R))$ for some fixed $l,r$ over every $L,R$. Let us extend $[l,r]$ to $[l,r+1]$. How does this sum change? Hint 3When we add a new node, it either extends a path or completes a cycle.
•  » » 4 months ago, # ^ |   +1 hintThink the contribution of all subsegment in a loop
 » 4 months ago, # |   +4 How to solve C and D?
•  » » 4 months ago, # ^ |   +9 For C : a[0],a[n-1],b[0],b[n-1] all must have connection with other row any element efficiently..
•  » » » 4 months ago, # ^ |   0 Lmao I can't believe I thought any 4 points work and now I see it's nowhere near correct. Thanks very much!
•  » » 4 months ago, # ^ |   0 I did BFS from the borders for D. Saving information in a map,...>
 » 4 months ago, # |   0 D can be solve by dfs.. right ? and C statement was confusing for me :(
•  » » 4 months ago, # ^ |   0 My DFS failed, needed BFS. What do you do if your DFS has back edges?
•  » » » 4 months ago, # ^ |   0 I used BFS (from neighboring points not in the given set) and still got TLE. Is this not the idea? Or is there a way to speed up my BFS? 149168850
•  » » » » 4 months ago, # ^ | ← Rev. 2 →   0 Phew, just looked at your code, it looks ok in principle, your times seem to be ~1.5-2x slower than mine. Guess you need constant optimisation here maybe? In your first loop, when looking for borders, it seems that you can push solitary points 4x into the queue. You can put a continue; there in your ifs. Also you can mark them as used then already. Yes, constant optimisation did the trick: 149174036
•  » » » » » 4 months ago, # ^ |   0 Ah, didn't think to optimize constants. That did the trick. Thanks!
•  » » 4 months ago, # ^ |   0 I passed the pretest with dp and a lot of sortings.
 » 4 months ago, # | ← Rev. 2 →   0 What was the 2nd testcase in C?? I tried all the cases can someone tell what I was missing?149157891
•  » » 4 months ago, # ^ |   0 In the last for loop of your code, you have written: aml=min(abs(a[0]-b[i]),amf); That should be: amf=min(abs(a[0]-b[i]),amf);
•  » » 4 months ago, # ^ |   0 Failing testcase: Ticket 1576
 » 4 months ago, # |   0 Is DP the correct approach for D? I tried to check the answers of my neighbours and pick the best for me but got TLE on test case 9 :(
•  » » 4 months ago, # ^ |   0 a topological sort to decide the distance of closest point
•  » » 4 months ago, # ^ |   0 You can use a DP-like idea. The intuition for me was, for a point, the closest point to [x, y] must either have the same x value (so it's on the same vertical line), or it must be on a path through [x — 1, y] or [x + 1, y].
•  » » » 4 months ago, # ^ |   0 I tried your idea. Is this way is correct to implement it? 149191779I got RE on testcase9.
•  » » » » 4 months ago, # ^ |   0 Failing testcase: Ticket 1708What's so special about the testcase? It contains all points in the interval $[1,5] \times [1, 5]$
•  » » » » » 4 months ago, # ^ |   0 Thanks
 » 4 months ago, # |   +1 How to solve D? I used dfs to transmit the nearest point but got wrong answer! :-(
•  » » 4 months ago, # ^ |   +16 I solved using multi-source BFS from all untaken points neighbouring taken points
•  » » » 4 months ago, # ^ | ← Rev. 3 →   0 I tried something similar, but because i am not pro like it took some time for me to code it well. basically what i did is pushed all the points that has an direct answer into queue. (more like pushed the outer layer of points in queue first). then i translated outer layer's answers to inner layers. but i got Wrong answer on Test 9. can you help if you find anything sussy here?Edit: no worries got accepted. it. was a minor issue :(.
•  » » » » 4 months ago, # ^ |   +3 That's unlucky. I was trying to debug for you but couldn't understand the candidates things. In principle your idea is correct, and equivalent to what I described as multi-source BFS, except starting with the outermost points rather than untaken points. Should achieve the same result.
•  » » » » » 4 months ago, # ^ |   +8 thanks for trying your code is sooo clean.
•  » » » » » » 4 months ago, # ^ |   +3 Thanks. Although actually your approach is quicker and less memory intensive as it doesn't require storing all of the neighbouring points (which can increase the size of the graph by a factor of 5).
•  » » 4 months ago, # ^ |   0 I used bfs and I got accepted
•  » » » 4 months ago, # ^ |   0 why does bfs not time out
•  » » » » 4 months ago, # ^ |   +2 200k points, at most 800k unused neighbours. Graph has max size 10^6. Why would this time out?
•  » » » » » 4 months ago, # ^ |   0 yeah it won't , i didn't think in the neighbour direction
•  » » 4 months ago, # ^ |   0 I have the same problem as you
 » 4 months ago, # |   0 What was the technique for B? I wasted so much time on A due to a stupid typo.
•  » » 4 months ago, # ^ |   0 any pair of i, j must satisfy Ai * 3 <= Aj
•  » » 4 months ago, # ^ |   0 Powers of 3 work
•  » » 4 months ago, # ^ |   0 Try writing the inequality that needs to be satisfied based on the pair you pick.
•  » » 4 months ago, # ^ |   +1 Suppose you'll do the operation on $x$ and $mx$ which are present in the array where $m \geq 1$. After the operation, $x$ and $mx$ will be replaced by $mx - x$, we want the change of the sum to be greater or equal to the previous sum, which implies $2(mx - x) \geq x + mx$ $\implies m \geq 3$So, you build the array like this: $[1, 3, 9, 27, ...]$ until the array size is $n$ or the value gets higher than $10^9$. For the later case the output is $NO$.
 » 4 months ago, # |   0 How to solve C? I tried with all 3 cases, but WA on tc2 149150730
•  » » 4 months ago, # ^ |   0 try link left and right end of A and B. either use optimal point in opposite array, or link two end points.
•  » » 4 months ago, # ^ |   +1 Failing testcase: Ticket 1579
 » 4 months ago, # |   +1 In C my idea was that computers at end must be connected to some other computer, so we must find a way to minimize this. But for whole 1 hour I was not able to implement due to edge cases of computer at ends being connected to each other. I knew edge cases(like first computer of array A can be connected to first computer of B or last computer of B or both first and last computer of B etc.) but was not able to implement and deal with those.Any tips on how to improve this ability? Sometimes I break down a problem in different edge cases and get overwhelmed by that.
 » 4 months ago, # |   +6 it really was an intersting contest i have solved C with topology :" which i have studied in CS
•  » » 4 months ago, # ^ |   +8 Same! Lmao maybe that's why they ask CS fundamentals in interviews
 » 4 months ago, # | ← Rev. 3 →   0 My approach for Cs1=abs(a[0]-b[0])+abs(a[n-1]+b[n-1])s2=abs(a[n-1]-b[0])+abs(a[0]-b[n-1])s3= For a[0], find the element in b such that difference is minimum using binary search and mark the index for that element. Similarly for a[n-1]. Now if both b[0] and b[n-1] are marked then the sum is already calculated, else if any of the b[0] or b[n-1] is unmarked, find the optimum element in a and take the sum.ans=min(s1,s2,s3)Still WA on test 2 :( 149166751
•  » » 4 months ago, # ^ |   +4 I calculated all 7 cases.  // Case 1: a1 to b1 and an to bn; ans = min(ans, a1_b1 + an_bn); // Case 2: a1 to b1, min bn, min an ans = min(ans, a1_b1 + a_n + b_n); // Case 3: a1min + b1min + bnmin + anmin ans = min(ans, a_1 + a_n + b_1 + b_n); // Case 4: a1min + b1an + bnmin ans = min(ans, a_1 + b1_an + b_n); // Case 5: a1 to min, b_1 to min, an bn ans = min(ans, a_1 + b_1 + an_bn); // Case 6: a1 to bn, an to b1 ans = min(ans, a1_bn + b_1 + a_n); // Case 7: a1 to bn and b1 to an ans = min(ans, a1_bn + b1_an); 
•  » » 4 months ago, # ^ | ← Rev. 2 →   +1 [Updated] Failing testcase: Ticket 1699
 » 4 months ago, # |   +1 How to solve D ? I got TLE on test 9.
•  » » 4 months ago, # ^ |   +1 I think it's BFS + Dijkstra. If a point in the input is next to any empty point, then the answer is either 1 or 2. The other points must border at least some point in the input. From there you just BFS + Dijkstra to get the point with the least distance to empty point and populate the neighboors
 » 4 months ago, # |   0 Can anybody help me figure out what‘s wrong with my code149129168 in problem B？ it runs well on my computer but kept failing the preset when I submit, I was literally driven crazy during the contest (╥_╥)
•  » » 4 months ago, # ^ |   +4 move "ios::sync_with_stdio(false);" to the very beginning of the main function.maybe you can try to use CUSTOM TEST on Codeforces platform when you meet questions like this.
•  » » » 4 months ago, # ^ |   0 oh got it! Thanks for your reply! I will try it next time!
•  » » 4 months ago, # ^ |   0 Put ios::sync_with_stdio(false); before reading the inputs, not after.
•  » » » 4 months ago, # ^ |   0 thanks a lot! what a silly mistake of mine！
•  » » » 4 months ago, # ^ |   -8 what happens if we put after, will it cause undefined behaviour
•  » » 4 months ago, # ^ |   0 In these cases, try to use custom invocation for testing
 » 4 months ago, # |   0 I got WA in C because of a typo :)
 » 4 months ago, # | ← Rev. 2 →   +9 In problem D, I used a similar idea as This Atcoder Problem. For each point, i binary searched for the shortest distance and used merge-sort tree to check the number of points that has smaller distance than a certain number with that point. I noticed that the distance should be smaller than sqrt(n), so after finding the distance i brute forced through all possible points for the answer.I think The final complexity is O(nlog^3 + n*sqrt(n)*log) but got TLE on test 9. This will be my 6th consecutive rating drop.(╥_╥)
 » 4 months ago, # |   0 my submission for D giving MLE... If anyone have experienced similar problem, please help me
•  » » 4 months ago, # ^ |   0 I got MLE because of stack overflow. Forgot to maintain a visited hash table when recursing.
 » 4 months ago, # |   0 How to solve D?
•  » » 4 months ago, # ^ |   0 I used bfs，put the unused neighbours in the queue，and calc the answer when you meet a node which is not among the given $n$ points.
 » 4 months ago, # | ← Rev. 2 →   0 Is $O(Nsqrt(NlgN))$ intended for problem F? I know that we can do pure $Nsqrt(N)$ if we solve for each block for each monster instead of for each monster for each block, but still, it sounds like a lot of work :(
•  » » 4 months ago, # ^ |   0 Model solution is $O(N \log N)$, but we decided to allow slower solutions with good implementation pass (although $O(N \sqrt{N \log N})$ sounds kinda scary)
•  » » » 4 months ago, # ^ |   0 Apparently even $O(N \sqrt{N} \log N)$ can pass :)
•  » » » » 4 months ago, # ^ |   0 Well, the time limits were kinda generous in this problem (the model solution runs in something like 1s on 32-bit C++ compiler). Although 3931 ms out of 4000 sounds very scary
•  » » » 4 months ago, # ^ | ← Rev. 2 →   0 Not at all mine runs in 1.5 seconds which is kind of okay also it's really easy to write and you can argue that if written with prefix sum the operations would be simple enough that you wouldn't be able to differentiate between $O(NlgN)$ and $O(NsqrtN)$ "unless you go full Chinese on the constraints/test data ig".
•  » » 4 months ago, # ^ | ← Rev. 2 →   0 I changed to unordered_map, so $O(N \sqrt N)$, and it passed, although 3899 ms. D:
 » 4 months ago, # |   +34 If you are/were getting a WA/RE verdict on problems from this contest, you can get the smallest possible counter example for your submission on cfstress.com. To do that, click on the relevant problem's link below, add your submission ID, and edit the table to increase/decrease the constraints. If you are not able to find a counter example even after changing the parameters, reply to this thread, mentioning the contest_id, problem_index and submission_id.
•  » » 4 months ago, # ^ |   0 Thank you so much! This is amazing!
 » 4 months ago, # |   +3 is cf_predictor gives correct predicted values ? i think that there must much increasing points in rating than i see in standings.
•  » » 4 months ago, # ^ |   0 Yes, I have found the carrot chrome extension to be more accurate.
 » 4 months ago, # |   +6 Screw C. Took more than an hr to solve it, and 40 minutes to finish D but not time to submit it
 » 4 months ago, # | ← Rev. 2 →   0 My approach for D:Rotate coordinate space by 45 degree by transforming (x,y) with (x+y, x-y). Now, for a point X, observe that all 4 points at distance 1 form a square, then next 8 points at distance 2 form another square and so on. Each square will have 4D points here D is distance from X.Since N is 2*10^5, D can be at most 350. So for each point, I checked number of input points in squares of length 1, 2...350. As soon as I get a square of length d which has less than 4d points, we need to search in this square.Then its simply searching all points at distance d and returning any one excluded point.Complexity: O( N*Sqrt(N)*log(N) ) Solution: 149223014But solution TLE'd on test case 30. Would appreciate if someone can take a look at my code and suggest scope for optimization.
•  » » 4 months ago, # ^ |   0 How did you check the number of points in a square?
•  » » 4 months ago, # ^ |   +1 I got TLE on test 30 149175151. I doubt the time complexity but also can't imagine what does the worst case look like.
•  » » » 4 months ago, # ^ |   0 Yeah, I think theoretically the number of operations in the worst case would be around 5.6*10^9 [(2*10^5)*(350)*(20)*(4)]
•  » » » » 4 months ago, # ^ | ← Rev. 2 →   0 I was trying to use openmp's parallel functionality, does cf supports it? i am not getting any error nor can i observe any significate difference in execution time.without pragma 149232334with pragma 149233499
•  » » 4 months ago, # ^ |   0 changed your unordered set to set, because i guess performance of hash structure is complicated to guess.can you find why it is printing an already existing point?
•  » » » 4 months ago, # ^ |   0 Thanks. Fixed a silly bug. Updated the comment.
•  » » » » 4 months ago, # ^ |   +3 You can eliminate use of set, you already have list of sorted point with respect to both dimensions.
•  » » 4 months ago, # ^ |   0 I have a solution which barely passes and it was possible thanks to IceKnight1093 optimizing my previous code by omitting pairs.It uses pragmas and shit so it's totally not safe for a contest submission.Perhaps someone else can bother to optimize it further.
•  » » 4 months ago, # ^ |   0 I solved D considering/using that answer can never be greater than 700 . I made pair of interval for each coordinate of x from 0 to 2*10^5+1 which doesn't have a y coordinate . Then for point {x_i , y_i} it's just binary search the left nearest and right nearest value of y for each z s.t. ( z>=x-700 and z<=x+700) It's ACs in approx 1 sec Solution
 » 4 months ago, # | ← Rev. 2 →   +8 My ideas for D (not AC on contest) : Pick a random unvisited point name it $p$ and mark this point as visited Do floodfill to another points adjacent from $p$ and mark them all as visited (adjacent : manhattan distance is equal to 1) So now from all the floodfilled points we form a connected component from $p$. Pick border points from this connected component (a point is called border if there's at least 1 adjacent point which doesn't exist in the input) From all border points, we can get their respective "source" (i.e. the neighboring point which is not in the input) Do multisource BFS from all border points and now you can get source for all non-border points by setting it to the source of their nearest border points My implementation 149198396I love the idea of the solution! Didn't quite like C (because caseworks), but this problem has a beautiful solution
 » 4 months ago, # |   0 chance to get specialist...
 » 4 months ago, # |   0 I want to know how people know this formula for problem B : (b — a) * 2 >= b + a
•  » » 4 months ago, # ^ |   0 We need the next sum to be no less than the previous one. If we change a and b to |a — b| then their sum will be equal to |a — b| + |a — b|. We can expand the module and get (a — b) *2. And by condition, we want this sum to be at least a + b. => 2 * (a — b) >= a + b
 » 4 months ago, # |   +3 when are the ratings going to update as it's past 12 hr already?
 » 4 months ago, # |   0 when editorial will come.
 » 4 months ago, # |   -6 When will the rating be changed?
•  » » 4 months ago, # ^ |   -12 some problems in the platform maybe? Even system check after hacking phase is pending.
•  » » 4 months ago, # ^ |   -8 Same question..
•  » » 4 months ago, # ^ |   -9 same doubt its showing in unrated constest for now
•  » » » 4 months ago, # ^ |   0 System testing just started
•  » » » » 4 months ago, # ^ |   0 How did you know that?
•  » » » » » 4 months ago, # ^ |   0 same question
•  » » » » » » 4 months ago, # ^ |   0 same question
•  » » » » » 4 months ago, # ^ |   0 Saw the standings and system testing was going on.
•  » » 4 months ago, # ^ |   +30 You did not participate in the contest, why do you care about it? This kind of question has been asked a ton of times for every Educational round.
 » 4 months ago, # |   +7 The rating is still not updated......
•  » » 4 months ago, # ^ |   0 yes
 » 4 months ago, # |   0 Auto comment: topic has been updated by awoo (previous revision, new revision, compare).
 » 4 months ago, # |   0 Hey does anyone know why I didn’t get rated yet is educational round ?
•  » » 4 months ago, # ^ |   +1 No one got rated yet. The editorial has just been released so I would assume ratings will be updated soon too.
•  » » » 4 months ago, # ^ |   0 The contest shows up as unrated in my rating graph. Maybe this one would go unrated.
 » 4 months ago, # | ← Rev. 3 →   -7 Good luck
 » 4 months ago, # |   0 Me here reloading the website every 10 seconds for +2 delta :/
 » 4 months ago, # |   0 MikeMirzayanov I received Plagiarism on Problem A of this round with some guy I don't know . Logic for problem A was one liner how can i get plagiarism on this problem . please check again . Its my request to you.
 » 4 months ago, # |   0 In yesterday's competition, my code for question B was judged to be duplicate with others, but my code was submitted earlier, so there is no possibility of plagiarism at all. Plus, I want to score 2100 for this game, so there's no way I'm giving away my code to anyone else. Moreover, the solution shown in my code is a very general solution, and it is very easy to repeat it in a small amount of code. I don't think it's reasonable. Please check it carefully!
•  » » 4 months ago, # ^ |   0 MikeMirzayanov，I could have been a master, but now I'm an expert. Please check it! Thank you!
•  » » 4 months ago, # ^ |   0 same with me bro i got plagiarism on A. how can it be possible problem A was one liner
•  » » » 4 months ago, # ^ |   0 I hope for a reasonable explanation. This accident reduced my rating by 300 :(
•  » » » » 4 months ago, # ^ |   0 I can understand bro but I don't think our issues will be responded
•  » » » » » 4 months ago, # ^ |   0 Thank you for your comfort. I hope there will be no similar problems in the future. Good luck.
•  » » » » » » 4 months ago, # ^ |   0 Good luck to you too bro :)
 » 4 months ago, # | ← Rev. 3 →   0 .Finally rating changes after waiting hours..
•  » » 4 months ago, # ^ |   0 Yeah
 » 4 months ago, # | ← Rev. 2 →   0 I had a coincidence of my solution(149124483) matching with solution(149099666) of Performanceartist for the problem 1651B . clearly it is a coincidence as he is a candidate master and i a newbie with no prior communication between us. He submitted his solution much before me. please look into it.Your text to link here...he(performanceartist) has commented for the same problem.
•  » » 4 months ago, # ^ |   0 A sad coincidence, isn't it? Haha. We need to work hard in the coming contests bro :)
 » 4 months ago, # |   -6 downvote me plsss
 » 4 months ago, # |   +5 The worst C problem I have ever solved.
 » 4 months ago, # |   0 This contest Is Very good for thinking