Ciao, Codeforces!
I am glad to offer you to participate in Codeforces Round 1002 (Div. 2), which starts at Feb/02/2025 17:35 (Moscow time).
The tasks were offered by me and Artem shfs. We hope you will find them interesting.
This round will be rated for all participants, with rating below 2100. Participants with higher rating can take part, but the round will be unrated for them.
You will be offered 5 tasks and 2 hours to solve them. The last task will be split in 2 parts.
I want to thank:
- shfs For help with creating tasks.
- Artyom123 for elegant coordination.
- our testers makrav, imirdy, IzhitskiyTimofey, sadness, djm03178, ArSarapkin, max0000561, Vash_nick, AndreyPavlov, 127.0.0.1, Solove, nichke, dope, etian6795, umimel, daniel071292, _DuMoH4uK_, LittleTiredShpinatik_TwT, CodeCraftsPhoenix, marzipan, commitery, jj_d.
- MikeMirzayanov for Codeforces и Polygon.
We wish you good luch and rating as big as the bottle in the photo!
Point distribution: 500−1000−1500−2000−(1500−1500).
UPD: Editorial
Congratulations to the winners:
Rated only:
Auto comment: topic has been updated by yunetive29 (previous revision, new revision, compare).
Wow!! 2nd author looks like water bottle.
he took "staying hydrated" to another level
speedforces?
The round turned out to be really cool, hope you'll enjoy)
Tester.
I am missing interactive problems hope there is a one in this contest
True , we do miss it
it looks like
Ciallo, Codeforces!Div. 2 with 5 problems? Not so common.
Ciallo～(∠・ω< )⌒☆
Ciallo~
Ciallo～(∠・ω< )⌒☆
Ciallo～(∠・ω< )⌒☆
Ciallo～(∠・ω< )⌒☆
Ciallo～(∠・ω< )⌒☆
Ciallo~
Yuzu chef everywhere.
As a participant, i will try to empty the water bottle :)
Hope you can success before geting a stomach like a ball.
As a tester, I have a proof that upvoting this comment will lead to positive delta. But the proof is too long to fit the margin.
Why is the water bottle so massive?
This is the smallest it can get, the bottle is on the horizon!!!
low taper fade
you know what else is massive?
No.
Dekubopdomdem dekubopdomdem
It is going to be a speed force, based on the rating distribution
That's a lot of offers!
I understand the logic, but why is this bottle so big? I've never seen a bottle this big. :P
道理我都懂，可是这个瓶子为啥这么大？我从未见过这么大的瓶子。:P
Italian announcement. It will be a fantastic round without a doubt. Ciao!
Ayooo tourist alt account?
Actually tourist is my alt account
We wish you good luch and rating as big as the bottle in the photo! *luck
I'm a simple man. I see Evelyn, I upvote.
where's the pfp from?
A custom character created in Baldur's Gate 3
Would it be advisable to attempt problem E before problem D according to this score distribution?
That's not how it works... E as complete is 3000 and divided into subtasks E1 and E2; that's why individually less than D!! Anyways, me struggling on B itself :(
E1 will be easier than D and have about the same difficulty with C, but E2 will be harder than D (in problems with subtasks, you need to add the scores of the easy and hard version to find the difficulty of the hard version compared to other problems)
Note that score != problem rating and that the distribution isn't always 100% accurate
Personally, I'll probably read D, and, if I don't have any ideas, I'll read E1, then decide which one to solve first (if I actually solve the first 3 problems :P)
It seems that C is more difficult than usual.
you're a legend bro all the respect for you but give me some tips if you see my submissions i have solved 410 problems something like that but i cannot be consistent because the school the exams are destroying me i cannot practice all the time just weekends
No,it's not sloved by myself,i just tested the Deepseek by asking it to slove the problem.
I can't do this by myslef,if i can,i will not be a Newbie.
Btw,Deepseek is great.(i have not used it to slove problems in any contest)
And, Never!!
what is "We wish you good luch"? seems to be a typo
Is shfs a water bottle?
No, a bottle can't make problems, LOL :)
cant wait to see how many (will avoid explicitly stating a group of people) will use o3 to win the div2 contest
Hope to tackle C
use o3
hope you get a positive delta
what is your ages guys lets see who's the youngest
I am 12. Nice to meet you!
As a tester, I have a proof that upvoting this comment will lead to positive delta. But the proof is too long to fit the margin.
i almost like the refrence to fermats last theorem
This bottle is so big, interesting! good luck!
wait, why E1 and E2 have same score as C?
ahh yes today i compete against deepseek R1.
I am trying to register but their is no button for register
I think regiter is closed
hj
a speechless round. we need special attention.
Why is the hack not accessible?
wtf, how D got over 2k accepted???
cuz it's not hard
no. it was solved by gpt in the first prompt not because people actually did it.
that's sad
YES, it is not ad-hoc so GPT can figure it out !!
It's a slightly modified Dijkstra Algorithm!! Would you believe I solved D but not C !!
dude you copy pasted gpt code. lol. Just like i did.
Hehe dumboo.. GPT never writes such codes..
Edit: Seeing your codes and that too in multiple programming languages ensures that either you are a pro coder or just a cheater who can write such well commented codes in the contests and that gives you no audacity to blame someone else !!
hey dumbass. i got the exact same code as you from chatgpt. You just removed comments. u r not fooling anyone. lol.
Dumbass, I always use multiset (easier to remove any element) !! I doubt LLM does that.. Anyways I don't owe you any explanation..
Prompt to GPT code on Dijkstra's Algorithm : The Link is not working though
Now, keep comparing the codes and crying over and over !! This platform is just for practice and fun; no one is here to make a statement by rating boosts !!
dude just stop. No one is buying ur bs.
you forgot to remove one remaining ChatGpt comment.
The code I have provided is provided by ChatGPT, not the one authored by me !!
@dummy16661 on CF using chatgpt in contests is not allowed. Dont do it next contest please
It's not that difficult, to be honest. You only need to realize that you will get an answer if both graphs have at least 1 common edge.
The other part where you find the shortest cost to reach a pair of nodes is quite trivial graph problem.
Also, this is the type of problem where AI shines.
Modeling this graph, and actually looking that $$$m$$$ and $$$n$$$ allows you to do that, is not the stuff, that allows 2k ACs
Obviously people cheated. Its nothing new.
I have seen more surprising ones compared to this.
Anyways, This problem shouldn't be in a CF Div 2 round. Maybe more suited to Div 3 or Div 4.
To me this problem is at best 1700-1800 graph problem.
"You only need to realize that you will get an answer if both graphs have at least 1 common edge." — It's not correct criteria, actually. You also need to check if graph is bipartite or not. Because there are cases, when answer would be inf if 2 graphs are bipartite. (Both graphs have single edge (1, 2), s1 = 1, s2 = 2). It is another interesting and easy task about this model. But it can be solved in time O(n + m)
Yeah my bad, forgot about it. But these observations aren't that difficult for a div 2 D.
After reading the problem, if you think about how to solve this problem in the most brute force way. You will realize soon that you only need to find a common edge such that you can traverse in the same direction at the same time. Also the whole thought process to solve this problem is quite linear.
seems plenty of AI generated codes left untouched even after plagiarism check.
some of the solutions that I think are clearly AI generated: https://codeforces.me/contest/2059/submission/304082660 https://codeforces.me/contest/2059/submission/304091441
speedforces
Today, a div2B send me straight to pupils again... damn why I can't even solve B problem properly?
If it makes you feel better, I think B was harder than C and D. Although I got B, I completely guessed it out of desperation. I was even a bit surprised I got AC. For C and D I was 100% sure it would pass when I got AC
For now, the next div. 4 is what makes me feel better (I hope)
Today I upsolved C. And it's indeed easy one (compared to B).
While D is still beyond me, I dunno how people called it standard tho. (I do know dijkstra, but the transition is not really clear to me)
No way D got 2k solves
Here is Why
Please. Somebody explain to me how the 4th test case in problem C is 3? I spent an hour and a half staring at it and could not figure it out
you might read the i and j wrong. I made the same mistake.
Omg... it doesn't help that all the other examples work with i and j flipped the other way.
Switch cols and rows, then it works!
Actually, I needed like an hour to figure this out. One could say, ok, beginner error. But given that the statements (also of A and B) where kind of hard to read, here in C we are presented with the first three testcases that work with switched col/row also.
That is bad statement text.
Serve queues $$$1,4, 2, 3$$$ in that order
Saaaaaame. Was staring at this forever and never figured it out. The fact that the 3 explained examples are all the same when transposed is quite bad.
B got me like 15 minutes, while C took me around 1 minute to figure out the answer. B >> C
Could not figure out C, I was thinking of some dp approach. How did you solve it?
To maximize the MEX, we need to have $$$0,1,2,..., x$$$ with x is large as possible, that's the whole idea. In fact with any sort of MEX problem, I always think about it at the first thought and it works mostly all the time.
To get $$$0$$$, just use the operation on any queue at the $$$n^{th}$$$ moment. To get $$$1$$$, use operation on the queue with the last element is $$$1$$$ at the $$$n - 1^{th}$$$ moment. To get $$$2$$$, use operation on the queue with the 2 last elements are $$$1$$$ at the $$$n - 2^{th}$$$ moment... and so on.
Primary observation is that
a[i][j] >= 1. Greatly simplifies the problem and allows you to make the observations that DuongNeverAlone made.Yeah dude, I reduced the problem to “place rooks in a n by n grid” and was like “how the hell can I solve this without using bitmask??”. Then I noticed that and completely changed strategy.
How to solve B ? Was stuck on pretest 2 with 3 tries.
Hint: split the problem into 2 case: k==n and k < n. The answer is very simple
Is there a faster approach than O(n^2)? I thought it would be TLE if it was O(n^2), but mine got AC with a pretty solid time.
Rofl just O(n) what are you talking about?
If $$$n = k$$$, calculate the answer naively, otherwise the answer can be either $$$1$$$ or $$$2$$$. Calculate it with this piece of code:
what if n=7, k=6 array a : 1 1 1 2 2 3 3 then the answer would be 4
[1], [1, 1], [2], [2], [3], [3]. so the answer is 2.
Nope, $$$1 | 1 1 | 2 | 2 | 3 | 3$$$, $$$b$$$ = $$$1 1 2 3 0$$$, answer is 2
I couldn't understand the statement of problem D until the very end. Why explain two degenerate (test) cases and not the one where you have a finite non-zero value? Reading comprehension got me I guess.
Now I see, I've missed part of the stopping condition in my solution. :sadge:
Wtf is C? How to solve C? 4k solves, is it that easy?
There's only one way to get required mex. There's a pattern here. Pattern exists coz he can only serve one row at a time. Take a look at samples again carefully, big hint hidden in them. Or maybe try to construct mex in reverse direction.
this contest is so trash B is shit C is weird D is free bad contest
how did u guys solve C ?? I tried using suffix sum but couldn't solve
same i got the idea of row suffixes but cant able to determine how to avoid overcounting of suffixes from same row ,how to select the better suffix to include from that row.
Yes I tried to maintain that in a priority set but it was a guess so couldnt get accepted
Anyone solved Problem C with prefix sums and DFS?
I thought of that, but the complexity would be too high.
I used suffix sum for each queue and check if the sum of values equal to the index
then sort them according to how many checks are correct
finally, calculate the $$$MEX$$$ (if the current value of correct checks is greater than or equal to my $$$MEX$$$, increase it
Interesting, I sorted the suffix sum arrays in reverse lexicographic order and then did greedy.
It actually simplifies to only counting longest suffix of all 1s in each queue.
304097874
why these values does not matter such as last 2 values are 2 and 1 so this should contribute to suffix 3 as there might be case when there does not exist any other row with suffix value 3?
Intuition comes when you actually begin to maximize the MEX.
Observe the events from last round to first.
In last round, we emptied a queue. So 0 would surely be available for our MEX.
Now we need 1, observe that in each round, each queue increments by atleast 1.
So the queue that gives us the 1 must look like [.... 1], but since all elements are greater than 0, we must empty the queue before last 1.
Now we need 2, again an observation here is we used our last 2 operations to get 0 and 1 to our MEX, then the queue which gives us the 2 must look like: [.... 1 1] , there is no other way. And hence we would use an operation here to clear the elements before and get 2 to out MEX.
Similarly, the queue which gives us 3 will look like: [....1 1 1] and so on.
Rest is implementaion.
for problem $$$B$$$, is the answer always equals to either 1 or 2 ? (except for the case of $$$n = k$$$, you have to check it manually and it would be higher than 1 or 2)
My solution was brute force and it passed, that is why I though the answer will be either 1 or 2
Yes.
It's always 1 or 2 if n!=k.
How is the answer to this sample test case of B equal to 1?
Can't we divide array into
{1}, {1, 1000000000}, {2}, {2}so that arraybbecomes{1, 1000000000, 2, 0}and hence firstindex != iis2(a2 != 2). Output should be 2, not 1?{1, 1}, {1000000000}, {2}, {2}
We can split into {1, 1}, {1000000000}, {2}, {2}. Now array b becomes {1000000000, 2} and hence first index != i is 1.
one can devide in ~~~~~ {1,1}, {10000000}, ... ~~~~~
cooked
D is so old :(
Did anybody think of doing C with O(n^3) DFS with suffix sums ?
Since t*n*n <= 2e5
So , t*n*n*n <= 6e7
seems acceptable to me, but i could not complete that code correctly.
But why? Observe that a[i][j] >= 1. So, the longest suffix of 1s is all that matters.
"the longest suffix of 1s is all that matters."Well, I still do not get it. Can you please elaborate?
The problem can be seen as picking suffixes from lines, since that's equivalent to picking prefixes to erase. You'd never choose to pick a suffix that contains an element >1, since that'd mean that you wouldn't get value matching that suffix length in the final xs list. And that would break the mex chain. Just try making up an example where you'd pick >1 element and you'll quickly see it's impossible.
https://codeforces.me/blog/entry/139012?#comment-1243218
at first i also thought n^3 could pass but at the end i choose n^2*logn
did you do DFS?
no, binary search
I tried, got TLEed.
yes i saw your code. but i reckon it is possible. we just don't know how to code it.
Yeah, I hope so
I've not really enjoyed the contest, but at least the problem statements were quite clear.
Overall, not the worst contest but it could've been much better in terms of balance if D was appropriate.
Completely agree for D. It's also sad that random grays can type so much faster than lgms. My typing speed is much slower that o3 and r1, unfortunately.
Yeah, I hope plag check will find some of them.
I've just checked some known cheaters that I added to my friend list. Their solutions all look very similar.
LMAO, 6th place seems very suspicious. The infamous D problem.
Look at submissions by harsh__h and the spaces between keywords like "for" and parenthesis. In D he has them, whereas in other problems he doesn't have the spaces in between.
No accusations, just my personal opinion about specific submission. But if you cheated, it will be hilarious.
Bro I solved E1, E2 too T_T
You used chatgpt to save time. Which is still cheating.
Don't know man, i didn't use it. Any proof is welcome. I don't even know if gpt can solve this question, any reference will be helpful.
LMAO, "I don't even know if gpt can solve this question,".
Are you serious? The most streamlined problem I've seen for a while.
Also, the fact that you ignore the questions about codestyle and still ask for proof, only exarcebates the situation.
Now, I am sure that you cheated and that's a shame.
If he indeed used chatgpt, then it will be another case when relatively high ranked indian problemsetter is found cheating. If you remember Psychotic_D
I thought only low ranked people would be affected by LLM, as they are usually not able to solve past B; but damn, even high ranked people? This is extremely demotivating. So sad I did not reach Master before it got to this point ;( now it seems much harder to achieve and even if I do it won’t be the same cool achievement if I get placed in the same group as a lot of cheaters. Strongly considering just retiring; but let’s do this for another 1/2 months see where I can get :\
It is sad but I don't think that you should be upset about the cheaters to the point of retiring.
Say, if it is a situation about a job, and in this situation the high rank matters, then probably a cheater-high ranked will be exposed easily during the interview. (maybe just my copium, though)
So, my point is that if the environment can distinguish high ranked people, then in this environment cheaters will be exposed. And if the environment does not distinguish, then why should I even care about it.
You has AC only in E1, which is easier than E2.
Also, mind explaining why you have such different codestyle across submissions? In some you use spaces betwee brackets, in some you don't. In some you use endl, in some '\n'. Very suspicious.
Submited E2 just a small modification in the AC submission.
Used '\n' in E1, E2 too ig. Am just saying don't ruin my name without some solid proof, its sad seeing this after doing CP for 4-5 years.
I've not accused you of anything in the beginning. I just noted that the codestyle was suspicious. But you started avoiding the discussion and asked for "some solid proof" (proof of what? I didn't accused you at that point).
Bro, you are ruining your name yourself by cheating and making attempts to cover it.
The gap between him submitting B and D is just 3 minutes. He for sure didnt type all that long ass code for D in just 3 minutes.
gpt killed both c and d in just 2 minutes. Hence the high submission rate.
Any hints for C? i was thinking of storing all the suffix sum in min heap, but could not solve it further...
Think how to get the maximum mex.
a[i][j] >= 1The worst div2 I have ever seen in this platform.
Coming from a Grandmaster, why you feels so?
D is too easy and gpt-able and E2 is meaningless.
I respectfully do not agree with you. For me problem D was challenging and I tried to solve it this round but I could not come up with an idea. I usually solve Ds though and even right now I am still not sure the ideas I have in my mind work. maybe for a grandmaster it was trivial but for me it was not.
I suppose you haven't ever participated in tracker rounds?
anuj23334 leaked the solution during the contest on Youtube YT, please ban him MikeMirzayanov, yunetive29
Not sure if I understood E correctly. Because it seems so simple.
If we see all the i arrays as one long array, then we have to go from left to right, and foreach a[i]!=b[i], insert the b[i] at that position in a[i].
So it looks like a two pointer like construct. Also E2 is simple then.
What did I miss?
You cannot insert elements in arbitrary positions. only at the beginning of some row
ah, ok. Thanks.
history.
The worst div2 I have ever seen in this platform.
Does anyone have a "nice" solution for B? Apart from my horrible implementation, in retrospect it still feels harder than problem D. Maybe I'm just missing something.
first element has to be in first subarray, then after that the only case when we're restricted by the test case is when n = k, because then we have to choose one element for each subarray. in that case we can easily find the answer. but if k < n then we can always make the 2nd subarray (first subarray of b) have a size > 2. so now our answer depends on that. if all the elements from i=2 to i=n-(k-2) (leaving last k-2 elements for the rest of the k-2 subarrays as 1st subarray was element 1 and 2nd is going to be within this range) are 1 then our answer will be 2 otherwise it will be 1
That is a lot easier than what I've done, thanks!
If n=k, go manual.
Else, answer is either 1 or 2.
First try to get 1. (Just try to find an element greater than 1 which is not the first element and if you're able to make K subarrays)
If not answer is surely 2. To prove this only consider the 2nd and 3rd element of array. If both are equal to 1, you can choose [1,1] (since n > k, atleast one subarray can have greater than one element) as your second sub-array to get 2 as your answer. And actually there is no other case since if it was not the case, you would have found 1 as your answer already.
The problem ordering for B and C in this round is not very reasonable; perhaps C should come before B.
Not really, I think we're just bad. (Even in my case, if I'm not choke myself in B, I may have the chance to solve til D)
I agree, I do think C is easier to see than B. That being said, they were both very simple. Personally, I'd be more concerned about grays solving D in < 15min. It's so blatant, just look at this guys submissions: https://codeforces.me/profile/abhi23100.
Gave up on C++ and started writing GNUC11
A gray player spent 6 minutes solving Problem D, and the code is 160 lines long. WT????
They are obviously just really really fast at typing. And they just happen to be more skilled at speed coding dijkstras in C than C++.
And he coded all that in 6 min :D. Codeforces is dying
It's the GNUC11 that's killing me. Actually, so funny. Can't wait to see this dude's Oscars speech on linkedin for this phenomenal performance.
It's already dead. I just do virtuals of old contests now. No point in competing against LLMs.
Did some silly things in C. Tried to calculate suffix sum and transform the problem into picking one number in $$$[1, x]$$$ per row and per column, and check if we can pick $$$x$$$ numbers. Initially I thought this is a simple Bipartite Matching, but it failed because there are actually three constraints(row, column, value each can appear at most once)
I know this is not the intended solution for C, but I'm still curious about is this triplet(?) flow problem solvable?
I'm curious about this as well, if this is some kind of max flow matching ? I came up with this same idea as well and wrote a brute force DFS which got TLE ofc :(
I did same as you and reduced the problem to place n rooks in a nxn grid. Unfortunately I am pretty confident that using this strategy the problem becomes NP complete, from other “place rook in the chessboard” problems I have seen
As a person who submitted matching without spotting that it can be further optimized, triplet problem is probably unsolvable, but I've spotted a fact that number $$$1$$$ can only appear if some queue has $$$1$$$ in the end, so we already know that on the pre-last day some queue will be cleared. Then, $$$2$$$ can only appear on pre-last day, meaning there need to be two $$$1$$$'s at the end and so on. So we only need edges from number $$$x$$$ to row $$$i$$$, if the last $$$x$$$ elements of $$$i$$$ are $$$1$$$.
In order to solve the problem now in $$$O(n^3)$$$, we run dfs from vertices, that correspond to numbers, in increasing order, and if we didn't find the augmenting path, current number is the answer
I've also failed to notice intended logic for C so I solved the way you described: 304114466
Hungarian (adapted from e-maxx) does exactly what we need:
-v[0]stores current answer forirows. So I just transposed the matrix and overkilled D2C with hungarianIf we have to maximize the score in B what can be possible solution ? Unable to think of one
I think the following solution may work, find a way to mark the indices in the following order, let the index i is marked and the element at index i.e., (v[i] is x), so we have to get index j such that (v[j] = x + 1) and (i + 1 < j <= n) and j should be minimum among all such (v[j] = x + 1), so after getting all these indices, we can say that we marked the indices of elements from [1..p](also ensure index of 1 is > 1) and there are gaps between each pair of elements, so that we can take those subarrays at even positions to get sequence (1, 2, .., p) (assume that indices stored in vector A). Now, all we have to wish is that the condition over k should be satisfied. So, to ensure that binary search on the marked indices, to see that if a answer exists from [1..mid] as the number of subarrays that will be formed here is (2 * mid + n — A[mid])(we can see it is monotonic as A[mid] — A[mid — 1] >= 2). So, see if answer for the current mid exists or not and shift accordingly (if the no of (subarrays >= k), we can select at least mid number of elements).
sorry, but, where if solution?
are Russian have no access to ChatGPT due to sanction?
seriously, I have no idea how problem like D is acceptable
It’s been so long I haven’t seen a Dijkstra on codeforces I was going for a completely different solution. When I finally realized my solution was wrong with 26 min remaining, I stepped back, thought for a while and realized it was literally just a Dijkstra (which funnily enough was much simpler to implement than my previous strategy). I don’t know why so many people complaining, I thought it was a cool question.
I think people complain because you look at the start of the round and you see screenshot.
Like <2000s solving D in 15m is just terribly egregious. Need to increase verification. Otherwise, virtuals would be more fun.
you can't increase increase it beyond catching the lamest of cheaters who will copypaste the whole solution, only way is trying best to make soln not possible by LLMs and accept they will be used
I mean phone number verification + account lifespan/activity would greatly reduce it, I feel.
At the very least, we can start removing the brainless copy-pasters who are getting 5 compile time errors across 4 different problems in 10 minutes. But there's not much of a point if they just keep making accounts.
For the input in Ques 2
1
8 8
1 1 2 2 8 8 4 4
answer should be 3... But codes outputting 5 get accepted For example this
Submission
Verified. What sorcery is this? Can you hack submissions?
EDIT: PS: Please post in way that is readable for others. Like put code in Code tag so we can drag in/out etc.
Done..
Did you hack!?
No, I got my pretests passed at first with this wrong code then I changed the solution realising my mistake... But to my surprise my previous wrong code is still getting accepted
when will the ratings be updates?? system testing is already finished.
can anyone explain what problem D asks us to do ?
it ask us to find the minimum cost.
the cost can sometime be infinite as we are performing the steps infinite times
oh ok ty
does E2 soultion depends on E1 soultion?
Check please this submission. https://codeforces.net/contest/2059/submission/304088287. I think this is ChatGPT and people didn`t delete comments. It is suspiciously. Because Green complete D on 30 minutes.
Too many downvotes incoming? :)
For the first problem A, I have found a corner case that should give me the output "NO" but is giving me a "YES" output for the editorial solution. The corner test case is: 1 3 1 1 2 1 2 1. Could someone please check if I have misunderstood the problem, or if there is an issue with the editorial?
None of those arrays ([1,1,2] and [1,2,1]) are good; read the problem carefully
C is great!
Although I became Master in this contest. But I still think using an ad-hoc problem as a distinguishing problem is bad.
Bruh, it's just your personal taste
Sure, but I don't think there's anything wrong with saying that expression (other than English grammar issues).
Why does the max rating update so slowly? It's important for me because it's my first time to become expert!!!
Now it's updated
007WSsr
Yes, yes! I see it! Thanks
how did the ratings of tianbincheng and FHQY change although they were master before the contest?
It seems, the test set for problem C is weak. My 308010308 was ACed but it does not pass local stress testing. Could somebody please check and confirm that?