Привет, Codeforces!
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В 23.06.2025 17:35 (Московское время) состоится Educational Codeforces Round 180 (Rated for Div. 2).
Этот раунд будет рейтинговым для участников с рейтингом менее 2100. Соревнование будет проводиться по немного расширенным правилам ICPC. Штраф за каждую неверную посылку до посылки, являющейся полным решением, равен 10 минутам. После окончания раунда будет период времени длительностью в 12 часов, в течение которого вы можете попробовать взломать абсолютно любое решение (в том числе свое). Причем исходный код будет предоставлен не только для чтения, но и для копирования.
Вам будет предложено 6 или 7 задач на 2 часа. Мы надеемся, что вам они покажутся интересными.
Задачи вместе со мной придумывали и готовили Адилбек adedalic Далабаев, Иван BledDest Андросов и Максим Neon Мещеряков. Также большое спасибо Михаилу MikeMirzayanov Мирзаянову за системы Polygon и Codeforces.
Удачи в раунде! Успешных решений!
UPD: Разбор опубликован
Competitive Programming in 2025 – A Broken Race?
Two coders. Same leaderboard.
Result?
The problem: CP is turning into a “pay-to-win” model.
Impact:
The spirit of CP is at risk !!
The ask: Platforms like Codeforces, CodeChef, LeetCode must act fast. Cheating isn’t just breaking rules — it's breaking the community.
Let’s keep CP clean. Let’s keep it fair. Agree or disagree?
What a cinematic last week we had on cf, this edu will feel good
last night only I guess :)
Yes, and then I hope I can reach 1750 in this round
Hope to see no Newbies and Pupils winning the contest....
Yeah, man, I am waiting for the rating changes, it feels like eternity for this round. When will i get to see rating changes......
Did it become unrated ? or the rating would change because showing in unrated section in contests .
ive been opening to see the ratings changes since morning its taking an eternity
Until the ratings are updated,it is shown in the 'Unrated' section.Once the ratings are updated,it moves to the 'Rated' section.
nobdy knows wht they r trying to do ...
too
did it change for u now??
no
I am still waiting……
.
.
yeah totally agreed man!!
It's one thing to obviously AI a post and another to copy paste it to every contest announcement blog
.
You submit approx. 5 questions(Like E,F) daily in 5 minutes ...
WTF is that ??
acting like not cheating, cant invest too much time into acting , lol
It would be better if the question stem could be more concise!!!
Cheater, how dare you show your face?
cheater? Let's see it tonight.Don't accuse one casually bro.
Evidence? Honestly, I'm a bit surprised you have the nerve to contradict me.Firstly,see this https://codeforces.me/contest/2118/submission/324122164, Wow, was this code written by a human being? Usually,you have your code tends to include some macro definitions like using i64 = long long; using u64 = unsigned long long; using i128 = __int128; using u128 = __uint128_t; Why did you stop using it from https://codeforces.me/contest/2121/submission/324852523 on? https://codeforces.me/contest/2121/submission/324822011 These two pieces of code come from the same competition. Why? Based on my profile, you were stuck on Problem C and used LLM for all the subsequent problems.
You are right bro, I used llm to obfuscate my code in[submission:324122164] and cheated in R1032. The goal is to make the rating reach a certain target quickly. It is promised that LLMs will be abstained from in the subsequent matches, and I feel sorry for the adverse impact brought to the community.
I can be certain that you have used LLM in more than one competition. If you can turn over a new leaf, that would be good after all, rather than just rewriting your own code to make it look less like it was done by AI.
The above apology is also written with AI if you see clearly fate
Yes, absolutely, using AI translation is also using AI,you are totally right bro, then everybody cheats do you guys agree with this? (just for those who with baseless meanness)
lol should apologies in your own words ... not copy apology letter from gpt
this shows how much ai u use .... in silyly matteres
You two gugys have destroyed the meaning of comments, this is not a place for arguments,but for discussing issues, nobody wants to see both of you barking at each other endlessly.
who the hell are you ... we are not using any bad language or topic on a bad language ... if you have any problem go eat grass somewhere else.
How do you get such high rank with 0 code submission?
those are skipped submissions, basically they cheated
feels good to see an edu round, high hopes from it
feels good to see an edu round, high hopes
r1033 still makes me feel disgusted...
Can we please get hints in Educational Editorials?
As a participant, I hope the managers can exclude cheaters from the standing during the round!
and most of them are going to cry's basement
i think cry's basement is running out of space
dont worry hes updating it
i hope i can be a specialist and then free of this
As a participant, I just wanna enjoy the problem statements and wish everyone goofy luck!
Hope everyone moves up to a new title!
Unfortune I think that in order for people to gain rating someone has to lose rating
BledDest orz!
Good contest
Sadly, in Indonesia, the city level qualifiers for its CP nationals will be held the day after this contest, and this contest would be held at 9:35 PM to 11:35 PM the night before, so unless I want to mess up my performance tomorrow I can't participate :(
Educational Codeforces Round 180 – Quick Info
Good luck and have fun!
.
Hope to enjoy the problemset and solve at least ABC.
So we are just ignoring all cheaters last round or is useage of LLM now allowed?
don't they usually have rollback few weeks after?
Hoping for a better round then the previous one!
feels good to see an edu round, high hopes
Good luck!
hope you to get a good grade .
After a few bad contests, I hope this one makes me feel better.
I WANT TO PLAY DIV.1 ROUND
Codeforces should publish cheaters list after every contest and ban them. This may create fear among potential cheaters.
I hope there won't be many cheaters in today's competition.
Am I the only one that has tendency to lose rating on Edu rounds?
Therefore I am registering unrated.
I didn't know I had the tendency, but after giving the contest, I sure have one.
I spent 40 minutes solving all I know,
next question I immediately know there is 90% chance I couldn't solve it.
Logged off early, really glad I'm not registering rated.
For me it's the opposite, I tend to gain huge +ve rating in Edu and Div. 3 rounds, but lose in Div 2(non-edu) rounds.
Same here. Something always goes wrong in edu rounds dont know why though.
This is going to be my first contest. Some of my students recommended me to try this website, after I failed to convert them to the atcoder side.
Just clarifying that I am not a beginner and if I'm violating any rules by participating today I sincerely apologize. Make sure to have fun today :)
just don't use LLM help and/or other cheating ways.
I think that's the most important rule here.
Nyemot thank you so much for clarifying on this.
If you dont mind answering, am I allowed to post my own solutions and editorials here, so my students can refer to them, or only the officials are allowed to do that?
Have a great day.
You're allowed to post your own solutions here (as comment), or as a blog.
Just don't do it during contest.
Good luck to all participants!
Educational CodeForces Rounds haven't disappointed me for a long long time..!
But, the rating updation delay is disappointing me..!
how to solve D? I was root->child<-grandchild->xxx<-yyy...... kind of thing it was producing n-1 good pairs then was trying to change direction of any one leaf node whose depth is greater than 2
In fact, you had to find any vertex of degree 2 and go left and right from it. Your solution says "no" for those cases where the answer exists.
shit why did I not think of this now I think about it taking 1 as root always might not be good choice
I had the same error at the beginning
It is possible if and only if there is some node with degree 2, and for that, we make one edge into it, the other one out of it. Then we basically two-color the subtrees of the 2 nodes that degree 2 node is adjacent to.
I undertsand now ;/
thanks. i literally messed up the basic premise of the code. just saw degree 2 in your comment and rushed back to code it up. just messy, trying the same incorrect logic for 1.5 hours
Choose any node $$$x$$$ with degree 2. Select an edge $$$(x,y)$$$ and disconnect them.
Do your algorithm (generate $$$n-1$$$ good pairs for a $$$n$$$ node tree).
Finally connect $$$(x,y)$$$ back.
... ...
Find a vertex with a degree of 2 as the B of "A->B->C".then let all edges with A start at A,all edges with C end at C;then let all edges with the vertices next to A end at them,all edges with the vertices next to C start at them,then you will get a graph like:
the part containing A,B,C makes 3 good pairs,and other edges only make 1 good pair each other,and the total number of good pair is n.
it is same as coloring graph such that no adj vertices have same color and have atleast one vertex of 2 degree
How to do B?
You can perform at most 1 operation if it's possible.
I got stuck in this problem for about an hour, too :(
why is that?
for this case:
1 4 6 1
isn't the required amount of operations 2?
1 4 4
remove
6 1and insert4No, you can replace 6, 1 -> 5. Thus we get: 1 4 5
If the array is either monotonically increasing or monotonically decreasing, and no pair of adjacent elements has an absolute difference ≤ 1, then it's impossible to make the array beautiful.
If at least one adjacent pair has an absolute difference ≤ 1, the array is already beautiful, and the number of operations required is 0.
If the array is neither strictly increasing nor decreasing, then exactly one operation is sufficient to make it beautiful.
This is because, in a non-monotonic sequence, there must be a point where the direction changes from increasing to decreasing or vice versa. At this turning point, at least one of the surrounding elements will fall within the range defined by the other two.
Example: Consider the array [1, 4, 6, 2] The first three elements are increasing with no adjacent differences ≤ 1. But at the turning point from 6 to 2, observe that 2 lies within the range [4, 6] This ensures that applying one operation is sufficient to make the array beautiful. In general, any such directional change guarantees that at least one element from either side of the turn will lie within the other's range [min, max].
I hate constructives.
What was the solution to $$$D$$$? I thought that you had to have a leaf node with a parent with degree $$$= 2$$$ in order for the situation to be $$$YES$$$ (and then switch around that edge) but that didn't seem to work.
That's basically correct, but it doesn't need to be a leaf.
Ah okay I see what you're saying. Basically we have a root with degree $$$2$$$ and start the edges pointing out of it, then the children have the edges pointing in and the children's children have the edges pointing out and so on. And then we just flip one of the two edges of the root and then go down to the child of the flipped edge and flip all of the edges in that subtree, right?
yep, that's what I did.
Leaf node is unnecessary.
close, you need a node with degree 2 only, no need leaf whatsoever.
seeing 3 comments with the same idea is quite funny.
same here. actually only a node with degree 2 would suffice.
Bro, I got the exact same idea, but turns out you only need a vertex with deg 2. Check this image:
gotcha, yeah that is quite a tricky generalization to spot
why colors
Because it's a coloring problem. The problem is basically bicolloring a tree after choosing a node with degree 2. The blue nodes only have out edges, the red nodes only in edges and the orange node has 1 in and 1 out edge
I guess that's another way to think about it.
You can use dfs to traverse and maintain a bool dir (direction of the edge). Keep changing the dir when having a dfs traversal. However, if you meet a node with degree = 2, you do it special for once only (It will cause an increase of 3 in answer with two edges costed. As there are n — 3 edges left, the answer will be exactly n)
Here is my submission: 325811538 (I did check for the starting node to be degree = 1 then do dfs but that's unnecessary, just did it for safe)
I also made the same mistake :(
You just need any node with a degree of 2 and have one as out, one as in, and then keep alternating.
The scoreboard of this contest is way more normal than the last one standings. I guess it wasn't the ChatGPT's fault yesterday, something smells fishy...
I think they were just banning ChatGPT users during the contest this time. I was seeing my rank go up without doing anything.
overall a good contest. was only able to solve a , b ,
was working on problem c , but not able to pass on the case.
6
2 3 3 4 5 5
mine was giving 14. I used sorting, two pointers , + maths. to count the valid triplets.
Please let me know, if you guys have followed any other approach. not submitted out of fear
Iterate over two balls and use binary search to find the possible indices for the third ball for
O(n^2log(n)): 325767781in such cases you can simply use, upper_bound and lower_bound, that saves a lot of coding time, and is also easier to debug with them.
Cool E. At first glance, the statement seems intimidating, but after some thought, the solution turns out to be surprisingly simple.
My Solution:
According to the definition, it's not hard to see that all green nodes form a connected component.
Let’s remove the green connected component. The remaining tree consists of several subtrees. From the problem’s definition, we can deduce that each subtree must be either entirely yellow or entirely blue.
If $$$m$$$ is even, there is no solution. This is because, apart from the coloring scheme where all nodes are green $$$(g, \dots, g)$$$, all other colorings come in pairs. For a coloring scheme $$$X$$$, we can pair it with coloring scheme $$$Y$$$, where $$$Y$$$ is obtained by flipping the colors in $$$X$$$ (yellow $$$\leftrightarrow$$$ blue).
If $$$n$$$ is odd, there is always a solution. For example, the number of valid colorings for a chain of $$$n$$$ nodes is:
Let’s remove the root node. Let the number of valid colorings for the $$$i$$$-th subtree be $$$f(T_i)$$$.
Then, the total number of valid colorings is:
This is because we can also color the entire subtree as all yellow or all blue.
From Observation $$$5$$$, we have a DP solution.
Let $$$dp[i]$$$ represent the number of valid colorings when $$$m = i$$$. Then:
Where $$$d$$$ is a divisor of $$$i$$$ such that $$$d \gt 2$$$
Code:325789361
What's the idea behind $$$dp[i/d]+dp[d−2]$$$?
You can take a tree having $$$dp[i/d]$$$ nodes and attach a tree having $$$dp[d-2]$$$ nodes, so the number of colorings now will be $$$(i/d) \cdot (d-2+2) = i$$$.
Shouldn't this be $$$(i/d+2)(d-2+2)$$$ since we are taking subtrees $$$(i/d)$$$ and $$$(d-2)$$$, and the total number of valid colorings is $$$\prod{(f(Ti)+2)}$$$?
No. We are attaching the tree having $$$dp[d-2]$$$ nodes as a subtree to the tree having $$$dp[i/d]$$$ nodes. The root of the new tree is the root of the tree having $$$dp[i/d]$$$ nodes so you don't need to add $$$2$$$ to $$$i/d$$$.
Ah I see now. Thank you!
We can also update dp[i] = min(dp[i] , dp[i / d] + dp[d] — 1),325839794 where we subtract 1 because no additional edge is added — we simply remove the root of one subtree and make the root of the other subtree the root of the entire tree.
Is it possible to have a binary search on the number of nodes and then a greedy algo to solve the problem of "Given x vertices how many different colourings are possible"?
Certainly can't.
I absolutely missed that I can colour subtrees entirely yellow or blue. What a miss :(
dude this was wonderfully written kudos!
In problem E for m=5, Given sol is :- consider a bamboo tree with 3 vertices with a root at the 1-st vertex. There are exactly 5 beautiful colorings: [g,g,g], [g,g,b], [g,g,y], [g,b,b] and [g,y,y].
Why [g,b,g] and [g,y,g] are not beautiful ?
sorry:
they've considered a straight chain of 3 nodes in the explanation i think it should have been mentioned
because [g,b,g] contradicts the statement "consider all yellow and green vertices, they are reachable from each other without passing through any blue vertices" since there is a blue vertex in their path
NVM
The leader board in this contest is much normal now !
Why N is freaking low in B :)
UNABLE to open test cases for C problem. Only first test case is visible.
Note : awesome problems. Getting Huge negative delta, but still, awesome the problems.
My C almost TLE, my D was 5 minutes too late... sad performance again.
why is this failing for C? https://codeforces.me/contest/2112/submission/325767247
Integer overflow corrected: 325820072
damnn!! i thought since they were not asking for mod it should be storable in int, costed me a lot of rating points, btw thank you
Problem B is really funny. First, check if it is "beautiful". If yes, output 0. Otherwise, if there is a maximum or minimum value, output 1. Otherwise, output -1.
What do you mean by if there is a maximum or minimum value
he means local maximum or minimum
Could anyone explain why this code ran a tle https://codeforces.me/contest/2112/submission/325747110
Maybe because of python. My logic & time complexity are similar but got accepted 325789337.
Also, in your submission 325809550 you are not using binary search for finding k which leads to a TC: O(n^3).
I misread the Problem B and realized my mistake after wondering about 10 minutes,
but out of curiosity, if we change the condition of a beautiful array from:
there exists 2 consequtive elements with difference <= 1
to:
all consequtive elements has difference <= 1, and the length of the final array >= 2
at what time complexity is this question solvable?
Cheaters again. Take a look at this submissions by Yagnik_Dhameliya, lakha_luffy, AbdullahIshfaq: 325754179, 325788858, 325771032
Notice anything strange with those solutions? They have the same "implementation" for generating divisors.
its probably just a template
firstly they are not same, second the function for generating divisors is already available on the whole internet like here : https://www.geeksforgeeks.org/dsa/generating-all-divisors-of-a-number-using-its-prime-factorization/, so there is a high chance that the function maybe same and online resources are allowed(as per the contest rules).
where does this approach for C overcount? in some cases its overcounting,
27 32 35 37 45 57 65 65 71 87 96.You have made exactly same mistake that I did. and here is your answer.
[30 , 40 , 57, 69 , 138 ]
so, when your a[i] = 30 , a[j] = 40 , your
kwill point at 69, so you will simply add indexof(69) — j = 2. Which is wrong. You can't take (30,40,57). that's an overcount.You also need to find lower bound of (a[n] — a[i] — a[j] + 1) and find its index. (lets call that
ind2) .Than your addition to the answer will be
k - ind2.Look at this code : https://codeforces.me/contest/2112/submission/325822074
By god's grace, if we both would have GOTTEN AC. somebody would suspect we plagiarised... LOL
someone please explain why does O(n^2 * logn) works for C? its almost 3e8 operations and time limit is 2.5sec
Because binary search is really fast.
BTW: python may fail.
but how would you analyse this in a contest, by writing you know O(n^2logn) is bad but practically you know binary search is fast, but i would not know how fast it is, so do i just guess it works and code or look for an optimisation? its consufing as sometimes a constant factor really matters and gives TLE but in cases like this here its doing well with binary search too
I didn't write binary search in contest. I wrote two pointers.
Expected Rating for D?
1600-1800 ig
1500-1700
1700
Why for problem E is a tree (root)g — b — g invalid?
Because the green nodes are not connected.
I'm sorry for the follow up, but why must this be necessary because we can still go from every blue vertex to any blue or green vertex without passing through a yellow one
From the problem statement:
To clarify, if you have a pair of green vertices, they also must be reachable from each other without passing through any blue vertices, so (root)g — b — g is invalid.
You can neither have any blue or yellow vertex between two green vertex
in c how n^2 is working isnt it around 1e8 .?
Its O(n^2logn) which is about 3 * 10^8 operations. This is relatively slow but still fast enough for the time constraint given.
In 'B' statement, whoever decided not to put the "sum of n over all test-cases won't exceed 1000" is absolutely genius, lol
I saw lots of people from my friendlist got one rejection there so i thought n^2 might not work. but it did!
Was able to solve the first 4
visualize a number line if x <= a <= y, then it's not possible to choose b or else it is always possible
325723799
first check if array is already beautiful if not then check if it is strictly increasing/decreasing. In this case, we can not make it beautiful or else we always can in 1 step — just pick any element which is lower/higher than both adjacent, and here one of the pair will work
325734602
total possible triplets are nc3 — if we try to check for all possible triplet then it will be n^3, so we need to optimise first sort the array, for every pair i, j where i < j, try to find two indexes using lower/upper bound (logn)
k1 such that a[i] + a[j] + a[k1] > maxElement
k2 such that a[i] + a[j] > a[k2]
and here we all indexes from k1 to k2 will form a valid triplet with i and j
325770998
we just need to find a vertex v with degree 2 — if there is no vertex with degree 2 then not possilbe
now let's say we found v with degree 2 and 2 edges from v connect to x and y, we do x -> v -> y
the idea here is that using these 3 vertices, we got 3 good pairs. now we do bfs from x and y(mark v visited) separately and at every level flip the direction of edges,
325807734
For third, could explain k1 case? Why does a[i] + a[j] + a[k1] > maxElement work? If I have chosen a[i] and a[j], don't I need to find max excluding a[i] and a[j]?
Because in 2 3 4 5, we can use the triplet (2, 3, 4).
Whereas in 2 3 4 9, we cannot use the triplet (2, 3, 4)
Hii
I'm new to hacking. If I do hacking now, and my attempt is unsuccessful, is it going to affect my rank in contest??
For Problem B,, for this input:
1
5
1 5 9 13 1
the author's answer is 1,, why not 3??
i can remove 13 and 1, replace them with 9. That's just 1 operation.
Just understood the answer is actually never more than 1
Feels good to see GMs at the top of the standings again
Why 325802678 does this fail ? I believe it is (n^2)*(log(n^2))?
st doesn't have all required pair sums at the moment you're doing the query Worse, insertions are repeated — the same a[i] + a[j] is inserted multiple times unnecessarily for different is And since st is not cleared, its size keeps growing → order_of_key() becomes slower → TLE
I have a different solution for $$$C$$$ : instead of doing binary search, you can use an cumulative array / prefix array of length $$$10^5$$$ that represents the $$$a_i$$$, and then get the number of $$$a_i$$$ from $$$l$$$ to $$$r$$$ by doing $$$cum_r - cum_{l_1}$$$. So the original $$$O(n^2 log(n))$$$ becomes a $$$O(n^2 + 10^5)$$$ per testcase, so it gives at most $$$10^3 \cdot 10^5 + 25 \cdot 10^6$$$ operations, which is faster than the $$$O(n^2 log(n))$$$ !
325733800
finally a good contest with classic CF style problems
Can anyone provide me failing testcase for my submission 325842292.
Or point out my mistake. I first generate output for graph like ...->child<-parent->child<-grandchild->... and then flip one edge between degree-2 vertex and leaf node.
It need not necessarily be a leaf node
I understand it now. Thank you, friend.
The logic for qs B was much at the easier side , only we need to check if the number is b/w the two adjacent number or not so that the condition can be satisfied and there is just 4-6 lines of code needed , but indulge in much longer code. And surprisingly it was not me only during the contest many other top coder does the same.
Has profile ratings updated?
Please, please, don't go down again!
Why was CF down for so many hours? @MikeMirzayanov Was there any specific reason?
Why was CF down for so many hours? @MikeMirzayanov Was there any specific reason?
Can I ask for help? I don't have a rating for this competition because I'm not a trusted participant. I admit that I once participated in div1, and div2 is not my level, but I really forgot my previous account and email because I haven't used them for many years.
Rating will be updated for all participants with rating <2100, no matter whether trusted or not (except for those who opted out of getting rated). All submissions will be rejudged with updated test sets, and rating updates will happen after that. Just be patient.
ty
it shows unrated for me, why?
(except for those who opted out of getting rated). -- i didn't know about this, when do we select this
ya me too.
Apparently Codeforces does not distinguish "unrated" and "not rated yet" in those screens. Do not worry about that.
Regarding opt outs, you should have seen a checkbox when you registered for the contest. If you checked that box, you will be not rated.
Is this contest being shown as unrated for all of you or only me?
All rated contests will show as unrated until ratings are updated.
ok, thanks for the info
what happened to this contest, guys ?
its finally in system testing
My standing section showed asterisk in the starting of username . I joined rated . What is this issue and the submission are in queue rn . Is this server issue ?
Asterisk means you are not a trusted participant (you need to attend 5 rated rounds to become trusted). You'll still be rated -- do not worry about that.
All submissions are currently rejudged with updated test sets. Rating updates will happen after that.
Why my B AC became In Queue??? And also why can't submit problems?
Solutions are testing with main test cases. During this you cant practice thin round.
not main test case, it's hack test case (from the people for the people).
When do we expect our ratings?
when will the rating publish?
In problem D, how to calculate the number of good pairs in a given directed tree? UPD: Got it
good pairs starting from node n = sum(good pairs starting from every child for node n + 1)
calculate the above recursively for all n, but tbh we dont need to do this in the solution
Use dfs on node have no edge in, if a dead end is reached then res += depth*(depth+1)>>1
What I did was: Create a chain of nodes A->B->C (B must have degree 2 ) Now I have is 3 nodes and 3 pairs Now for each node added to above structure should give 1 new pair We can do this if a node has only out going edges or incoming edges I.e A-> k or k->c because A has outgoing edge and C has incoming edge Keep adding nodes until u have used all the nodes
Drop editorial please. I want to know D and E approaches ;)
In D, find a node with degree 2 and make the directions as A->B->C (B is the 2-degree node). Now do dfs from A and C alternating the edge directions as we go. this ensures the ABC triplet contribute 3 good pairs and rest
n-3nodes contribute 1 good pair each making the total good pair count3 + (n-3) = nIf there is no node with 2 degree, it is impossible to do this arrangement
Hello, this is my 1st comment.
ahh, Man!
what?
this is your second comment:) FPythonX
can anyone be kind enough to share where can I find problems like E where we work around tree structures to count answers, really facing hard time around these problems.
thanks a lot
think combinatorics + tree filter works ?
Do update rating fast... we are all waiting desperately
When u have a bad contest, rating gets out next min.__ When u have a good contest, it feels like rating will take eternity to get updated.
Auto comment: topic has been updated by awoo (previous revision, new revision, compare).
Still no rating changes? Has been 2 days.
I came here to see the updated ratings. Its weird that they are still not there.
Can say bro... What is admin doing don't know
Does anyone know why my rating hasn't been updated yet due to the competition?
No one's rating has been updated yet.
Why hasn't the rating been updated yet?
It seems the ratings have not been updated yet after the contest. I’ve never encountered such a delay before — perhaps system testing is still ongoing or there’s a temporary issue on the server side. Hoping it gets resolved soon.
when will the rating publish?
Why hasn't the rating been released yet?
To day pro
Does this contest became unrated?
Why hasn't the rating been updated on Codeforces? Is this contest going to be unrated? Can someone share any information?
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues. I don't think it will change to an unrated contest
My code is failing for 2nd question 2112B - 11 - Shrinking Array. Can Anyone help me ? My submisson : 325964066
in your code, it is this part that might be the problem:
when N >= 3 and the array is not originally beautiful, you have to check if the array is sorted in decreasing or increasing order. if it is, then it doesn't work. if the array is not sorted in decreasing or increasing order, then it does work (check the tutorial for the explanation). so you gotta change that part
The bug is from
if(abs(res[i]+res[i-1])<=1), try (-4, 4, 2), that is, abs(8 + (-2)) > 1 and the branch gets false, but actually it should be true. The key to this problem is to find a peak or a valley value, that is, a local maximum or a local minimum value. You may replace the branch judgment withif (abs(1LL * res[i] * res[i - 1]) < 0), it works.when the rating will be given
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues
rating .
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues
from yesterday until tomorrow i entered codeforce thousand times
we should be patient, since the server crash might have messed some things up
When is the rating going to come?****
mabye in the next year hahaha
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues
How long does it take to update the ratings? After the disasterous previous weeks contest, why so long updating this constests ratings?
it's better. I hope there is more proper checking for cheaters cause i saw some guy above in the comments who had an ai code.
When will the rating changes roll out?
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues
my rating was not added
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues
why's the rating not being updated?
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues
You took the same sentence from the last reply you did
I mean, it's the same answer anyway
Prabhu woh thodi rating aa jati toh sahi rehta..
why my rating is not added?
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues
Cheater wants their rating, please give the cheater his rating :( Sad state of affairs man..
Me waiting for rating to update!
Please update the rating Can't wait more
When will the ratings come?
it will probably be updated soon. with the server crash and everything, they may still be resolving some issues
can you tell the estimated time for rating arrival donutmaster?
I'm not sure, I hope they will be updated soon or at least today, but idk
Stop spamming the same f*king comment everywhere.
unrated???
Bro even Jesus woke up on the 3rd day. Plzzzz do the rating changes.
3 days and still no rating change. They should publicly announce if there would be any delay.
it has only been 1 day since the system testing, but there should be an annoucement
In question B test like : 4 1 1000 10000 1 I think the answer here will be 2 if we take 1 with 1000 and replace them with 1 the array will be [1, 10000, 1] again take 1 with 10000 and put 1 the array now [1, 1] so the answer is 2
is that right or I misunderstand something
в данном тесте ответ будет все же равен 1, ведь ты можешь взять 3 и 4 число, то есть 10000 и 1, и заменить их на 1000 и тогда массив будет выглядеть так: 1 1000 1000 и разница между 2 и 3 элементом будет равна 0
Why is it taking so long for ratings to be displayed ???
f*ck men, why is it take too long??, I'm so annoyed
I think it's cause cheaters are still being caught in the last div 2.
I can accept that the rating updates are slow, but I don't understand why there hasn't been an announcement explaining this issue.
been rapidly checking to see the rating changes, WTF are they doing bro
How did you, as a 1000+ rated, solved 3 problems during the contest and also later upsolved D? why do your submissions have comments, and why are you using big variable names? Pretty suspicious, isn't it?
what the hell maybe you should stop going around accusing basically everyone of cheating! and yeah thats exactly how i code because it gets confusing without comments and proper names it's not like im grandmaster or sth. and as of AI i only use it when i encounter problems with the syntax not for generating the entire code!
nice defence but nah still you cheat lol , you cheaters
yeah well it really doesnt matter how others see it, im preparing for the IOI in the future years i use codeforces and attend contest for practice so it really defeats the purpose to cheat. the rating is just the cherry on top its not like they reward me if i bring it higher.
ok dm me I will give you a basic question prove me in video call ? and IOI lol bro you cant even pass school level exams by cheating you cheater
dude it doesnt matter that much for me to prove myself, I explained and you feel free to take a look at all my other submissions not made during or after contests, dont see a reason why i would cheat there all my codes look like this, nevertheless if you still believe im cheating, then honestly i dont care
Will the round be unrated or what? Because there has been 48 hours and the rating changes still hasn't been released.
Will the round be unrated or what? Because there has been 48 hours and the rating changes still hasn't been released.
At least make an announcement about the expected time for the rates to be updated or if it's gonna be unrated since there are an unusual delay
not(Hardworkworking and competent). 3 days no rating changed.
No one is lazy and incompetent. They are doing their job. It's just taking time.
hello everyone, I just wanna know that was this contest unrated like I haven't get rating so but I see there was written that rated for div 2
Sorry
Hello Cheater You give contests in multiple programming languages , dont you?
So smart, great.
Please see towards this code forces
Just using multiple programming languages doesnt mean he is cheating,even i use java and c++,java sometimes give tle so i write the whole code in java and ask chatgpt to convert it to c++ but every line resembles my own java code
Yeah,but by his codes it doesn't seem so. Anyways he admitted below that he cheated.
I think this round is unrated.
I think this round is unrated.
They would make an announcement about it if that were the case.
I am a cheater but for a purpose
I think your purpose is to demotivate others who work hard to get a positive delta. Nice purpose.
It's my personal but how could you say i am cheated
what purpose , to appease the cheating gods ??
is this a unrated round?
if it was an unrated round, there would be an announcement, but since there aren't right now, I don't think it will be unrated
I guess they forgot that this contest is rated.. I don't mind tbh
Hi, I participated in Educational Round 180 (handle: ombelose1304), registered before start, solved 2 problems during the contest, but my rating hasn't updated yet even after 36+ hours. Can someone look into it?
It's not just you, bro. My rating didn't update either. I think there's some issue with the contest.
you gonna get your rating after 61 days. and me after 32 days.
and those who dont have number next to their name they are doomed.
Apparently, I’m supposed to get my rating in a day, lol. But I’ve waited 12 years to win a Div. 2 round—what’s a few more days? As long as it updates eventually.
Ain't no way we are getting GTA-6 before the rating update of this round.
I'm going to sleep now. Hopefully, the rating gets updated by the time I wake up.
How it feels waiting for the rating changes
Hopefully they’re doing a thorough investigation and purging all the AI cheats. The hacking round needs to be extended to contain a cheater catching phase
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Why do you care bruhh? You cheated in it lol atleast remove the comments duhh! Bruh you are from an IIT arent you ashame of yourself..? Cuz of guys like you indian cp coders are getting so much hate!
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nice try diddy! Bruh just look at your code you think that's written by a human? I mean ur code has more comments than code can you tell me the reason why? also who keeps variable names that big and complex when ita all about speed in cf contests haan?
bro his last contest was skipped too .
lol these guys should get banned instead of this skip and all
you know that everyone can still see your previous comments right? there's no point editing it. also anyone with a little coding experience can tell you submissions are directly copied from llm. it's just a matter of time before an admin sees you profile and bans you. nice try!!
Why it's taking too much time to show rating??
At least they should make an announcement to clear everyone's confusion.
Why have the rating changes not rolled out yet?
Been 3 days :(
The hell is going on?... atleast admins should clarify what's going on... we all r waiting from 3 days yet don't have any clue ...
I think there are many cheaters
What a facinating problem the E is!!!
Besides, why the editorial is not connected to the competition page?
I solved first 4 probs why is my rating 581 only @organisers
Bro, it was just your first contest... take it easy hahaha you'll be going up after a few contests, good luck!!
Great performance
Hello Codeforces team,
Handle: khanna_dhruv
Submission ID: 325731757
Hello, I received a system message saying my solution to problem 2112B in Educational Round 180 coincided with another user's submission.
I want to clarify that I wrote my code entirely on my own using Sublime Text on Ubuntu, without using any online IDE like Ideone or sharing it with anyone.
If the match is real, it may be a coincidence — or perhaps the result of using a similar idea for the problem. I'm happy to provide evidence that helps verify this.
Please let me know how I can assist further.
Thank you for your time.
"Please let me know how I can assist further??"
Lol, this line says a lot about you.
Dear Codeforces Team I don't know why you skipped my submission of probelm C of this contest, and messaged me that my solution coincides with other people solution, even though I did that problem myself and used standard technique of two pointers to select first two no and iterated in a loop for selecting 3rd number. I also viewed submission of other people whom you said it coincided and they were also using it. But there was no fault of mine for using a standard technique used by many other. I hope you see into this matter and resolve this.
Ban all these cheaters ^^^
Hello CODEFORCES, I just got message that my submission match with another guy I have added comments to explain the way I think and I just resubmit them please check them
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