Hello Codeforces!
The series of Educational Rounds continues thanks to the support of the Computer Science and Artificial Intelligence (CSAI) program at Neapolis University Pafos, with scholarships provided by JetBrains.
Educational Codeforces Round 186 (Rated for Div. 2) will start on Dec/29/2025 17:35 (Moscow time).
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 6-7 problems and 2 hours to solve them.
The problems were invented and prepared by Adilbek adedalic Dalabaev, Ivan BledDest Androsov, Maksim Neon Mescheryakov, Roman Roms Glazov and me. Also, huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.
Big thanks to the testers FelixArg and KIRIJIJI for their valuable advice and suggestions!
Good luck to all the participants!
And on the occasion of the New Year — Happy New Year!
We wish you clear ideas, strong solutions, and many ACs in the year ahead.
UPD: Editorial is out
first
The real Good bye 2025
Bro is a forest
so many cheaters in last round, it's last contest of the year, I hope we don't see another round where 10000 people solve problem C.
It wasn't just because of cheaters, the problem was probably the level of Div2 B.
It was a dp problem and I have seen myself that AI can explain and solve DP very easily that's why I am saying there were a lot of cheaters, even a master level cheater was caught buddy in the contest.
Not a Dp Can be solved by greedy.
I solved it using a dp approach — https://codeforces.me/contest/2178/submission/355396319 and like I said if a problem can be solved using dp then AI easily cracks it because essentially it becomes an optimization problem which is fundamentally how machine learning and AI models work.
AI working that way in the backend does not mean that it's necessarily good at reasoning that way
there is literally an entire paper published by Deepmind (the first pioneers in Reinforcement learning algorithms) which basically tries to prove that Graph Neural Networks are basically dynamic programmers.
Here is the source — https://arxiv.org/pdf/2203.15544
I am myself working on RLHF at the moment and I can argue that dp is involved everywhere in RL due to policy optimization, monte carlo methods, etc. These types of methods are used in AI technologies as RLHF (Reinforcement learning through human feedback) and a lot of dp is involved to optimize the policies due to high amount of trajectories (Data essentially). This is why I argued that dp is better for AI compared to other types of problems.
All RL algos are Bellman equation, which is used to solve dp, so it's not entirely true to say dp is used everywhere in AI. Also RLHF in practice barely uses pure DP, but rather approximate DP like value iteration. I think you meant "DP principle" instead of "DP".
Anyways, "dp is better for AI" doesnt inherently mean "AI can easily solve dp problem".
Yeah sure, I just wanted to give the idea that dp is involved in optimization and RL algorithms.
The whole point of the previous source is that the general dp related problems are easier for AI because of their own architecture resembling an optimization compared to other types of problems. This increases the chances for cheaters to get correct results compared to other types of problems.
Never heard of this paper thanks for the reference going to give it a read.
Hope you will like it!
I am still learning dp, but the first thing that came to my mind was to use prefix and suffix sums on that problem.
I also did the same thing!
It was combinatorics...
I solved it in my first codeforces contest because I also do olympiad math, also I have given atcoder contest.
I have a question how to get the right and useful solution with the explanation for a problem that I couldn't solve؟
Editorial is up!
there is no way he just did the 67 joke...
67
L
lol re u a death note fan too
For Problem 148A A. Insomnia cure, should I study the inclusion, exclusion version that was in O(1) time. Like is it common to see things like that, so I should study it? Do discrete mathematics concepts like that appear sometimes? My solution was really brute force compared to that one.
Will score distribution be released soon?
All questions are of equal weightage in educational round. There is no score distribution.
Ok, thank you for telling me!
Why 6-7
Cuz six sevennnn...
Hope you didn't say 6-7 problems just for the meme
Now you know it was for the meme...
What does it take to AK this round ?
i think 2100rating is very enough.（If I wasn't particularly tired during the competition, I feel like I could AK this round too)
Solving more than just 'E'
Skill issue :(
i really hope to reach new max :(
Where unrated register?
Wish you all have great ending of 2025 with this Educational round.
hope for all a year full of new Ranks
As a femboy, I am NOT looking forwards to dropping to Pupil this round!
embrace the :3
Hope to solve at least 4 problems, thus increase my rating.
I want just 2 or three problem to reach pupil
i hope i don't get clicked this time ^_^
Hope i can solve ABC
how did it go?
Solved ABC but looking at the ranking i think i should practice hard to go for D next time..Overall i enjoyed the contest...
Nice! I took so much time on B that I only had 30 minutes for C so i just didnt try.
+1 Wasted so much time in B for no reason but somehow just managed to solve C as well.
haha wtf lol 4k solves on $$$D$$$ and 2k on $$$E$$$?
too easy or all cheaters?
E too easy, D too hard
I guess. By the time I solved $$$D$$$, I had 5 min left. Couldn't solve $$$E$$$.
I solved E in 15 min but no D in 75 min.
Lucky that I saw this
4k on D, in 2 HOURS, is unreal... I barely passed first test
D felt easier than normal Ds, and E felt easier than normal Es. I think it felt that way to me because both problems relied heavily on logic rather than complex data structures and algorithms, which favors me as I haven't studied any of the advanced competitive programming data structures and algorithms.
Yes, prerequisite for both problems were not that much
D >>>>>> E
I'd agree with E is easier than D, but not by that much
How did you solve E?
greedy + multiset
priority queue is better.
Ask yourself, what is the optimal allocation of the box with the smallest prettiness?
No wonder you managed to solve it in 4 minutes...
C should have been better.
how 4k people solved d ?? is it that easy ?
.
D >>> E. Shitty speedforces :( Please make a better contest for the end of the year.
please tell Idea for
EI did greedy.
Step 0: for all Y[i] , k -= Y[i]; remaining value of k is your final budget...
Step 1: You sort the boxes by the beauty...
Step 2: You process all the friends in sorted order, and check which all friends could be fullfilled by this box, we keep pool of such friends. among set of all such friends, find a friend, for whom the DELTA of Z[i] — Y[i] is highest and fulfil that friends requirement ( we give him the i'th box, which has beauty more than X[i] ).
Step 3: After processing all the boxes, now among unmatched friends, try to fulfil their requirement by paying extra from your final budget. for this, we will choose the smallest delta (Z[i] — Y[i]) , (opposite of what we did in step 2 , in step 2 we try to match the highest delta ).
I did similar like giving boxes first then increasing
(y-z)but my bad in step 2 .. I think I was giving boxes randomly in decreasing order ofDELTAor somethingthanks for detailed answer
I don't see why we want to remove some Y[I] if we potentially don't want to give anything to that friend.
because in question its mention it is mandatory to gift each friends regardless if he becomes happy or not
oh I missed that
Thank you!
I somehow managed to get through the math for D and couldnt understand E. Guess I got kinda "scared" only because it was a div2E. But looking back now E was really easier than D, should've solved it :(.
Tks for clarifying
Why is it necessary to find the maximum DELTA in the second step to satisfy?
Focus on using boxes for friends with highest z[i] — y[i]. After you've used up your boxes, simply use your remaining k coins to pick up as many of the leftovers as you can, starting with the lowest z[i] — y[i].
yeah I was allocating boxes for
z-y, it seems I need to pickz-yfor a box ... thanks for the idea !!!What if z[I] — y[I] is the highest but the y[I] is so high it doesn't make sense to buy anything to this friend?
The problem's constraints gurantee that we will always have enough coins to take all y[i], and we must gift every friend.
lowkey have to quit cp ig not improving little bit
same condition here :(
There is still hope. I got $250K+ SWE job despite sucking at CP.
I think you are improving, but due to cheaters, your improvements might not be as visible as they should be.
My suggestion, Go to past contests, like in 500 to 600 series, and attempt Virtual contest will full focus and see what rank you would have received with your final score... that will help you judge your performance more accurately.
C was so bad!
I don't think so. C was a good problem in general.
yeah opinion can vary I found c to be bad
How so many solves on C and D? Cheatforces!!
Thank you for D and E. amazing problems.
I felt E < D in difficulty.
Also, GAP between C and D was little more than expected, but it was worth it. ( I can't believe so many people solved D !!! ).
Great contest! (Though I got a million WAs for no reason on A nd D XD)
All the problems I solved (A-E) where interesting and more or less the right difficulty, and the other problems seem nice, too. I especially loved D!
Only one small problem: E was much easier than D, though I don't think that it affected the contest quality or my performance, so it wasn't a big issue for me
2k rank for 4 solved XO
People seems to got faster in a few months or I am just getting worse
llms are getting better :(
HOPE CODEFORCES DO CodeChef LIKE THING IN EVERY CONTEST.
D felt way harder (combinatorics) than E (simple greedy strategy)
can u please tell the strat which works for
Eand how did you it was correct ?I think D was on the very easy edge of combinatorics ngl
Fair enough. Looking back, I overcomplicated the solution. Although, the margin of error is still probably much higher on D.
couldn't solve
E, tried some greedy thing but failed :( , like give boxes to people who might need more coins and then give coins from small requirement to large ..please give hints for E
Your idea for E is correct, maybe try to swap set with multiset?
I was wrong a bit I think .. explained here https://codeforces.me/blog/entry/149654?#comment-1336505
I used a greedy approach with priotity queue by first matching boxes to eligible friends who have the highest cost, and then spending the remaining cash on the lowest cost friends left which maximizes the count.
Submission (Rust 2024): https://codeforces.me/contest/2182/submission/355740776
yeah thanks for the idea !!! I missed some part of it
Big clutch for me ig, I forgot that the contest existed until 20 min into the contest
How to solve C ?
I used DP ... length of longest valid subarray from
ab[i][j]... meaning u start at indexiinaand indexjinbonce you have this .. then it is not so difficult .. but I still feel this was tricky
please visit mine for "c" i feel i have solved it very clean and without dp
oh wow!!!
can you please tell what that loop is doing .. although it looks cool .. it is mystery to me
see you should not calculate individual i ,j and k rather give some importance to gap sequence like 0,1,2 ..n-1 what i mean is if gap of 0 for a and b exists then for a and b 1,1 2,2 and n,n all exists now it makes sense to see a<b and b<c in a separate way so i tried a little combinatorical approach proving that a<b choices and b<c choices are independent so why not multiply them and eventually multiplying with n cause if one gap is valid then total of n new pairs gets generated
ok thank you !!!
I will try to think / understand your approach
certainly!
Yeah makes sense this is the simple solution
For i = 0, you can calculate all valid j(ignoring k for now) in O(n^2). The number of valid j can be called
val_j.Note that if i = 1, each valid j is also incremented by one. Therefore, the total number of valid (i,j) pairs is
n*val_jsince i can be any value and there will always be the same number of valid j.Next, we can repeat this logic with j and k. For a given j, the number of valid k is
val_kand can be calculated in O(n^2).The number of total valid j is the number of valid (i,j) pairs which were calculated above, so the answer is
n*val_j*val_k.if you want a valid pair for i,j,k
u can break it as number of valid pairs for i,j and number of valid pairs for j,k
since its a transitive relation
to compute number of valid (i,j) pairs
you pick an offset value from 0 to n-1
and need to check if for all i = 0..n-1 , the j = i + offset, satisfies the condition a[i] < b[j]
if it satisfies then this offset contributes to a valid pair,
similarly find for (j,k)
now the answer is the product of these 2, but these are just in the form of i, j = i + d1, k = j + d2
but i can start from 0 to n-1
so answer is n * (number of ways to get pair (i,j)) * (number of ways to get pair (j,k))
Great it was actually this only, I couldn't think it this Way. Thanks for explaining.
I solved it in a stupidly simple way. I don't know if it's correct. for string $$$a$$$ and $$$b$$$, I rotated $$$b$$$ by $$$0,1,...n-1$$$ positions and counted the number of positions by which if $$$b$$$ is rotated, $$$a_i \lt b_i$$$ for all $$$i$$$ (let's call this $$$numCompliantRotations(a,b)$$$.
the answer is $$$n*numCompliantRotations(a,b)*numCompliantRotations(b,c)$$$ because the three rotations are independent.
4K rank for 4 solved.....;(
Why is $$$n \le 50$$$ for D when a solution in $$$\mathcal{O} (n \log MOD)$$$ exist?
red herring
+1, re-read the problem statement multiple times to make sure I wasn't missing anything.
It is not given that $$$\sum n \le 50$$$, so the time complexity is $$$O(n \cdot t \cdot \log \text{MOD})$$$ and t is large.
Although $$$\log \text{MOD}$$$ is not required as we can precompute all $$$\binom{n}{r}$$$ for required range using the formula $$$\binom{n}{r} = \binom{n-1}{r} + \binom{n-1}{r-1}$$$
D absolutely kicked my ass. Can I have a pointer on it? I was thinking it was something DP related which is a big weakness of mine.
I was thinking of sorting the private boxes, then somehow making dp[k] being the ways to pick k people fairly?
Why makes a permutation unsuitable? Given that there is a valid permutation(answer > 0), what differentiates permutations?
Once you figure this out, the rest becomes intuitive.
A permutation is unsuitable when there isn't enough in the public box to cover the 'debt' difference between the private boxes and their maximum; to put it loosely. Permutations are differentiated by their orders.
So say we make the maximum private box our K'th box, then our permutation count is the possible permutations of our 0th to K-1'th box, multiplied by our K+1'th box to N'th box.
Add these permutations up, and we get our answer.
I think I get it now, thank you so much for your help.
I did it this way: first balance box 0, then to make it fair, either all elements must be x or x followed by x-1. Since you can only add to balance, x at least has to be the largest element in the boxes, so we can divide it into two zones, the quantity x and x-1, find the quantity in different ways with permutation and multiply them.
Was F1 like a binary digit dp?
Yes, it was.
From where to read the solution editorial there is no stream for it's solutions i solved A was stuck in B was very close to solving it couldn't no gpt rule so where can i read the solution ?
Miseed F1 by 5 minutes, feels bad man.
The only Diff between wa and ac
Yes, if the number of digits shifted to the left exceeds or equals the number of digits of the data type, the result is undefined.
Great round!
How 9k successfull submissions for C ?
It was quite a simple problem, your answer is |a|*f(b, a)*f(c,a); where f(x, y) is the number of cyclic shifts of x such that y_i < x_i.
You can get to this by realising that you can only produce all n snowmen if a_i < b_i for all i, and likewise for b and c, given a specific cyclic shift.
WA on D because I think when n = 1 the answer can be 0...... ;(
Hope we will all get good ratings before 2026 ! :)
4000 AC on problem D? CP nowadays is becoming a nightmare for legit contestants
Yeah the AC count today is more like a div.3 contest ngl
I couldn’t understand how come such a huge number of participants came up with a bitmask solution for Problem A, most of which got hacked. But when I copy-pasted the problem into ChatGPT, it became clear.
bitmask for A??
isn't the STL find function sufficient
Huge cheating on d and e
May someone explain why is n not greater than 50 in D?
awoo constraints ;)
cause there isn't obvious non-deliberate worse time complexity solution, maybe.
0_0
I actually solved it in O(N^3) using dp. I thought it would tle but apparently it is fast enough.
Can you please explain your DP solution.
I would recommend waiting for the editorial because I think my solution will turn out to be completely unnecessary/missing something, But here you go:
I started by looking at the condition that makes the permutation fair.
First, let $$$S = \sum a_i$$$ (summation of all $$$a_i$$$ from $$$0$$$ to $$$N$$$).
Each player will play for $$$\lfloor S/n \rfloor$$$ rounds. Except for the first $$$S \pmod n$$$ players, who will play an extra round.
Each player will try to use only his box until he finishes it, and only then will he use box $$$0$$$. So he will consume $$$\lfloor S/n \rfloor$$$ gifts in total, $$$a_i$$$ of them from his own and the rest from box 0.
Additionally, He will consume an extra decoration if he lies in the first $$$S \pmod n$$$ positions, he also greedily consumes it from his box if possible otherwise box 0.
If the sum of those consumptions is $$$\le a_0$$$, then the permutation is fair. Otherwise, it is not, because if it exceeds $$$a_0$$$, then box 0 will run out too quickly and one of the players won't be able to play.
Now the problem becomes: who will I assign in the first $$$S \pmod n$$$ positions, and who will I assign in the following $$$n - (S \pmod n)$$$ positions?
This is what I use DP for. My state is $$$dp(pos, sub, amnt)$$$: * $$$pos$$$: current player being considered. * $$$sub$$$: how many players from the first $$$S \pmod n$$$ use a decoration from box 0. * $$$amnt$$$: the amount of players that I need to select to be in the first $$$S \pmod n$$$ spots.
The transition is just: either put them in the first $$$S \pmod n$$$ group or put them in the other spot. For the base case, I just multiply by $$$fac[S \% n] \cdot fac[n - (S \% n)]$$$ to account for all possible orderings of this grouping.
Keep in mind most people had simpler solutions, I haven't upsolved yet but they probably exploited something I haven't noticed so that they don't need DP.
Thanks for explaining!
Sorry, I was incorrect with my previous judgement, I misread the bounds
$$$t \lt = 5e3$$$ and $$$n \lt = 50$$$ though.
I could have implemented my solution in O ( N ) I believe .
Looks like if I want to reach expert then I need to consistently do D :(
how can i get the editorial?
There is a 12 hour open-hack portion, I believe after hacks end (11:35 PM EST), then the editorial will be posted.
ok
does anyone know if we can solve problem F1 with the help of segment trees, and if yes, then how?
i tried to do it but couldnt, the type 3 query was making it problematic for me. i would really appreciate if someone could tell if there is a way to solve using segment trees.
seg tree?!? may i suggest using static array instead sir.
I also tried and failed, but I think my general idea was close.
Think of it like digit dp with dp[pos][tight][started]. DP is not necessary, but the ideas are useful. Considering started as an implementation detail, and pos as each bit, you can iterate downwards through the bits. Let's also define
tas the number of bits greater than ourposwhich are in thex. Let us also assumetightis always true.If the previous bit present in
xisp, we can do some casework. If we select the next number to be at a bit in(pos+t, p]then the number will be greater no matter what the next bits are. For each number we can choose, we addw * 2^zto the final answer wherezis the number of integers in [0,p] excluding our current number.wis the number of ways to reach this position(will be defined later).If select the next number to be at a bit
[0,pos+t), the number is smaller resulting in no solution.Finally, the next number is equal to
pos+t, the number is not guaranteed to be big enough but also not guaranteed to be too small. I definedwas the number of ways to reach this position. The transition is justw *= segtree[i]. This does not add anything to final answer, it just increases the number of ways for the case where the next bit is greater.The final case(which I ran out of time for), is applying
wat the end of a number. The value ofwafter going through all digits expresses the number of ways to hit exactlyx. Based onp, you can consider every subset of future numbers(2^zwherez= size of ([0,p]-1)), and then add it to the final answer.I didn't get the problem, so I'm not sure, but this was my idea.
How to calculate the CP can be very confusing sometimes, but I don't think that really makes it harder to solve D than E, since the method is already there. Sadly, I didn't manage to solve E since I'd wasted to much time figuring out D =(
For some reason, I found D (4000+ AC DAMN) to be much harder than E, Was able to solve D in last minutes, and post contest, just took 3-4 minutes to come up solution for E.
I basically fixed the maximum element first. Every other element (those < Max) must use box 0 at least
Max − a[i] − 1times to “catch up”, assuming they all sit to the right of the Max element. After subtracting these mandatory uses from a[0], if R becomes negative the answer is 0.With the remaining R extra uses of box 0, I choose how many of those right-side elements can be moved to the left of the Max block. Let F be the count of elements equal to Max. Then for
i = 0…R, I place(F + i)elements on the left, compute the number of ways using factorials + nCr, and sum them upE is impl heavy, logic easy
D is impl easy, math harder
all in all I think both have similar difficulty
agree.
I think, E was very easy to implement. Some of the people I follow ( who are just CM , coded it under 10 minutes )...
I also finished solution + code in 25 minutes...
I have a question how to get the right and useful solution with the explanation for a problem that I couldn't solve؟
it's in the editorial, usually it's out less than a week after contest.
it's there now thank you :)
how to solve c ?
what i had in mind, is to to fix array a and try all allowed rotated array's for b and c
but i couldn't get how i,j and k's can be counted.
we should learn from these coders , rank 17,20,18,19 they solved all 8 questions in their first contest , they should be awarded by codeforces @MikeMirzayanov
A quick (scientific) note to the authors pertaining to F1/F2:
The carrying capacity of a population (a group of a species) is the amount of organisms that can be sustained in the environment. It was pretty clear towards the end you were referring to carrying gifts, but just a FYI.
Good Bye 2025
I want to report cheater wegokic234 with using LLM
Here https://codeforces.me/contest/2182/submission/355745999
And you can see this code has if (!(cin >> ... ))
By when the ratings are supposed to change ?
A great goodbye for 2025!
me too!
What's the expected rating of Problem C ?
Based on solve count about 1000-1100.
I think it was much more harder than that.
Hello Contestant, My name is not showing on standing table — What's the reason behind this, please anyone can tell me?
one needs to be a trusted participant (means one must have participated in atleas 5 contests and solved 1 question in them ig) but you can go to your profile, friend yourself, and then see your standing in friends standings
Another way to solve C
numValidTriplets = numAllTriplets - numInvalidTripletsnumAllTriplets = n^3We can compute
numInvalidTripletsas follows:numInvalidPairsAB= No. of pairs in A and B which violate condition (Pairs with same drift count to one pair)numInvalidPairsBC= No. of pairs in B and C which violate condition (Pairs with same drift count to one pair)However, these are just pairs across two arrays and we will need to exclude duplicate triplets while deriving invalid triplets from these invalid pairs This can be computed as
n*numInvalidPairsAB*numInvalidPairsBCnumInvalidTriplets = n*n*numInvalidPairsAB + n*n*numInvalidPairsBC — n*numInvalidPairsAB*numInvalidPairsBC
Hence solution will be: n^3 — (n*n*numInvalidPairsAB + n*n*numInvalidPairsBC — n*numInvalidPairsAB*numInvalidPairsBC)
https://codeforces.me/contest/2182/submission/355838323
Is this contest unranked?
67
So let me get this straight, this guy https://codeforces.me/profile/Ananya_CodeNova has a Youtube channel https://www.youtube.com/@CodeNovaYT where he is live streaming contests and displaying solutions directly and pasting them into YT comment section, Telegram group and google drive.
Not only he is not banned, his submissions are not even skipped.
Additionally, there are at least 10-15 contestants in the top100 of this round that have basically identical code and not even they are banned.
This is an absolute joke.
I mean there's nothing that can be done cheaters would return from alt accounts best we can do is enjoy problem solving and let those cheaters rot
Doubt they would keep returning if they're banned after 1, max 2 contests each time. But since even the blatant cheaters like the one I mentioned above are not banned, it's obvious that nobody in the CF leadership cares about these issues.
He solved till E in just 20 mins. Were Questions leaked?
I have checked his submission and I believe that all of this code was written using chatgpt. I can only be grateful that I have solved more problems than chatgpt.
when will ratings be updated?
even the editorial hasnt come out yet, ratings will come after that. hopefully today eod
You sure about that?
Where rating?
Why did i not get any rating from this contest even after solving three problems? Is anybody else facing this problem and does anybody know what to do?
rating in edu was slowly updated..
Tutorial for D:
Prerequisites: Factorials, combinations $$$n\choose r$$$, stars and bars, inverse factorial modulo $$$p$$$ using Fermat's little theorem
There are two important observations to make.
First, we can assume that every person takes from their own box first until it runs out before taking from the common box. If this were not the case, then someone had the choice of either taking from their own box or the common box, but chose to take from the common box instead. We can only be worse off in this situation, since that person could have potentially taken away a gift that someone else needed whose own box was empty, so we have less chance for the permutation to be 'fair' in the end.
Second (this is the crucial observation to make), the number of 'rounds' that the simulation lasts (that we care about) is $$$\max_{i=1}^n a_i$$$. To think about it easier, assume that the common box had an infinite supply of gifts (or simply some large enough number to ensure that anyone can take from it when their own box is empty). Once the last person (or last few people with $$$a_i=a_{max}$$$) finishes taking gifts from their own box, then everyone would have an empty private box, and whatever is left in box 0 can simply be taken until it runs out.
If at any point in time before the last round, someone is unable to take from both their private box and the common box (both $$$a_i$$$ and $$$a_0$$$ are 0), then there are no fair permutations at all and we can return $$$0$$$. Since it does not matter what permutation the boxes are in if there is at least one person who always runs out of gifts to take before the last round regardless of the permutation. To be clear, one 'round' is when we loop through the persons from $$$1$$$ to $$$n$$$, and each person takes from their own box if it is not empty, or otherwise takes from the common box. In every round before the last, we only care about the value change in each person's $$$a_i$$$ as well as the change in $$$a_0$$$, not the order in which they change, since they all result in the same change regardless of permutation (non-empty $$$a_i$$$ decreases by 1 and $$$a_0$$$ decreases by number of $$$a_i=0$$$ which are forced to take from the common box).
Let's assume there is enough in $$$a_0$$$ to support everyone until the last round (otherwise we can simply return 0). We can split into two cases, one of which is easy. Let's assume there are $$$m$$$ people with one more gift in their private box, and $$$n-m$$$ people who have nothing in their private box and are forced to take from the common box. Then if $$$a_0\geq n-m$$$, there is enough in the common box to support all those with empty private boxes. This means that regardless of whichever permutation we chose, there will always be enough in the common box to support everyone through all rounds, so the answer is $$$n!$$$.
Otherwise, $$$a_0 \lt n-m$$$. In this case, only $$$a_0$$$ of the $$$n-m$$$ people can take from the common box. The other $$$k=(n-m)-a_0$$$ are forced to be the last $$$k$$$ in the permutation. They cannot come before any of the $$$a_0$$$ people who are taking from the common box, nor any of the $$$m$$$ people with one more gift left in their private box. This is so that the process ends before they have to take a gift from $$$a_0$$$ (now empty), so the permutation remains fair.
So we have to choose $$$a_0$$$ out of $$$n-m$$$ people, permute the people who take from the common box, and permute the people who are forced to be the last $$$k$$$. There are $$${{n-m}\choose a_0}\cdot a_0!\cdot k!$$$ ways to do this.
For the $$$m$$$ people who have one more gift in their private box, they can be placed anywhere in the first $$$n-k$$$ positions. An easy way to think about it is, count the number of ways to insert $$$m$$$ people into $$$a_0+1$$$ bins (the $$$a_0$$$ people are 'bars' or separating dividers and we can insert the $$$m$$$ people anywhere beside them), then permute the $$$m$$$ people. An even easier way to think about it is: we have $$$a_0 + m$$$ people in the first $$$n-k$$$ positions; choose $$$m$$$ of those positions i.e. $$${a_0+m}\choose m$$$.
So the final answer is $$${{n-m}\choose a_0}\cdot a_0!\cdot k!\cdot {{a_0+m}\choose m}\cdot m! \mod 998244353$$$. You can simply precompute the factorials and inverse factorials and use them in the $$$nCr$$$ formula.
Submission: 355849613
When the rating score will come?
It says the round will be "rated for participants with rating below 2100". I'm a newbie with rating 382 (just started cp), I solved 2 questions, but in my profile it shows this contest as unrated, and no rating change has happened..can someone pls explain.
Rating updates for the round have not been done yet.
Ratings were recently published. (Congrats on CM)
Thanks. Congrats to you on ranking 1st lol.
Thank you! :)
Hello everyone. Have a nice day. Where is our ratings??
well i had fun
good bye 2025 , finally I reached pupil
I have a question how to get the right and useful solution with the explanation for a problem that I couldn't solve
THANKS :)
Read the editorial in the announcement https://codeforces.me/blog/entry/149733
Thank youuu
hey everyone, just started doing codeforces again, anyone have any idea how many ques i need to do in div 2 to reach pupil and any tips to reach fast?
Today, I received a message regarding my solution for problem 2182E (New Year's Gifts) in this round, which has resulted in a plagiarism flag and a significant rating penalty, resetting my score to 420. I want to clarify that my solution was developed independently using standard competitive programming techniques. The structure of my code specifically the use of a suffix-based greedy scan combined with a priority queue is a well-documented algorithmic pattern. This logic is almost identical to the optimal approach for CF 1140C (Beauty of Playlist), a problem I have studied extensively to understand how to manage resource constraints efficiently.
**https://github.com/sourabhkanojia/CodeForces/blob/master/1140C%20-%20Playlist.java**
The variable names I used, such as pq, cnt, grp, and cap, are universal abbreviations taught in standard tutorials and found in public educational repositories like Prakash-sa/Competitive-Programming. My implementation follows the standard "Greedy with Heap" logic used to solve resource-limited selection problems, where one iterates through a sorted constraint and maintains a pool of candidates in a heap. I believe the similarity to other submissions is a result of this being the most direct and efficient way to implement the greedy strategy required for this problem. I have always worked to improve my skills honestly.
I deeply respect Codeforces and the rules that keep this community fair, but I must express how incredibly frustrating it is to receive this notification. I worked very hard to arrive at this solution and to reach my rating, and seeing it drop so drastically due to the use of a standard algorithm is disheartening. I request a review of my submission based on these standard practices and my original submission time to restore my account's standing.
I received a notification that my submission for Problem E was flagged for plagiarism. I would like to clarify that I wrote the solution myself.
I have a complete local development timeline in VS Code, including autosave history and timestamps showing the incremental development of the code from scratch. I did not copy the solution from any external source or other participant.
Could you please let me know what kind of additional evidence would be acceptable for verification? I am willing to share editor history, intermediate drafts, or any other information that may help clarify this.
Can someone help me with C and D. Please, how did you got idea and how to improve implentation.
Reporting a cheater, used LLM to write the code.... the submission : https://codeforces.me/contest/2182/submission/355725936 reason, --> No consistency of coding languages that are used... --> there's the various inconsistency in coding style of this person...
This is very wrong. I mean what is the use of doing cf if you are there just to copy.
I hate these people.this is not fair to them who genuinely use their minds to solve