How do you even think of C? I can understand the editorial and follow along, but the idea is just so unexpected.

you can just replace that element, with something that will not afect your answear like for sum 0, min a bug numebr etc

"have violated rules of our community"? whose community's rules has he violated?

just IMO, the deadline is too soon. 90% of people saying "I'm in" do not realize that April is like 2 weeks from now

Perhaps you are just saying this to get a reaction, but you are actually wrong. All soccer players do is kick a ball around. There is dexterity involved, but no IQ. Football players have to take an actual (well, kinda) IQ test before getting recruited. A fun fact is that some of these football players are quite smart. One football player scored a $$$143$$$ (or $$$99.8^{th}$$$ percentile) on his IQ test, and after his football career, he went to get a math PhD at MIT. Many have scored in the

Me too!! Can u pls share your plans u gonna follow ,I am also a struggling to go beyond pupil from 3 months now, would love to know about your plan

In Mike's post he says it's allowed to use autocompleting tools

what happened here, i have just came but CodingKnight comment was erased.

if you dont know,its common here,ppl reach certain rating let say expert,cm and start selling solutions for money,ppl here also selling referrals too,they are this low,ppl bashing us(indians) on international platform for some reason,its ugly,its shameful,

More contests will be added, so just wait and prepare for the next contest :>>

maybe you can't do it, but i know many that can

Probably got something close: #7 0x00007ffff7c8c4b1 in std::terminate() () from ./libton_crypto_lib.so #8 0x00007ffff7c8d033 in __cxa_throw () from ./libton_crypto_lib.so #9 0x00007ffff7981713 in vm::load_cell_slice_impl(td::Ref<vm::Cell>, bool*) () from ./libton_crypto_lib.so #10 0x00007ffff7981a00 in vm::load_cell_slice(td::Ref<vm::Cell> const&) () from ./libton_crypto_lib.so #11 0x00007ffff78a14e4 in block::gen::BlockInfo::cell_unpack(td::Ref<vm::Cell>, block:

So I saw this comment and was like "OMG harshith_04 is CM" and then I realized that he is actually CM

Maybe add option for Excluding problems from $$$Div$$$ $$$\ge$$$ $$$X$$$ , because div3-div4 problems are inflated

Already set this goal for myself. but instead of touching grass, I would waste my life on leetcode instead

Thank you so much!! I wasn't able to understand the tutorial solution due to storing alot of information like id, start, end, etc.. But your solution seems very much intuitive.

template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;// find_by_order -> iterator of kth value, order_of_key -> number of elements less than k template <typename T> struct OrderedMultiset { ordered_set<pair<T, int>> s; map<T, int> m; void insert(long long element) { int count = m[element]; pair<long long, int> p = {element, count}; m

Oh, it doesn't meet the requirement. I feel very upset that I only tried 1013, as the next prime 1019 is right behind it, which can be the correct answer.

How about sharing your journey (problem ideas, practice methods etc.) as well so that it can become a guide to them on reaching M

and do not forget the legend Abhipsita 1_2_3_4_5_9 Das

there are very few genuine guys doing cp from india,

3 4 3 2 3 3 5 8 you can swap cols like this and the path would be 3+4+3+5+8 = 23

A structured plan with progressive checkpoints. So how do you prove your plan structured without even reaching Expert yourself...

In problem A,I use 1013 as p,but I got WA.I can't figure out why? Submissions

it is not just about money. If you don't believe me try it yourself, it is a great start for 2025. It is only $5 and with discount even $4

I have the same target but many of my days goes without laying hands on keyboard and just doing some random things. I can see you're regular, you will surely reach expert or above with this much consistency. Can you give some advice on how you can do it regulary.

Thank you so much! Will there be an api for random problem under certain restrictions?

This is how we show text font in a fontless environment lol

damn touche, I kinda walked right into that one

These soccer players got nothing on true football players. Tom Brady would rip ronaldo and messi's arms off.

pfp checks. As a messi fan, CR7 is definition of hard work and grit

I just thought a number like 142857 had very good property for the problem, so I searched on Wikipedia . There, I found a link to Cyclic numbers —everything you need is on that page. I can't imagine how to solve the problem without the Internet.

So I saw this post and was like "OMG JLP IS GM" and then I remembered that magic exists...

You forgot about North Macedonia TST

It would be useful to have build with debug symbols too! Not cool: (gdb) p *root.ptr $7 = <incomplete type> And can't go into std_boc_deserialize std_boc_serialize. Why not?) UPD. Found CellSlice in another .zip, but still, debug symbols are pretty useful!

if he genuinely helped you, why do you care what he does with his own personal cf account?

GYM -> MUSHUP -> create a new mushup -> click the "Add random problems" under the problem list

This feature is only for mushup, you can access it through: GYM -> MUSHUP -> create a new mushup -> click the "Add random problems" under the problem list (seems unfriendly for those who are unfamiliar with MUSHUP? (Including me))

How can I remove a list from "Shared with me"? I wanna keep it empty.

Wow, amazing What would be the reason for such a random visit ?

I am in , but I will set a more realistic goal of reaching Expert by September. Hopefully I can do it this year.

1180B - Nick and Array — 1500 rating.It's really good question(one of my favourites).For this problem you should have a little math knowledge.This is constructive algorithm :D

It was discussed on stream

dp[i7][i6][i5][i4][i3][i2][i1] where ij is no of yet to be decided losing teams in jth round. ij <= 2^{7-j} $$$O(2^{21})$$$ states per testcase, 7 transitions for each state. We start with dp[1][0][0][0][0][0] Transitions looks like following dp[i7][i6][i5][i4][i3][i2][i1] -> dp[i7][i6][i5-1][i4+1][i3+1][i2+1][i1+1]+(1+2+3+4+5)*aj , notably we do not need index in dp state.

Challenge Accepted! My team will qualify ICPC online round this year and Reach the Regionals.

Final rank list : here

https://amritaicpc.in/final-rank-list

Can someone explain how to solve question D ?

In problem C, will anyone explain why is it necessary to presence of dm in b? I mean if it is not present then we can assign it to some value which is already present then it will be overite?

Trust on allah , he never leaves anyone hungry for sure he is the one who will give you food

When you get 3 2700+-rated problems in a row from Asia-Pacific ICPC finals, that's an automatic 5N overhead. Maybe you're so good that you instantly solve all of these, most of us are not...

I can see your nationality with my eyes closed

Soo many cheaters,It's kinda funny that how all this cheaters is connected to regit top cpers from india,first that kakadiya guy and now this,hope this stops

Challenge accepted! But first, aim for Specialist or Expert.

Request for a filter of "exclude problems solved by users:". MikeMirzayanov

This whole Acade thing is just a nice way to exploit impressionable CPers — it's almost impossible not to get overhead there, and they set the withdrawal limit way too low, probably deliberately so that they don't actually have to pay anyone... Also the you can only use the "Reduce overhead" button once in ten "levels", and getting up from level 5 to level 15 takes so unbelievably long that you will have obviously made 10N overhead in that time. It's all a scam, don't wa

This is really hard for me to find it... especially when I felt desperate because of the verdict in the contest. :( Anyway, thanks for your help! Appreciate ^_^

I have a question for the editorial of the problem B. Suppose $$$n = 2, p_2 = 1$$$ and $$$G = \{1 \rightarrow 2, 2 \rightarrow 2\}$$$. Then $$$f(G, \emptyset) = 1, f(G, \{1\}) = 0, f(G, \{2\}) = 1, f(G, \{1, 2\}) = 1$$$ and the sum will be $$$3$$$ instead of $$$2$$$, what am I missing ? For $$$S = \{1, 2\}$$$ "consists solely of several vertex-disjoint cycles." does not work.

Problem C can also be solved somewhat easily if you google the right keywords and dig out this paper . Solution Thanks for the contest! This is probably the highest delta I've ever received for solving a single problem.

Bro just killed my CLI Tool . Seriously though, this is a better solution GG!

Bro ask for upvote but only got downvote :v

During the contest I simply guessed that lcm worked and I couldn't come up with a counter test case so I went with it

Something like April Fools contests or Kotlin Heroes probably.

Stack size in codeforces is 256 MB (and in problems with lower memory limit, you will sooner get MLE than runtime error from stack overflow). So that answers why you got Runtime Error. As to why Codeforces does this, no idea, but it has been like this for a long time.

You can just see the first line + first summation of second line above, that's how we get it.

What made you think of decimal expansion when solving A? I OEIS'ed it :( I brute forced when X and 2*X have a lot of digits in common, and searched the digit sequence in OEIS. Found decimal expansion of 19 It also referred me here, to try 383 after 19: magic-square-something Then I wrote the checker to see if all multiples are present, and indeed 19 and 383 worked. Next item on that list was 32327, which is too big. So I just guessed that maybe the first primes after 1000 would work

Waiting for how to reach GM in 4 trivial steps

Can you add multiple random problems at the same time?

What I meant to say was just to solve above your current rating level (not like abusing some 800 rated probs)

Can you please list all the most important algorithms/ideas to solve div2D??

It's easy you already made the 2 steps of +107 and +100, you only need the third one

I no longer need a discord bot to give me training problems, great news

Couldnt make it to the regionals this time but next year for sure!

Can anyone debug my code on problem E? I kept TLE on test 3 and I discovered that a lot of people TLE3 changed their solution at last but I don't want to change my solution

this is rare,epic ultra super update

is there a way to filter old problems? i think this is very important

Congratulations and good luck for the next year!

unusual problems are automatically excluded. What type of problems are deemed "unusual"?

Anyone who had a bad performance will send that he was affected by I2, and thus, all the people who had negative rating change will be unofficial which will affect the rating of other people, for example, being the 300th on 20000 participants is better than being 300th on only 5000.

I'm surprised how simple the solution to B is. During the contest, I created a list of segments to check the validity of the ranges. I have completely overlooked the easy solution.

I guess you need to have some relatioships with the staff...

Thanks, I indeed hope to reach GM soon... I will try it yesterday.

Congratulations, hope to see you in GM soon

Would you make me your patner of your dear journey. Pleases!!!

// int n; // cin >> n; Even if you simply put this at the end of the code it says segmentation fault. It's because you are using n in your code and then kind of redeclaring it even though its commented out. To check this simple change the variable n in the above commented code to some other variable which you did not use in the code (some thing like "var") then it works. TLDR : using the variable name again (even though commented out)

if you choose the complete array, the triplet $$$(9, 9, 110)$$$ can't form a triangle. Note how the elements of the triplets do not need to be distinct.

[image] I found that it was your first time participating in AGC as a rated contestant. It showed that you actually attached importance to the contest. Wish you good luck!

Because of some serious problems, this contest might be the last atcoder contest this year. So please treasure this precious oppotunity!

I quite liked 1973D - Cat, Fox and Maximum Array Split . Another one would probably be problem C from Serbian OI 2024 Day 1.

Question Problem D-Diagonal Separation Sample Test 1: 4 3 4 1 B 3 2 W 1 3 B If a cell is painted W , then can we change that cell to B if i+1 cell is black B ? If a cell is painted W , then can we not change that cell to B if the i+1 cell is white W ?

Consider the normal edit distance DP with time complexity $$$O(|S||T|)$$$, i.e. $$$dp[i][j] := \text{minimum edit distance between } s[1, i] \text { and } t[1, j]$$$ $$$dp[i][j] = min\begin{cases} dp[i - 1][j] \\ dp[i][j - 1] \\ dp[i - 1][j - 1] + [s_i \neq t_j]\end{cases}$$$ If you analysis it carefully, it's unnecessary to consider all states with edit distance $> K$, thus for each $$$i$$$, we only need to consider $$$dp[i][j]$$$ where $$$i - K \leq j \leq i + K$$$, which reduce

"you can check what is the minimum number of one is required to make it divisible by 7. I got that minimum 6 1's should be present so if n>=3 it is divisible by 7 regardless the value of d" can you pls explain the above part. If n>=3 then number is divisible by 7, can I read proof of this somewhere?

My final solution of G passed with a maximum running time of 9874ms(9624ms actually when testing pretests), which has a great risk of getting TLE during system testing. After I found that, I soon told Error_Yuan to rejudge that(yes, FSTs on pretests can be avoided if you told the staff). However, the rating calculation is done before that, and now I'm waiting for the rating rollback. Update: you can't.

You never have to consider these cases explicitly. The point is that we don't need to process [m2,r] , so we split only [l,m1] and apply our observation recursively.

Thats why i have requested to give some downloadable file as like in gfg or cses

Is there any precedent to this? The closest I can think of is this problem and the entire div.1 was unrated.

https://codeforces.com/problemset/problem/2033/G

Why did GroupMatrix 's rating drop, while the performance is 3114 (way higher than the previous rating)? MikeMirzayanov

There are many other submissions which match the exact same solution just with some added macros and preprocessors is there any way I can report them?

https://codeforces.com/contest/2053/problem/E This was the problem in yesterday's contest. I am usually afraid of solving problem >=D (D,E..). However today I was adamant that I will sit at this problem no matter how long it takes. And boom... at first glance what seemed so difficult was doable in about 2 hours.. Loved it. Wish to solve tougher problems in 2025.

there are t = 10^4 test cases at max, and for each test case you are initializing a vetor of size 400001, which makes overall time complexity to be O(t*400001) which wouldnt run in given time limit of 1 sec

Could someone help me understand why my code for E 298930084 is giving TLE? I am probably doing something stupid somewhere but can't spot it. Edit: Never mind, I found the mistake.

why you took 1.5 hour for A bro, and solution same as pastebin code?

I hope Codeforces bans me too, so I can enjoy life once again

The tests aren't shown fully when they are really long

Divide the 50 pile into 2 different stacks one will have x moon y-x stars other will have 50-y-x moons and x stars now flip the other stack we have equal number of moons Looking forward for more puzzles.

[image] Divide the coins into two piles, one pile of ten coins and the other pile of forty coins, if the pile of ten coins has A moon, then the pile of ten coins has 10-A moon, and the pile of ten coins will flip over and both will have 10-A moon. Please forgive the retarded translation software.

Would be shits and giggles if we kept only 8. :P

The solution comes down to a single observation that if [l,m1] and [m2,r] are two halves of the segment [l,r] , then the values we add to the answer while processing [m2,r] are the same as the values we add while processing [l,m1] just shifted by (l+r)/2 .

Last year at GoodBye2023 there were also some mistakes by the authors, which seemed to me to be much larger. So I hope that this next year will be better than the previous one) I also wonder if anyone was able to come up with the right solution for I2 or is it impossible? P.S. Also here is my experience of solving this problem: looked closely at the code for I1 -> hmm this should be improved to I2 by simply adding a few structures -> started writing the code -> after 5 minutes rea

I might be biased cuz I just saw it yesterday, but I enjoyed Eliminating Balls with Merging: https://codeforces.com/problemset/problem/1998/E1

Congrats! Glad to be a part of your journey! (I will pass you sooner or later in contest perf)

Maybe they just misunderstood your question. For example, I read it as "can the first player only move the head and the second — only the tail?" at first

I still cannot understand what exactly problem is trying to say. Can someone pls explain me how are putting wi values and what is the criteria for a unique subsegment in problem B ?

Impressive! .. Keep it up!

But there is no word in the statement that says that $$$p, q$$$ are changing, so I asked to confirm my guess. Why didn't they answer with a word "yes", instead they referred to the statement ?

It's written "whenever p is a leaf, Nora wins, whenever q is a leaf, Aron wins. If INITIALLY p and q are both leaves, it's a tie". If p and q weren't changing then this statement wouldn't make sense.

you can first make the array equal to [3,3,5] in one move and then make it equal to [4,5,5] in the next, making the sum equal to 14. Notice that the condition is that a_i is different only to a_(i+1) and it doesn't specify that for example it has to be different from a_(i-1).

Again 2008 trash pfp, idk when will people stop putting pictures of Naruto characters for themselves and creating new accounts with 'cool' nicknames, it is not even a pretense of irony. Guys, I remind you that 2025 is officially the year of this user

I love the coordinator's (or anybody else who answers the questions) work for the problem E. Nothing in the statement says that we will change $$$p$$$ or $$$q$$$, only some sequence of vertices. So I asked the question about it, and their answer was brilliant... [image]

I feel like one positive delta doesn't mean much. For me I just have sudden realization that I am better after feeling I haven't improved at all. But I am very excited for what your future performance will be like if you continue your experiment.

As long as you are having fun, go for it. Don't feel pressured to hit certain number of problems each day or certain rating, just have fun!

If that were the case, I probs would drop to cyan and get back to blue when I'm like 80 (and after saying confession). But I would have to be careful to ensure that cf doesn't shut down while I am cyan, though, cuz I wouldn't want to make myself immortal on accident.

great round! better than goodbye 2023 :) ehehe

CM color is good, but imo there is nothing like sky blue (cyan, well it is close enough). I think that the red color is very distinguishing, but fsr it is kind of tainted with a black letter at 3000+ rating. Green and blue are both not super pleasing colors, but they are very popular, so I guess they had to own some rank. Anyway, congrats on red!

It probably doesn't, I indeed do use the fact we have the groups of same values. It is just the next (not obvious for me) step — that the array transforms without swapping elements.

I just like CM color. I remember really wanting CM a few years ago because I found purple nice :)

i think 20 questions a day is too much.....if you are doing doing 20 questions a day then you will not get anything new because you will be doing lower rating question.

And with this We are officially finished

I have to say, it is an interesting choice to still use magic after just reaching red

Yeah, it happens. One of the worst kinds of bugs. I had plenty of similar bugs myself. Here's a tip. If you want to pass by reference but don't want to accidentally change it, declare it with const in function. For example: void coolfunction(const vector<int> &v);

No... That's so unlucky. I'm sorry. Though I am sure you will reach red soon.

937ms is very close to the time limit. Maybe another submission in Java 8 will give TLE. The reason is that Scanner is very slow. Look for faster method of input.

now time to write blog with title "How to Almost Reach GM but Fail at Reaching it in 3 Simple Steps" for extra free contribution

C was very new to me, E was a very nice problem if I was not stuck on problem C, I could've solved E.

Yes, that's true. I'm sorry but I was crazy thinking that I could do that much in a single day. My limits are lower than I thought they were. As I said, I'll do as much as I can.

can i use gpt o3 in this contest? if not, what messaures are beign done to prevent it!?

I check your solution later , I am so tired , but thank you for reply

i don't understand editorial for C, who can help me?

8-20 problems why do I have a feeling there's gonna be 14 problems tomorrow O-O

the set{10,2} has 2 possible triplets 10 10 2 and 2 2 10 only one of these is valid

tried using dp to store the results but not working

Can it be tried using the recursion? If we can reduce the TC somehow?

Every time I participated goodbye round, my rating is also "goodbyed".

I feel I1 is overscored than its difficulty. Normally, the subtask score is given so that it is slightly lower than its difficulty, so I guessed the difficulty of I1 is about 3500pts problem in this contest, but actually F>I1 for me.

For a segment [l, r], you need to know whether there are (r — l + 1) distinct segments whose length is 1, and they all fall within [l, r]. While taking the input, whenever l = r, update prefix[l] = 1. After taking input, iterate over the whole prefix array from left to right in order obtain the cumulative sum. Then, if prefix[r] — prefix[l — 1] == r — l + 1 for l not equal to r, then answer for that segment is 0. Otherwise, 1.

There's actually no need of a Fenwick/segment tree. I used prefix sums since l and r can be atmost 2n. You can check my submission.

Fun fact: [image] PS: It's three minutes before the contest starts.

Oh, that's smart, we basically never swap elements. Then there exists a clean solution, you are right! Still not sure if it is easily identifiable on the spot — i havent thought enough during the contest because pbds was not that difficult to include.

My segment tree fell in MLE on test 4... maybe my implementation is really messy lol

Why is the second player in E named Aron?) Kind of confusing (at least for me)

ABC felt a bit too difficult, even though I enjoyed them. E is alright — maybe a bit more enjoyable if we solve for first player, but then it'd be too easy. Gap E-F is hard, but maybe it's my skill issue :). Overall felt kind of balanced. What I can't understand is the intended solution for D: whatever you do, you need to keep track of the two sorted sets, and be able to recover an element by index and vice versa. This is either a lot of pain to implement, or a no-brainer pbds. For div1 tha

how to E? i tried a couple of intuitively decent methods but they were failing that 171 tc.

In this case 1 4 1 -2 2 1 My solution outputs 2 because it thinks that the two occurrences of 1 1 -2 2 -1 1 are different...

Congrats on reaching expert! What was the approach for E?

This contest has made me think my life choices....

I thought my solution to I2 is wrong so I only submitted at the end due to desperation...

use greedy answer is always multiply elements of a, b in sorted order. then answer only change if a[x] == b[x] on the sorted order I mean, so update answer on those incidents

choked on C for 1hr 20 min and solved D within 25 mins :)

perform really badly but quality contest imo. probably one of best in 2024

Hey hope you are fine . To become a pupil, you need to solve first 2 in div2 quickly and the 2nd question in div 2 is min 900 and max of 1100 rating so try to solve these questions, and don't watch editorial too early, even think for one question for hours. This would be my advice. Wish you high rating.

Yes we also thought it would be really fun and motivating. But later we thought that people might abuse the system given that editorial and solution for all problem are easily available, making people easier to climb on leaderboard. So because of that we drop the idea of leaderboard of ThemeCP.

To get good grades in high school and get to ioi 2025

Kindly join the discord group and ping me there (@10zin)

The level suggestion of ThemeCP works on very simple principle and you may feel v. easy/hard. You can always explore the level sheet and choose a level which you think best fit for you.

I liked problem E. at first I created a boolean array to check the valid countries, which was pretty misleading and took a lot of time to realize my mistake. overall nice one :D

Let $$$S$$$ be the set of all possible graph, $$$W(G)$$$ be cost of MST of $$$G$$$, $$$T_{\leq x}(G)$$$ be the minimum spanning forest form by edges in $$$G$$$ with weight $$$\leq x$$$, $$$w_e$$$ be the weight of an edge. Then we have $$$\sum\limits_{G \in S}W(G) = \sum\limits_{G \in S}\sum\limits_{e \in G} w_e = \sum\limits_{G \in S}\sum\limits_{e \in G}\sum\limits_{x = 0}^{M - 1}[w_e > x] = \sum\limits_{x = 0}^{M - 1}\sum\limits_{G \in S}(\#\text{ edges in spanning tree of }G \text{

wait actually it's more like this: let's say you have a problem where n<=2e5. so you know your code has to be nlogn or better. then, as your are designing your algorithm, you just need to store a single boolean in your head: "is every part of my algorithm so far <= nlogn"

It doesn't. You can only move down or right, you can't wrap around the grid yourself (only the rows cyclically shift). So the only way to get to cell $$$(i, 0)$$$ is from cell $$$(i - 1, 0)$$$, and so $$$g(i, 0, x)$$$ depends only on $$$f(i - 1, 0)$$$.

spend too much time on D couldn't do E. Wack round

probably your not getting each answer fast enough

2025=45*45 probably the only perfect square number year throughout my lifetime

Although it may seen very distant right now, I believe that if I continue studying hard I can reach cm next year and be part of a team that reaches regional finals in ICPC

yes bro hope for a good contest before 2025

You need to store the answer for every node after finishing, here what I would modify: Code

Hopefully I can gain some of the 100 ELO I lost after getting hacked last round.

I think you misunderstood, we can move only to the down or right and thats not about shifts : we already know how many shifts we did and tmp[i] is the ith row AFTER shifts. so we can simply write : tmp[j] = min(tmp[j],tmp[j-1] + a[i][(j+shift)%m]) (if you still did not understand, proof is -> code, you can submit it)

probably it can be (over)solved by root rotating and mex on segment queries. supporse we know mex for each subtree initially (root=0). if r is root and c its child we can rotate this edge to make c as root. how will mexes change? mex(c) will be mex of all numbers and mex(r) will be mex of numbers except subtree of c (current subtree) so current subtree is defined by oriented edge. in tree with root 0 this will be subtree or... undertree (how to call it better?). if we concate euler tour of

No, 6 tasks is usually a GM-level performance, 4 and 5 tasks are somewhere in between CM and M (but a lot depends on the speed), fast 3 tasks is usually expert level and lower results are for specialists, pupils and newbies. Although this may differ -- some rounds are harder, some are easier

is there any correlation between atcoder and cf rating ?? Like a multiple or something??

Exit code is -1073741819. Это значит что в твоем коде присуствует выход за границы

Actually, you can easily make it with magic

Guess what will happen tomorrow: [image]

Because of some serious problems, this contest might be the last abc this year. So please treasure this precious oppotunity!

It's $$$O(n^2 \cdot log(n))$$$ actually. It'll be $$$O(n^2)$$$ if use std::unordered_set instead of std::set

True. And it's much harder to get the same performance at Div.1 than at Div.2.

This code seems to be for problem CSES — Distinct Colors . If you did small to large, the complexity would be $$$O(n \cdot \log{(n)}^2)$$$, but this code is $$$O(n^2)$$$

This would make sense if Div1 + Div2 had higher rated participants around 1500 rating, but why would that be the case?

we can do it with suffix array and lcp imagine the lcp array form a histogram where the height is the length and the width is the frequency then if we used monotonic stack to get the prev greater, next greater we will be able to know for each substr how many times it exists then for each length from 1 to lcp[i] it should has duplicates equal to the width in the histogram

i think these are good resources errichto's video blog

Can anyone elobaroate more on stress testing ? any thread or source to learn more ?

ive created stress testing script which i use to find failing testcases which can be used only in practice ill see if i can find a way to host it bascially what it does it, u need to submit 2 codes, a correct code and your failing code, and then it will run almost 100 small tests, and tell u which test had a different output from ur correct one bascially what we do manually will be done by computer under 1 min

This in regard for the my solution to problem 2049C of this round. I got a notification regarding the coincidence of the solution. This was due the coincidence of the algorithm used in the problem. I wrote the code myself. Also the style of writing the code matches with my previous submissions. These links were availaible previously of the contest link and link . Using these links it was obvious that the solution becomes same. I request the Codeforces Team to review my submission. E

This in regard for the my solution to problem 2049C of this round. I got a notification regarding the coincidence of the solution. This was due the coincidence of the algorithm used in the problem. I wrote the code myself. Also the style of writing the code matches with my previous submissions. I request the Codeforces Team to review my submission. Evirir

Yeah I think all of these things are things you can do through extension on SOS. To handle wildcard, essentially we add a new "digit" to every position that depends on every other digit. This sort of transforms the original linear dependence in SOS dp to a tree Normal: a -> b -> c Now: a -> b and a -> c Some very messy code that does this: (in base 3, 0 encodes wildcard, 1 is a '0', 2 is a '1'): ~~~~~ int div = 1; for(int i{}; i < n; ++i){ for(int m = amt-1; m &

Done. Here is it.

Upd:I mean this reply used AI for translation,not that blog.

Did someone encounter clang++: error: linker command failed with exit code 1120 (use -v to see invocation) for some code that can run locally with gcc and clang? For local testing I am using the docker image.

May I proceed with the translation and subsequent relocation of the content to Luogu, a Chinese online judge platform? (I can't speak English well,so it use AI for translation.)

It think it's refer to Near Protocol You can check current price (to usd) here : https://www.binance.com/en/price/near-protocol

I think that this is like a nonrandomized quicksort. It might actually work more efficiently since it doesn't need to use randomness. Pretty cool stuff. Also, I think it would technically be $$$O(n \log A )$$$.

while I do not like edge cases myself, I rather prefer it not be shown, because it allows you to push towards a near-perfect solution with minimal cues.

Damn, that's smart as hell. So this entire trie traversal would end up being something like O(L*2^L) right?

Sometimes when in test cases >4 where i get wrong answers then to check where can i change my code from this we can learn to optimise the code as well. else we are only getting that wrong answer on test case x,y... Because all the edgecases and actual logic of code is being checked in hidden test case 2..

In my experience, the length of the contest affects the calibration of the problem rating (at least in terms of implementation difficulty). I think it's because in a longer contest more people will have time to solve it so the user rating of the average solver will be lower

i can confirm he does know infinite techniques, many more than me

I encountered an incident where the code was found to be similar. Here are two submission records: https://codeforces.com/contest/2040/submission/295612672 https://codeforces.com/contest/2040/submission/295614629 BUT I'M NOT A CHEATER! Firstly, I assure you that I did not collude with him in the code, it just happened to be similar. Considering that this is a very typical problem and there are not many parts that need to be written, it is very likely that these two codes happen to be s

https://www.zerogpt.com/

Yes your approach is right, to implement this here is a simple idea — for example there are 3 elements in array: suppose — a1 has prime factorisation p1^q1*p2^q2*p3^q3 a2 has prime factorisation p1^q4*p2^q5*p4^q6 a3 has prime factorisation p1^q1*p3^q7*p4^q8 (p1,p2,p3,p4 are primes and q1...q8 are constants) Now total count of each primes : cnt[p1] = q1+q1+q4, cnt[p2] = q2+q5, cnt[p3] = q3+q7, cnt[p4] = q6+q8 Now these total counts basically represent the counts of primes of prime factorisati

Codeforces Round 994 (Div. 2) Same, I apologize for this mistake. I'll be more careful in the future. I wasn't aware of the rule, so I hope you can avoid any penalties or blocking my account.

Where did you get this statistic? Majority of voices in your head?

I guessed that it could work and the number of elements looks like it's $$$O(\log w)$$$ level so I just wrote it, without any proof of complexity. The solution just popped in my head and I thought "Hey, maybe this can work" and wrote it, and this really looks like a solution that should be tested during a full-feedback contest :) idk whether the solution is right or not but it passed all the tests.

This is not ragebait, this is an opinion shared by the majority of Codeforces users.

Mastering CP is a mix of consistent effort and deep understanding. Talent might give some an initial edge, but long-term growth comes from persistence , learning fundamentals thoroughly, and thinking critically beyond just solving problems. You must always upsolve, analyze, and extend problems ; this will definitely deepens algorithmic thinking. As Radewoosh noted, passion and immersion matter most. If you truly enjoy CP, improvement will feel natural over time. Instead of com

Well if you multiply all array elements, you may not be able to fit it in any data type. I have better approach lets say all the elements have some factors if we map all these factors together, we should get number of ith factor a multiple of n. This is same as taking nth root.

To unfuck my life and make enough money to afford a place, I wanna stay as far away as possible from my dysfunctional family. I always felt jealous of people who had the privilege of being born into a stable family. CP used to be my comfort zone, it sheltered me from the brutal reality and gave me a purpose to strive for. But I've realized that the more I indulge in CP, the worse life becomes and my lack of accountability will eventually catch up to me. CP is to me what alcohol is to an al

It can be proved that while adding new numbers in the knapsack, the minimum number $$$s$$$ that cannot be formed is monotonically non-decreasing, because all numbers from $$$1\sim s-1$$$ can be formed by the original knapsack and will not be affected by the numbers put in. Let $$$f_i$$$ be $$$1$$$ if $$$i$$$ can be formed and $$$0$$$ if can't. After adding a new number $$$s$$$ in the knapsack and updating the array $$$f$$$(for $$$i=n,n-1,\dots,s$$$ let $$$f_i=\max(f_i,f_{i-s})$$$), iterate

Sorry if I caused misunderstandings. We need to repeat the operation (finding $$$s$$$ and appending $$$s$$$ to the end of $$$a$$$) until all the numbers from $$$[1,n]$$$ excluding $$$k$$$ can be formed, instead of doing the operation once.

I just did the problem E found it easy and fun !! // this is code signed main() { int N; cin >> N; vector<vector<int> > T(N); vector<int> degree(N, 0); for (int i = 0; i < N - 1; ++i) { int u, v; cin >> u >> v; u--; v--; T[u].push_back(v); T[v].push_back(u); degree[u]++; degree[v]++; } int result = N; for (int u = 0; u < N; ++u) {

I changed the degrees, as well as $$$(1 « i)$$$ if written like this, it will contain a maximum of $$$int$$$. If you write like this $$$(1ll « i)$$$ then it will be $$$long$$$ $$$long$$$ here is the modified code Code

100% agree just give up if you can't reach high rating long time and you not talented. It's just wasting of time and effort in vain

As 2024 is coming to a close, I would like to wish everyone a Happy New Year in advance and good results in the upcoming competitions. I am very passionate about algorithm programming.

How long does it take you to solve them in practice? Of course if it's longer than 2 hours then you won't solve them in contest. If you can solve them quickly in practice but not contest then it's probably nerves and you can try some virtual contests. Otherwise, what tends to take the longest when you solve? Coming up with the logic? Implementation? Debugging WA/RE?

well, can be, i was in a time where i haven't free time or hang out with friend for doing cp (it was two month), i was training for a icpc. Since i starting programing a year ago and start serious cp like 5 months, i dont know if i making a good progress or wasting my time... i trained in different platforms like codeforce, vjude, uva, cses, geekforgeeks, leetcode, hackerank and some other. I was mainly practice dp, and i don't if it was worth, i can do some basics and no-clasic dp, but i ke

This sounds like an abstracted version of SOS DP: https://codeforces.com/blog/entry/45223 Or maybe I am misunderstanding the difference.

orz idea, but how would you implement this? As in, how would you write a trie traversal that folds the tree?

Help as many people as you can; this will naturally increase your contribution day by day ... (sory for my poor english)

OK, thanks, may be my solution get TL becasue the too big constant.

Oh, this function runs because there are at most $$$\sqrt(n)$$$ different numbers in one block, isn't it?

It's possible, check out the only solution in the contest time. The actual solution only improves the time of updates and queries in the $$$blocks$$$ array.

If you do block decomposition of size $$$O(n^{1/4})$$$ on the $$$blocks$$$ array, you could also attain $$$O(1)$$$ update and $$$O(sqrt(n))$$$ query complexity. I think it's more intuitive, but it seems a bit slower.

Can someone give me advice to solve problem G in $$$O((n+q)\sqrt n\log(\sqrt n))$$$ or it is impossible, thanks. 298614069

My solution was a bit more combinatorial in that I used complementary counting. Also, I used more tree DP instead of summing the DP values from the first half of the official editorial. Here's my submission. Feel free to ask questions.

[image]

codetiger927 orz

become candidate master specifically 2000 rated by august 2025.

I'm sure that any problem you do will help you both on leetcode and codeforces.

i can predict that 2024 will be in either question or constraint

Or another way to prove it, is that the maximum xor you can achieve by only using numbers in the range $$$[0; 50]$$$ is $$$63$$$, and the minimum one is $$$0$$$, so you can have at most $$$64$$$ different numbers. If you have $$$7$$$ numbers such that for every subset its xor is unique, then you will have $$$2^7 = 128$$$ different numbers; that is a contradiction. * If for subsets $$$A, B$$$ their xors are equal, then you can find $$$C = A \, \Delta \, B$$$ (symmetric difference) whose xor i

reaching ICPC world finals (this year we were so close to do it) reaching Master on codeforces

good that u quit... but the sugestion u gave is still childish. Not everything can be answered by google...

No? If you already have some set of testers and you add like 10 blue testers that would not make round worse. Quantity of good testers matter, not the percentage.

2025-2026 ICPC Asia Pacific Championship (I guess I'm going for 2024-2025 one, but my two teammates(and only other Masters in my university) seems to graduate. So It would be harder than this year.) or becomming GM (also hard to be)

I've just recevied this message: Message And when logging in Ton Society ID app I get this: "You are not in the list of the badge receivers". Could anyone help me? I've provided my wallet adress before deadline. (Already fixed. Check the wallet adress matches the provided one in codeforces)

Is it ICPC format (Penalty , etc) or Points format (Points per problem ) :D ??

This is incorrect. The correct rule is the following: Instead of doing the sums, we just consider the blocks of 3 as number together. For example, consider again 12,332,455. Then, the value is 455 — 332 + 12 = 135, which isn't a multiple of 7 so 12,332,455 is not a multiple of 7. If we consider adedalic examples, then we get 022 — 1 = 21 which is a multiple of 7, so indeed 1022 is a multiple of 7.

Wow there's English statement on Luogu contests! Finally I can begin Luogu!

It's weird, from my testing, it seems like: gcd(X % Y, Y) < __gcd(X, Y) < gcd(X, Y) in terms of runtime

Uhhhhhh, Griffith is the absolute worst character in the entire anime world

Interesting problem.I’m waiting for the answer.

So, can this make that much difference $$$?$$$ On doing private mashup, I got running time as 1312ms (more than twice).

Link Here someone had the opposite result than you.

I had a similar experience, and like you, I took a time limit with __gcd time limit and exported it with GCD. This is because GCD uses the recursion method and takes much less time than normal

À mà bạn có thể gửi code lên đây luôn tại mấy cái link nó hỏng rồi mình không xem được

Mình tìm bạn mấy ngày trời này :). Bạn có làm sol trên trang usaco đúng không ( https://usaco.guide/problems/ioi-09-mecho/user-solutions ) Cho mình hỏi sao khi chặt np thì bạn để thêm cả điều kiện: Time[start.fi][start.se] > current; Còn nếu thiếu thì lại không ac vậy ;-;?

It was a round which has got nearly 5000 downvotes. It had an OEIS-solvable problem on H.

Happy New Year Guys, and I hope you get a lot ACs this yeart!

I would recommend to make this blog more informative, there is no mention of the college name or the type of course (credit based or not), Is this college's initiative or will be done at club level?, and there is no info about you?. etc. These information can make the blog more legit and bring more attention to good CPers.

That ishaandas1 for green testing is more like specialist or expert , visit his profile his ranks are under 3000 and if on this basis if ranks were calculated than our rating can be deducted

I guess it will be rated as Div.1+2, because 3 hours .

Happy new year! Hope to gain more rating in the last contest of 2024, good luck and have fun!

In solution for f, shouldn't the condition $$$if (dp[i][val][fl].cnt > 0)$$$ instead be $$$if(dp[i][val][fl].cnt >= 0)$$$?? As we are modding and that may cause cnt to be zero. Edit : Now they fixed it

Looks like meth business isn't giving returns.

says the guy who cheated his way to become expert. you sure are quite proficient in using google, especially during contests

I hope I no longer need to use magic to change color :)

This contest is fantastic for beginners to practice and understand trees and their implementations. Absolutely loved it.

Hope that the round can be much better than Goodbye 2023

Why are you playing Clash Of Clans in a coding contest :)

problem — divisors analysis — think, there should be more explanation on how this is done — have to carefully divide the exponent by 2 case1 — all divisor count is even can simply find the square root of number by considering only half count, case2 — there is atleast one divisor with odd count — as this divisor occurs in half the total divisors, considering only half count means dividing total divisors by 2,

Reach GM in the next contest I do. (I attempted this year but I failed [not gonna participate in Goodbye 2024] so I want some extra time)

You can find three more red ones on the line "helping with preparation on Polygon (including testing)" :)

there isn't a thing called "catalogs" in case you don't know.

odd = int(input()) for n in range(1, 8 + 1): # factorial(8)=40320, if not found until 8, then forget it times = math.factorial(n) for d in range(1, 9 + 1): # Check every possible number if int(str(d) * times) % odd != 0: break else: print(n) # If n is greater than or equal to this value, it must be divisible by odd break else: print(None)

very educational solution. thank you my bro

https://codeforces.com/contest/1618/submission/298437227 can anyone help pls?

Oh alright, well you can ask if you ever change your mind. In the beginning, yes. But then I figured out that it was always the test's problem. If I scored too low, it was the test's problem. If I scored too high, it was also the test's problem. In fact, I've figured out that every single test is bad (even the US army test) except for one called the Stanford Binet 5th edition. And I haven't taken that one.

I am kinda scared to do an iq test, as i believe i will subconciously limit myself if i see a number that i do not like, but thanks anyways. Were you ever discouraged when you failed or got a bad score on the tests?

No, but there was a time when they had maybe $$$6$$$ or something forms of the test that they would give out to the recruits, and, of course, some "recruits" started to memorize the questions and sell them, so those forms are definitely out there somewhere, but that was in the $$$80$$$s. Nowadays, it is pretty much an online test, and its question bank is huge, so there is no point in trying to memorize all that. Anyway, there are some free tests out there that come close to its va

it's basically bivariate normal distribution with Talent and Effort. both important. e.g) people who put in the same level of effort will be mapped to a rating based on their talent. and vice versa but I do believe talent has a higher SD than effort. It's very easy to increase your rating if you are smart, and some rating is simply unachievable even if you push top 0.1% of effort with median talent.

Is that test publicly available? And also, a bit off topic, but have you ever taken an iq test?

Ahh I see. But I don't think it's the same as race. Race does not inherently impact someone's ability to do any particular job, while IQ does. Height is somewhere in the middle I guess as it depends on the job. It seems stupid to not allow IQ bases hiring, but I guess that they circumvent it by using problem solving. Based take ngl.

I should have been more specific — they are banned for hiring purposes (at least here in the US). Discriminating based on IQ is like discriminating based on race or height. All three would be allowed in a true free market economy. Also, the US government thinks it's all high and mighty for banning IQ tests but the military uses one of the best IQ tests of all time to sort out its recruits.

i have seen consistency surpass talent in cp world, so keep coding, i was first among my friends to become pupil but they were consistent they became specialist i am still trying

As a tester,I've lost a chance to gain rating in the end of 2024.

can someone please explain why it is giving tle[submission:https://codeforces.com/contest/2043/submission/298310318]

As a tester, wyrqwq is cute and problems are more interesting than Goodbye 2023. GL&HF!

For problem $$$D$$$, I miscalculated that if we consider all possible pairs of integers from intervals $$$[l,l+10)$$$ and $$$(r−10,r]$$$, we will find at least one coprime pair. But, the problem passed the system tests. Can anyone prove it or uphack the solution? Submission — 298478051

Hacked I guess it's quite difficult to make a set of test data that TLEs all possible choices of $$$B$$$, so $$$O((n + q) n^{\frac{2}{3}})$$$ solutions can get accepted, although they can then be hacked (including all but one of the in-contest submissions lol)

Yeah, this was an error, thank you. Will be fixed in a couple of minutes

If a prime is greater than the length of the segment (for example, $$$30$$$), there is at most one number in the segment that is divisible by it. So, if a number appears in $$$15$$$ or more pairs that are not fixed by primes less than $$$30$$$, every such pair should be "fixed" by a separate prime.

could you explain why though? its my first time seeing long long giving tle..usually i consider it only for memory issues

Since $$$a_i=2x_i-x_{i-1}+m$$$, this means $$$2^{i-1}a_i=2^ix_i-2^{i-1}x_{i-1}+2^{i-1}m$$$, so adding for $$$i=1$$$ to $$$i=n$$$ gives $$$a_1+2a_2+4a_3+\cdots+2^{n-1}a_n=(2^n-1)x_n+(2^n-1)m$$$, so you can solve for $$$x_n$$$.

Because the size of the vector basis can't be more than 6 because $$$a_i \le 50$$$.

Think about this: why do companies use competitive-programming-style problems in their interviews? Do they really want to hire people who can implement Dijkstra's or toposort? Those things are useless on the job. What they really want is people who are good at problem solving (aka have high $$$IQ$$$). The competitive programming interview is just a disguised $$$IQ$$$ test. The reason why they don't use actual $$$IQ$$$ tests is because of two reasons: $$$IQ$$$ tests are banned in so

ask yourself why do you need rating so much, it's worthless to run behind rating, why are you are making a loveable thing like problem-solving non-lovable? if codeforces platform didn't had rating at all, would you solve problems? if the answer is no, leave it all and first find the beauty of this hobby.

I know right? I just solved it performing necessary operations on each row and column once, and i repeat it 100 times. It passes all the tests.

aa... what are these (N) points at the end of blog?

If you do sliding window on the testers you can get various majority tester ranks .

you are asking for too much, be reasonable you are already orange there is no going back

else " quit cf, touch grass, and enjoy life " Count me In !

I got the Griffith Handle . Best thing i could have asked for !

Cool, looking for more collaborations with TheForces from you ;))

yeah agreed. why is it called Shoelace formula tho? is it because of zig-zag pattern it makes when multiplying coordinates?

2043F - Nim . Nah, I've forgotten the dp way for finding xor = 0, and I wrote meet in the middle, because we can remove all the elements except for 7. So I can test if I can leave exactly 1 element, 2 elements, ..., 6 elements. And to check if I can leave 6 elements, I can check a pair of (3, 3) elements. 298456482

reach master and, as always, hopefully not die

Wish is stable 2800. Dream is touching LGM(as always).

please god make me Master please!!! & obv ICPC with my team V. Imp.

To comeback to cp after university entrance exam :)

298450724 see this

Problem C can be solved in O(n) time 298450724

Thank God. Now Codeforces is a more secure place

I solved exactly like this, would be very nice if someone prove or hack it

Pass Turkiye 2nd tier olympiads with gold medal

To get an internship for 2026 and comeback to cp

bro what about choosing levels the level it suggests according to my rating cant give me required progress

People get more points in their first five contest(for the original rating is in fact 1500),to cover up these points.(Nowadays the rating is counted from seemingly 0 instead of 1500 to avoid just dropping ratings for newbies)

I saw a comment that said you only have to check the ranges $$$l, l + 5$$$ and $$$r - 5, r$$$. At least one of them will have a $$$\gcd$$$ equal to $$$1$$$. During the hacking phase, I tried pretty hard to find a counterexample, but I couldn't.

I've tried it a few times, and it's really helpful for practice!

I think this is a good thing , since then , the cheaters they submit all there solutions at the end of the contest , so the total penalty they get is quite high, thus not benefitting the cheaters.

I really don't know why they choose the weakest test cases for the pretests this contest pretests was extremely stupid, over many hundreds of hacks and system test failures happened There were so many, it wasn't normal. It was really awful the one who set the pretests owe an apology to every participants. AcidWrongGod BledDest Ferume

passing $$$a$$$ and $$$b$$$ as references to solve function indeed makes it faster 298429050 , it is near to TL tho.

Translation of the pinged comment by himself(by chatgpt, maybe inaccurate)： I originally wanted to clarify, but I decided not to; the more I describe it, the worse it looks. Generally, I don't pay attention to the rules of "low-value competitions" online, and as you can see, I can still achieve high scores even if I don't follow them. This is not against any regulations or laws in my country; at most, it might provoke arguments, which is quite interesting. However, I usually d

case-2 as you described, max and min subarray sums to the left and right of x, they contain x right?

$$$-90$$$ only. Carrot predicted $$$+32$$$ before my A got hacked.

THIS IS MY PERFECT VICTORY! I WIN!

import math odd = int(input()) for n in range(1, 8 + 1): # factorial(8)=40320, if not found until 8, then forget it times = math.factorial(n) for d in range(1, 9 + 1): # Check every possible number if int(str(d) * times) % odd != 0: break else: print(n) # If n is greater than or equal to this value, it must be divisible by odd break else: print(None)

i added the dark reader extension and changed it to filter for codeforces, got rid of it for me.

I hope that I'll manage to hack it faster than you

second solution, i'm waiting for the submission.

For someone who doesn't like doing casework (like me). Solution for B — 298416560

Which solution did you hack? I have made a test for the second solution, I am waiting for the end of the system testing to hack the submission.

Your solution is very similar to this: split initial array into $$$O\left(\sqrt{k}\right)$$$ blocks and calculate "MO segment" between each pair of blocks. Then for each query use appropriate "MO segment" to answer it in guaranteed $$$O\left(\dfrac{n}{\sqrt{k}}\right)$$$ and then return this "MO segment" back to initial state. Even though it works in guaranteed time, it is probably slower in practice then your amortized solution.

Dude I was almost wondering how you're still Pupil. :⁠-⁠D

Why though ? In D, I simply used binary search on every range of coordinates Santa would travel on the two maps is maintained. I just deleted appropriate coordinates just once. Here is the submission .

It is rated unto 5 stars only, I posted the wrong blog by mistake, really sorry for the inconvenience. Updated it now.

Cool technique! Looks like its complexity is amortized $$$O\left(\dfrac{n}{\sqrt{k}}\right)$$$ per query and $$$O(k)$$$ per update, where $$$n$$$ is the length of initial array and $$$k$$$ is the number of so-called "MO segments".

I would say use the idea of euler's formula, x <= 1e18 would have atmax 20 primes. Let the length of range be L, after using each of the primes to discard numbers, number of coprime numbers in range would probably be $$$N = L * (1 - 1 / 2) * (1 - 1 / 3) * (1 - 1 / 5) ...$$$ for first 20 primes thats about $$$0.12L$$$. So for about $$$L >= 10$$$ you should expect to see atleast one coprime number. I know this may be a totally wrong way of proving, but thats just my intution for this

u can look at the standings, almost half of all participants solved it in a way too complicated way, just have a look at their codes.

I mean that put such a naive problem at F is not proper. Maybe it's suitable for C or D i think, putting here will only make ppl confused and solve it with methods too complicated.

Who told you, you are funny? there's a thing called silence if you don't have an answer for a question, in case you don't know

I divided the problem to solving for every bit Then what i saw is if i want to correct a position i might need to change a whole row or column, which can trigger new changes in other places, I can do them but then it can change other people ect... So the question is given that i need to trigger changes in some specific places am i gonna be able to do all the changes that will be triggered, or will it go into an infinite loop? Thats when I understood i could model "triggering" a

I guess you arent adding the element X on both start and end try it

bro it’s your problem if you can’t notice it. calculating time complexity is also a part of competitive programming.

[image]

personally i didn't model it to a graph,instead i solved it by just in case here is my submission 298383591 my time complexity is O(nm*log(max ai)*log(n+m))

One of the ways to solve this problem is compute $$$10^x \, mod \, p$$$ for example for $$$p = 7$$$ the sequence of the answers of this expression is: $$$[1, 3, 2, 6, 4, 5]$$$ (sequence of answers is a periodic with this period) so $$$6$$$ should count $$$n!$$$ it means $$$3 \le n$$$

F is totally shit. the O(nv^2) sol is naive and should't appear at this pos actually. if u see the rankings u can find a lot of sols using very complicated algorithms (and some get TLE). if u swap D and F I bet there will be much more solves. (P.S Me myself wrote a sol using DP and FWT in contest, which is O(nV*log^3V). although it passed i'm so sad.)

have u seen system testing happening?, because im pretty sure that it shows "final standings" even before system testing for educational/div3/div4 rounds

Do a Binary Search on your code: WA -> range too small TLE -> range too big

чекай)

i think it doesnot make much sense since it's not visible for everyone, in fact many people doesnot install this challenge, so you are not a real magic tourist

2024 is meaningful to me.It is a perfect way to end this year to take part in the expecting contest!

5700 ms using 5 segments))) https://codeforces.com/contest/2043/submission/298383495

50 отрезков почти за такое же время работают

Oops.My bad.Hope this year will be better.Anyway,happy new year.

its gonna start in some time ig, but it hasnt been done yet

all the solutions will be rejudged in the upcoming system testing, dont worry lol

Thanks for the blog! My thoughts: For static range minimum queries, it is possible to solve without a segment tree (using a sparse table for example, which can solve the problem in $$$\mathcal{O}(n \log n + q)$$$) For range XOR queries, you are overcomplicating it. It can also be solved with prefix XOR. Let $$$pref_i=a_1\oplus a_2\oplus \dots\oplus a_i$$$, then $$$a_l\oplus a_{l+1}\oplus \dots \oplus a_r=pref_{l-1}\oplus pref_r$$$. This is true because $$$x\oplus x=0$$$ and $$$x\oplus 0=

Never seen earlier a contest's blog with downvotes.

As the only "LGM" "writer" of this "specialist round" (though I proposed no problems actually), I wish you good luck, positive delta and a happy new year! (I believe an lgm may make the round look better :)

As a tester, I went cyan, which is the same color as wyrqwq 's, to express my adoration for wyrqwq .

Aren't 2 $$${\color{black}{l}}$$$$$${\color{red}{gm}}$$$, 6 $$${\color{red}{red}}$$$ and 10 $$${\color{orange}{orange}}$$$ good enough, how many do you want.

Okay but why does searching first 50 mutiples of G and last 50 multiples of G in bound(l,r) works?

As someone that uses codeforces entirely zoomed in (because i don't see a pito), I wouldn't be able to see the snowflakes at all, so I don't really like this idea. Also it completely kills the snowflakes for mobile version.

but then it's not clear what's happening with gcd, so it may well be TLE

As a tester for the first time, what can I say. The other testers are very good at expressing, but I can only say that this contest is the best one of the year.

it seems that p = a + i*x can exceed long long, maybe that's the problem

As a tester, the problems and the problemsetters are cute and wish you all have fun in this contest!

As a tester, wyrqwq is cute

Well then the birth of Jesus wasn't in fact the greatest Christmas present of all time, since it didn't happen on Christmas? Just a small thought.

Bro i just realized what you talking about. I though all of the example base64 can only converted to BufferSlice, but base64 on lz4 example can be decoded to regular string, and when decode it using caesar cipher algorithm with shift 3, its become 24 random words. And its same length with TON Wallet recovery phrase.

Congratulates!!!!! No 74TrAkToR RedMachine-74 for Goodbye again!!!!!!!!

As you may know, the c++17 compiler is 32-bit, and the c++20 compiler is 64-bit. In the 32-bit compiler, size_t is uint32_t and in the 64-bit compiler, size_t is uint64_t . Therefore, when you attempt int i=a.size()-1 on Line 127 when a.size() is $$$0$$$ (consider sample case 4), although both versions encounter underflow, the outcomes differ. In the 32-bit version, the underflowed result is $$$2^{32}-1$$$, which is assigned to a long long (note the #define int long lo

and 251Sec for being invited to test but have no time to do virtual. :(

Read Luke 2:7-8. Why were shepherds out on the fields looking after their flocks if he was in fact born on December 25th, in the middle of the cold winter?

[image]

As a parent, I can confirm that wyrqwq is cute!

After a small amount of hacking I now have first solve on G... although my solutions will most likely fail system tests to my own hack cases lol (they are vulnerable to being hacked in the same way)

So basically, ( B , A ) + (an imposter $$$x$$$ sneaking in among the $$$1$$$s, with a sprinkle of $$$G$$$ on top) = ( C , D ).

Hello, I recommend the page Cses for training some clasic problems. Also I recommend you to do all the contest as possible and to upsolve all the problems that you didn't solve in yours contests as possible, because in these problems were where you haven't some tool or some knowledge.

I think ABC is the normal. I don`t no why people hate B, this is not hard mathematical problem. D is very random. AC for proof for me and i think for many peoples. Round is the normal, but i was very stupid on D (+5).

Instead of divide and conquer, you can keep a sorted vector of pairs {block index, minimum in block} which you can update naively after each query with a binary search.

I feel like problems E and D should have been swapped,E is way easier than D.

You have a point, there have been some difficulty issues in the past. But I don't agree that today's B is really as intellectual and math-involved as your (and many others') comment suggests. Yes, it has a solution which is very math-involved (checking all divisibility rules for $$$3$$$/$$$7$$$/$$$9$$$). But this is not the only way to solve the problem. There are other, much less "mathy" ways. I will illustrate one of them in the official editorial.

I am one of the pixels on the mamali 's laptop and I can confirm I was part of the sentence "pls give me the code".

im in a very similar situation too, looking for advice

Holy shit dude, i thought i had worst contest.

Remember, 99% of newbies quit just before they reach LGM performance

Yes, I also fixed C = 10 but I don't understand why....I just made a guess... I want to know why this will work.

Max prime gap up to 1e18 is around 1600, Using that bound the solution will be too slow (1600^2 * 60 * 1000).

Can you please tell me why and what is correct way to get to 5 instead of 1?

yeah it works for any polygon and the vertices should be in order either from 1st to last or opposite

$$$10^{x} \mod 7 (x>=0)$$$ has a period of 6: $$$1, 3, 2, 6, 4, 5, 1...$$$ The sum of the first 6 numbers is 21 which is divisible by 7 therefore $$$dddddd$$$ is also divisible by 7 and hence n-digit numbers (where n is divisible by 6 and all digits are the same) is also divisible by 7 because the period is 6.

$$$\displaystyle \underbrace{ddd\dotsc dd}_{n!} = d \cdot \sum_{k=0}^{n!-1} 10^k=d \cdot \frac{10^{n!} - 1}{9}$$$ $$$\displaystyle d \cdot \frac{10^{n!} - 1}{9} \equiv 0 \pmod{7} \implies d \cdot (10^{n!} - 1)$$$ since $$$\gcd(7,9 = 1)$$$, so we have two cases: $$$d = 7$$$, trivially divisible by $$$7$$$ $$$10^{n!} - 1 \equiv 0 \pmod{7} \implies 10^{n!} \equiv 1 \pmod{7}$$$ We note that $$$6$$$ is the order of $$$10 \in \mathbb{Z}_7$$$, by Fermat's Little Theorem (or Eu

In Problem D Can anyone explain the correctness of finding the co-primes within very less range i.e 10 iterations in this solution of mine .. sadly it got accepted after the contest :( https://codeforces.com/contest/2043/submission/298300293

Problem C is a bit harder to be Div.2 C also, in my opinion D is much easier than C (may be because I am a math lover XD ). However, Great contest as we used to from BledDest !

[image]

this is called Shoelace formula, if someone wonders. https://en.wikipedia.org/wiki/Shoelace_formula

If you find two prime $$$p, q$$$, it's definitely good, and the distance between two prime is around $$$O(lg^2 C)$$$ so if you enumerate like that, it would stopped very quickly. btw, it's known as prime gap

For D, what's bounds do you have to iterate on?

One of the most enjoyable edu rounds imo IMO edu rounds.

It does not actually work, like a lot of stuff from geeksforgeeks upd: for example, if you take $$$14$$$ and $$$36$$$: $$$gcd(14, 36) = 2$$$ $$$gcd(15, 36) = 3$$$ $$$gcd(14, 35) = 7$$$ $$$gcd(15, 35) = 5$$$

First reformulate the problem into counting equal pairs rather than unequal pairs. Split into blocks of size $$$B$$$. We can maintain block_ans[i][j] which stores the answer for all the blocks in the range $$$[i, j]$$$ (there are $$$\frac{n^2}{B^2}$$$ such ranges), and block_prefcnt[i][j] which stores for each value $$$i$$$, its count in the first $$$j$$$ blocks. They can be updated in $$$O(\frac{n^2}{B^2} + \frac{n}{B})$$$ and using those values you can do queries in $$$O(B)$$$ (the i

woah woah woah bro there has been much worse

Never knew that Div2D could be solved with just two nested loops.

As we know that GCD(N, N + 1) is always 1 and we need to find the maximum difference of the pair with GCD 1 so GCD(L, R – 1) and GCD(L + 1, R) will be one as GCD(L, L + 1) and GCD(R – 1, R) will be 1 so there cannot be any common factors between (L and R – 1), and (L + 1, R). How is this True? What about L = 15 and R = 64

Actually, prime numbers are pretty regularly distributed. So you can always find a prime number on a long enough segment. That's why yours and similar solutions work.

https://www.geeksforgeeks.org/find-a-co-prime-pair-with-maximum-difference-in-a-given-range/ D ^ Edit: doesn't work oops

I have a proof for 50 x 50 bound, but unfortunately it gets TLE

thank you for giving worst ever-experience on Christmas in my life.

i'll never participate in edu. rounds again

https://codeforces.com/blog/entry/133293 Once the round starts, you cannot change your registration type.

As a testuwuer, I was surprised to find an ICPC contest with div.1 quality problems. Really high quality stuff! American ICPC, step it up ;-;

this is a very simple explanation of how maps works . at your level just of it like this .Maps are just array but the index can be anything. a good exemple is you can declare map<string,int> mp. mp will have a key of type string. so you can do for exemple Spoiler but you can't do this with arrays . Also wikipedia was right about find .There is this thing in c++ called iterators. you can use find on everything that has iterators like this map<string,int>mp; mp["fi

shenfe1 (won't win IZhO 2025) Wansur d imash1 Der_Vlapos Issa mispertion20 Tima5 Some Dudes from Kazakhstan

Oh good point, thanks for pointing it out! I have edited it

Dedication you put into a website just to entertain others is just insane. Website is also very smooth and vibrant. Thanks for this amazing project!

Hello guys, merry Christmas(also the Eve) and the New Year is around the corner. I also hope you can be happy all the year!

Is there any way to mirror it on cf gym or make problems public on codechef (for online judging) :)

Probably not master strength right now, but I should be able to easily maintain CM if I simply stop leaving halfway through contests.

I'm trying to gain 100 rating points to finally be green (for real without magic)

where can i download KFU onsite olympiad 2024 problems ? I couldn't find it online. I just want to practice those problems before round starts. Thanks!

a nhphant help me help me

Hey I really like the style and animation of this tool, but I couldn't help notice the content similarity to this project I built three weeks ago. Since I open-sourced it under the MIT license, anyone is welcome to clone and modify it :) But I would still greatly appreciate any acknowledgements made when you explicitly mention "I've created"! Spoiler

Hope chennie can reach Specialist after the contest.

I had the same question and here is what I think the writer meant: Consider choosing just doing inclusion exclusion on $$$I$$$. You get a factor of $$$(-1)^{I+1}$$$. Then you have to do inclusion exclusion on $$$J$$$, there you get a factor of $$$(-1)^{J+1}$$$. You multiply then to get $$$(-1)^{I+J}$$$.

i will , thanks a lot but what is "cses"?

I was just commenting on how the original post was weirdly phrased. I don't care much about CP anymore

Really Liked this CF wrapped. Excellent work.

Attention: the contest uses some problems from the onsite stage of the KFU Olympiad, so if you participated in it, please refrain from taking part in the round. Means ??

I really apologize to everyone and I promise not to do this again

I believe at this point word choices are just a matter of personal preference (and I tend to randomly pick one when many options are present).

although my rating sucks now because I don't take contests seriously anymore [image]

Brother, the person who cheated has posted an apology . I just thought you should know.

Their souls leave their body when they get WA on tc 2 that is why they consider it soulless

Can you give a nickname or his contacts please, I want to contact him to ask a couple of questions

I wonder if binary search is included in the list of obscure algorithms XD

Funny that they are praising the thing being way more likely to replace them themselves than those they dissed.

Try solving problems on Usaco Guide. You will learn a lot of stuff on it.

I’m glad I was able to help! One more thing: none of this is my advice; it was shared with me by the previous round’s problem setter, as I mentioned earlier. I also took his advice and explored Polygon, and it was a lot of fun to set the problems. The goal of our coding club was to organize a coding contest for our college’s technical fest, so we decided to host it as a Div. 3 round on Codeforces. Unfortunately, we didn’t have any Masters in our circle either, so we followed the same advice

Thank you so much for your detailed reply! Unfortunately, I don’t have any masters in my circle, but we do have experts and specialists who are ready to help with problem preparation. We are planning to organize a Div. 4 round, where the hardest problem would be in the rating range of 1300–1900. Your advice about Polygon and mashup contests sounds like a great starting point, and we’ll probably try hosting a practice contest first to get familiar with the process. In the future, once we ha

I had a similar doubt once and my college coding club also wanted to organize a Div 4 so we contacted one of the problem setter of previous div 2 round. This was his reply: "Afaik, there's no open way to propose a Div 3 contest on CF. There is a blog by KAN explaining the problem setting process for Div 2 and Div 1 rounds (even that isn't open to all either; you need someone who has achieved a master+ rank or has been a problem setter in a past contest). Note that you can still prepare

Not a new way,already forbidden,written as the tips of message.

just two contests before new year to get blue asodifjaoidfjaiodfjo (without magic)

Definitely for test 8 it should be 49! and not 48!

Instead of prices, i kept track of negative reviews, if negative reviews is more than k, i changed my high to mid-1 else low = mid+1, whats wrong with this approach. 298136423

Hi, this is my first time participating in an educational contest, so how is the contest organized?

bro refuses to believe that we do this as a hobby i mean he's right that this is soulless but swe is more soulless and i dont have a soul anyway so

Merry Christmas in advance my fellow CP'ers

oh darn I should've thought of Caesar cipher originally when I decoded that b64 string a few hours ago LOL. Good for you though!

Dehumanizing alienating and soulless activity? that’s exactly what I’ve been looking for.

"This set will consist of blocks from the TON mainnet that will be generated after the end of the contest." Kinda sus

wait till he finds out that AI can also change background color and read mails 😂

Insane that they think their work can't be replaced when their work involves very little thinking/problem-solving, and very much coding lol.

Been facing this for a while too, gonna try out the solutions mentioned in comments hope it works ; )

why American like this? redditor particularly. do they seriously believe they get paid 200k job in the US as an SWE because they are smart and creative?

The Hacker : Spends 30% of the contest solving the easy problems and the remaining 70% stress-testing others' solutions.

yea lmao cp is "soul less", not swe where each project is just copy pasted bits from a bunch of tutorials from which you understood nothing. how stupid can one be

The original post is heavily exaggerated too: https://www.reddit.com/r/GoodAssSub/comments/1hirq5t/openai_new_o3_model_is_equal_to_the_175th_best/ Getting 2727 performance in a single division 2 contest != having 2727 rating and being "#175 best competitive coder on the planet"

For Problem E, simply use line sweep: keep track of total a[i] and b[i] at each point. Then line sweep them.

Hi everyone, do we know when the test input/output data for this ICPC contest will be available, or where we can work on upsolving the solutions?

This might be a silly question, but why is the solution for the first problem the array in sorted order when the output under the question literally says "You are allowed to not change the order."

Solved the fifth problem after seeing your hint. This hints make your editorials the best than the official one.

Here a super-easy and clean implementation of problem E using upper_bound to find number of positive and negative reviews without creating new arrays (just sorting the original ones with negative values): 297888481

It's a creative and interesting solution. Thank you for sharing it with me.

There's a $$$\mathcal{O}(m + q)$$$ solution for C. Case where $$$n - k$$$ is 0, or $$$n - k > 1$$$ is handled similarly for yours, however, for $$$n - k = 1$$$, we'll do things a bit differently. Since we know that $$$q$$$ contains of $$$k$$$ distinct integers that should sum up to $$$n$$$, but there's one missing integer, the missing integer will be $$$x = \sum\limits_{i=1}^n i - \sum\limits_{i=1}^k q_i = \frac{n * (n - 1)}{2} - \sum\limits_{i=1}^k q_i$$$, so the answer will be 1 for $$$

Think about that,you will have part of tourist power next whole year!

Sorry, I mean, how many times can I submit at maximum during the contest.

Is there any submission number limit for this contest?

:((( I am still unsure where you missed it out. but it has to be either registerTestlibCmd , registerInteraction , registerValidation , or registerGen in the first line of a main function that needs to use testlib. In testlib's source code it is like this: ~TestlibFinalizeGuard() { bool _alive = alive; alive = false; if (_alive) { if (testlibMode == _checker && quitCount == 0) __testlib_fail("Checker must end

I think if you print output for every test case separately, it will be much faster.

Be careful guys, new Legendary Grandmaster is in the town.

Your continuity depends on your passion and what you want to achieve. how can we help you with this!!! no one knows but you.

If so, more of us will tend to post funny contents to farm contribution, which, will be no good. (Not everyone will become adamant )

This round really preserved the beauty of Educational Rounds as the authors are the same... sadly, I could not attend it live and had to participate virtually.

Could you share some background about this contest, or at least what should I know? I would like to know if CPers are a good fit for this problem, or is this contest not for me? I only have math sword and algorithm gun.

The FBI told me that this round will indeed have problems.

As the brother of the same father and different grandfather of mamali , I can confirm that my little brother did cheat on exams frequently.

Erm, how do you send messages to each other. I thought it wasn't possible.

Here's something more challenging: skip problem statement and solve the problem only by reading input/output format and the samples. You are allowed to see the tests you get WA on and analyze the jury's output.

Seem that Santa's magic is not enough for us to become a tourist XD!

[image]

This contest have a system testing, then leaders' prizes are based on provisional standings?

Should have thought about your new handle a few months in advance

298058215 Can anyone tell me why my code is failing?

Yes, I can change it again next year but I promised a friend of mine that I will change it only if in the meantime I will have achieved the International Master title.

Hope chenlinxuan0226 can reach Expert after the contest.

I guess you can change handle only once, is it like you can change only once in a year and next year you can again change if you have changed this year?

Wish you all bad luck will be in the handle and only good luck left in contests.

F is so hard. I can't imagine F being a question about segments.

In the future could be a interesting type of round, but rn noup

I am mamali 's boss and i can confirm he's a cheater because he copies codes from stackoverflow

I am mamali 's pizza delivery guy and i can confirm he's a cheater cause he orders Pizza Hut (and i am a Domino's guy) Spoiler

I'm also mamali 's neighbor, and I can confirm that every time their power goes out, he comes to us to borrow codes instead of studying!

In the test above your code literally accesses index greater than m, isn’t that a problem on python?

I am mamali 's ex-girlfriend. I also can confirm he's a cheater and has to be banned permanently from codeforces.

I am the father of mamali and I can confirm he cheated in several contests. I'll punish him with my belt.

lol you could try some cases on your own but okay, here is the test: 1 3 1 2 1 1 2

I guess pretests contained only n = m cases, at least my solution that swaped n and m in some places passed the pretests.

Oh, thank you for helping me this far. I didn't think of binary search because I just brute forced the solution.

How I mean I looked at the editorial but I didn't make sense of it? We need to find the number of pairs 'i, j' such that if we remove i and j the sum of all elements is at least x or at most y. How can I use binary search? and thank you for helping me!

Yeah, but what about cases after test 3? Would it get TLE or WA or AC? Currently, I can't submit because of system testing.

What about my approach? Are there any flaws?

You can have another chance Spoiler .

Yeah, my bad and thanks for pointing that out.

Exit code is 3 means there is a path problem.So it's probably a problem from the judge's side . I tried resubmitting and got Memory limit exceeded on test 2 Try rewriting the code

I am the boyfriend of mamali and I have myself saw him cheat on me several times

I think the tmp array is not keeping correct frequency of elements and try to maintain a map instead of vector

I think it's a good tool for someone who wants to study on their own, but don't cheat during the contest (sorry for my bad english)

Why my brute force solution doesn't work for D (I skipped C)? 297948999 . It doesn't give TLE but it gives Wrong Answer on Test 3 .

yes, its because the hashmap would take up space of 10^9 ints in the worst case which is around 4gb of space whereas the memory limit is only 256MB.

I mean 4000. That rank is called tourist right?

Where is the next part of this?? This is not completed yet

I won't because I don't have ChatGPT coding for me

It is optimal only to consider ai and bi as prices

Hello I saw a comment that mamali is looking for money like others and... I know mamali very well He did a great job for us. He taught for years without receiving money and was a very important element in the development of people who did not have a good teacher (which led to several of his students winning gold in the Iranian Olympiad). He taught us that we can help others regardless of material things. And we continue his work I said this in response to someone who said that this per

He changed his handle to mamali

Using Frequency Arrays instead of Dictionaries and Sets and Character Arrays instead of strings for PyPy is something I have learned after many Hacks and TLEs

The image is cut off. Clearly, that says legendary grandmaster, 8407.

I participated as rated but why did I get under the unrated category please tell me.

But it would discourage most people who smurf for that top 5. Think about it, what kind of psychopath would deliberately go through the trouble of throwing 15 contests in a row just for a single mention? Might as well use your main account to do the contest at that point.

Because tourist beats magic.(whatbeatsrock.com)

Pupils I'm assuming Most newbie cheaters reach pupils, after pupil they are mostly banned for getting multiple skips, so a lot of them don't make it to specialist.

https://codeforces.com/contest/2051/submission/297939657 in e problem i have done binary search on binary search , can you tell me why this will not work , basically what i did is first i did bs on profit , now to check if we can get profit x or not i have done binary search on price that what price we can keep so that it can give desired profit ??

i sure do wonder why everyone is picking rock

this problem use " quét bơ đậu phộng " algorithm and "free fire chế độ sinh tồn , chiến đậu thật phong cách" trick ... #j4f

I think E problem has some ambiguity because if the price advances bi ，that someone may dont buy it and also dont leave a negative review

It doesn't matter—SSDs weren't a common choice for the average user 15 years ago. Remember, technology develops exponentially. The cost of chips and electricity isn't the main issue; the key point is that it's possible. btw even cost of running that thing is 1 million per task, if it can solve open problem like P vs NP then people will pay even billion.

my rating is several hundreds higher than yours, stop yapping bud coding while commenting and doing it so fast? 297834073 Vladosiya please take a look and tell me if this guy is cheating or nah

even tourist couldn't sustain tourist, how could mortals like you withstand that apex form of energy?

Binary search wont work because total earning w.r.t cost of the tree is not monotonus or unimodal.

Because even with the help of magic, you won't become a tourist.

Cuz who TF cares about this in a problem solving platform

Anyway the wind goes, doesn't really matter

In chess ai engines outperform humans but that does not mean that people have stoped participating. Similarly, the world of competitive programming will adjust.

You can change it now.go to your profile and handle tab and change your handle.

I did it abit differently. My idea was to use something like difference array Submission

There are a lot of details in this code that don't make sense to me. Maybe you should try rewriting it with a clear mind.

Best problem setter! Best board game player!

I know that i'm not so good at cp, but I have some advices that can help you: Not to cheat. As some people (almost everyone) here noticed, your code is suspicious (but if you wasn't cheating except one contest that was skipped, I apologize). Not to write comments at cf rounds. It is unnecessary and it is a waste of time (unless you used AI). Not to worry so much about cf rating. Just enjoy problems. Solve more cf problems. Not to write strange lines like you

What kind of problems can we expect to see here? Any examples??

#define id main(){} id it`s so fun!

The E problem is really well designed. Initially, my understanding of the scanline algorithm was focused on processing points within a two-dimensional range. Therefore, when analyzing the E problem, I enumerated all the values of $$$~a_i~$$$ and $$$~b_i~$$$ and treated them as prices, which we denote as val. At this point, the number of buyers among those who would not leave negative reviews is the count of $$$~a_i~$$$ that are greater than or equal to val. The number of buyers who would lea

The vector pair approach treats each event as independent, but in fact you should only update the answer after all x s which are the same have finished processing.

Yeah)) Also he/she has skipped round) But it's really strange that such suspicious code wasn't banned

Try this input. Spoiler Your answer gives 100 , but 100 should cause 2 negative reviews, which is not allowed.

Your approach is correct. I didn't read the code that you submitted in contest, but I fixed the brute force implementation you submitted after. Your mistake was using std::set instead of std::multiset : https://codeforces.com/contest/2051/submission/297964342

Someone from telegram thought it's a special edge case.

you might also keep the method binary search, but do it way more fancier. There's many submission out there to read how they works.

I think the problem is in this line (at least at this contest): if(t==567) cout<<"yayayayayyaya"<<'\n';

To all downvoters for no reason : If you can't help then don't spread hat so much. Keep codeforces clean please.

I explained the trivial binary search on best price or the trees sold won't work because the value function doesn't continuous. Imagine a linear line function that is increasing/decreasing, but a few values got drop to 0 because of some condition. Then it's impossible to binary search the normal way because it will make your left/right decision being wrong. (when the mid value is pointing to that 0 value, your code bugs)

You can refer to CLIST (which hasn't finished estimation yet). The official estimation will be days later.

Ok so it took me 1 hour to realize the number of trees sold is not continuous by price (because the negative review condition break it), so binary search on it is wrong... sadly I got into wrong direction twice in this contest. Outplayed, it took more to be an expert I guess :(

Yes, I used it 297849167

An Advice (in my opinion):Do five problems everyday and on weekends when you get time Touch some grass

Update: even top 4 on the scoreboard ( a mhdaimm ) was using it, but he participated unofficially

Best of luck! Just be sure to stay healthy :)

I managed to upsolve E with your idea! Thank you!

In E, a third method to optimize would be using PBDS, 297960644

Multiple contestants with ChatGPT, they didn't try to hide it either, crazy: bdyby10001 NeVeDlE dikshit_barla Ak.24 donshaaab ynotme 028 budarin.028472

https://codeforces.com/contest/2051/submission/297947353 can anyone point out the reason for TLE here, is it with the manipulation of segment tree(for sums) i did or something else? ps. i know segment tree was overkill but back then during contest i was getting this approach

I've failed to seek why my binary search solution is WA sadly. Just now I read how others solve using intervals is very eyes-opening to me, learned and accepted with that method.

lol, you don't need ordered_set for anything to become red.

My flaw was while considering the price to be some $$$b[i]$$$, I need to have all the $$$a[i]$$$ corresponding to same $$$a[i]$$$ in my set. (I still don't know how so many stresses missed these TC).

could you share your solution? from what I see it's a line sweep/greedy question

If tutorials for some problems aren't loading, they should be up in about 3-4 minutes.

okay so 20 rounds of stress tests with $$$n$$$ being 2000 were not sufficient to debug my E. 297881571

sorry was too lazy to solve E(and i wanna sleep, i slept only 2 hours this night. and i wanna eat). 3700 place, i think that very small, but positive delta.

Did E require binary search? I couldn't form the predicate function but will upsolve

You can compute for all pairs of of snakes $$$i$$$ and $$$j$$$, the minimum separation needed between the two snakes assuming snake $$$j$$$ will be placed after $$$i$$$ in the strip. You can then run a TSP-like DP.

I've been thinking about problem G and I'm curious now, did someone solve it using TSP?

On the bright side, if someone makes the same mistake, they will get an error message instead of being renamed :)

The contradiction is that $$$a$$$ must be a subarray of $$$b$$$, but $$$b$$$ cannot contain $$$a$$$ (i.e. $$$a$$$ cannot be a subarray of $$$b$$$). The two bullet points contradict each other.

I mean that's what the rating of my 1 year older friend and he has reaally good results so I'm kind of happy with that rating. Thanks for that promise!

do you think it will remain 40-100 ? it will improve by the time and within 3-4 years there will be some another version may be o9-o11 and it's rating will be near Tourist and will be able to solve question within minutes.

Spoiler

orz leovl48

I think because Mike want to keep the respect ppl pay to Tourist because the title is so hard to achieve and it's aura is uncreatable.

Honestly your growth would depend on how much your own mind could digest throughout your training (and what kind of problems would you digest in the first place). Still generally speaking, I can kinda promise a rating of 1700 at least if you've done exactly as you planned.

22024 ICPC Kunming Invitational How did you have 20000 years of problemsetting experience? I'm envious, ngl.

Are there some typos in the Contest section? I see "22024 ICPC Kunming Invitation"

Since both queries have the same length, their results will be both correct or both flipped. If the 1 is in $$$[1, n/2]$$$, the only 1 in $$$a$$$ will be in exactly one of the queries, so they will have different results. If not, the two results will be both 0 if not flipped or both 1 if flipped.

Note that $$$p$$$ must be a permutation, so elements of $$$a$$$ being in $$$b$$$ implies that $$$a$$$ must be a subarray of $$$b$$$.

tourist is a genius, and it is almost impossible for everyone to reach his rank

Oh! The interactor does need to terminate with quit / quitf as well (same types of verdict as with the checker). In interactive the Interactor and Checker must both return a verdict.

does it work for you all the time? as soon as I refresh the page, the snowflakes come back :(

Problem G is OEIS-able . Why writers do not check it (especially for problems with few input parameters) before preparing the contest?

BECAUSE THEY ARE VERY HARDWORKING PEOPLE, THEY REACH LGM VERY FAST, WHY CANT YOU SPOT THIS IN THE FIRST PLACE??? Also LGM sir, can you teach me CPP please, I am only yellow and i am very sad. Please teach me because I am poor, I have no money to find teacher sir i am so sad. Please teach me some UFO tricks Amongus tricks stuff. If i know those I would probably full solve Div 1 contests in 30 minutes and reach tourist rank soon.!!!!

Ah I understand now, thank you! I am curious as to why the editorial limits the search space even further, to (8,8). how might that be working? because for that I am getting an AC.

Hi, thanks for this proof. one follow up doubt- how did we determine after 9 times it would be pointless to check? Also in the editorial it says it is useless to check after 8 times i.e replacing both 2 and 3 eight times. Am i missing something? would also kindly provide a small example for that, it would be easier to understand. Thanks!

No, I do not know why really, and it is not only me, but many users also face the same issue.

Thank you Mike!I finally become Legendary Grandmaster!

[image] so i'm pretty sure this wasn't stolen from last year

If you use struct to hold the number as fraction $$$\frac{a}{b}$$$,it will be easy to only output a number like $$$\frac{a}{b}$$$ at the end,so that the precision requirements is not that harsh.

https://dmoj.ca/problem/ioi24p2 great problem. final observations are really nice.

Then their (grandmasters') life choice is wrong lmao

Why am I getting wrong answer if I use binary search in E problem as if we can see from H height, then we can see from all heights greater than that?

Oh this is a fun contest,but I think D is kinda complicated so that I waste much time on it

No one is more deserving other than tourist himself

According to the account which the gpt-o3 use, it participate in just 10 contest and cross 4 years. And currently in codeforces, if you do not submit any code during contest, the contest will unrated to you. So if there is a another strong person who monitor the gpt, and gpt finish the code first, and if it not perform good, it just not submit the code, it will be easy to get the high rated. Maybe should wait a more reasonly benchmark, like continously 10 contests that it perform good.

297766203 This submission got wrong answer 38th testcase 297765680 This submission got tle. What am I doing wrong??

As a Grandmaster myself, I can confirm this.

Bet: he will break his rules on Saturday and Sunday. Agree: Upvote Me Disagree: Downvote Me

what happen to chess will happen to cp too,but ppl still enjoys chess played by humans not some mere machines,

Thank you Bhai , I checked 2*n solution worked.

[image]

Ok. Thanks, But still ,If it is having correct TC according to given constraint , It should work I think, Because 2*1e7 operation are performed.

[image] It's a huge mess now

I think my comments are going to get lot more downvotes than before...

If you count for each $$$k \geq 1$$$ the number of integers $$$n$$$ such that $$$k = \lceil log_2(n) \rceil$$$, the sequence generated would be $$$1, 1, 2, 3, 5, 7, 13, 16, 24, 33, 47, 60...$$$. Such numbers $$$n$$$ are all odd and mostly prime (probably for obvious reasons, that's how $$$\phi$$$ works). Searching the sequence above up on OEIS returned one 1-1 match which was very interesting to me until I looked at the definition of the sequence (it was created specifically because of this

Ohh I see. I guess they will make it available later. Thanx ^-^

ARC186A An amazing constructive problem.

yeah but i think CF made it special for him aka tribute still we dream about being tourist one day, right?

Thank you. Could help me with this please? When I try to change my pfp I get this msg 您尚未进行实名认证或者已经被禁止使用图床 You have not yet undergone real-name authentication or have been banned from using the picture bed

here! This feature will be available until January 10 (inclusive).

Because tourist is insurmountable and there is only one Tourist in the history :)

Happy new year! CodeForces will be better!：）

Evacuation plan (IZhO18_plan) It uses very interesting tactics, that I've never seen before, it opened me to the advanced part of cp.

Problem F: "Now, consider the effect of an operation on a column C. The beauty decreases by Ac*SumB". When an operation is performed on a column C, shouldn't the beauty decrease by Bc * SumA? Please help.

This happens when you used a testlib function but didn't call register* in a source; it might be your interactor, validator, checker, or generator. I am guessing probably checker or generator.

We still got the 2nd best thing though (enjoying ours days as grey coders without the stress of learning algos)

does it have ratings like codeforces, codechef, atcoder...?

But why there is no rank of "tourist" in magic?

tbh cost will become irrelevant, it's all about chips and cost of electricity. chips will be cheaper and they will build dedicated nuclear power plants for training. then it will be cheap as o1.

Fixed in 297729300 The issue is that when this is executed: if(s[j]=='0') {cnt++; j++;} j may be out of bounds.

Although I might certainly never be close to winning one, can you consider issuing the prizes in some other stablecoin ?

Yes! Thanks for the gift sir, now I can temporarily change my color to match my target for this year. Also a small change to handle is nice, thanks a lot.

Shorter code for C. for _ in range(int(input())): b, c, d = map(int, input().split()) a = 0 f = 0 for i in range(62): ch = 1 << i if d & ch: if ch ^ (b & ch): a |= ch f |= ch & c elif b & ch: if ch & c: a |= ch else: f |= True print([a, -1][f!=0]) Explanation

damn this comment contains 0 truths and 3 lies, you should become a politician

Yeah, I will participate! It's just that I have plans for the next 7 years so I can't. I am sorry, my friend.

good luck, wait you not get frustated when you feel stuck in a problems, is better take your time in hard problems for a better rating than solve a 100 easy problems :p

Nah I just really love programming and math

No, that'll not happen, but keep trying, my friend.

I thank some random Bangladeshi guy for saving the username for me!

Hmm, seems it's indeed weird, not sure why change of that flag would cause such difference

From Romania CNMB: AndiR Alex1298

Alex1298 will win IZHO 2025!!

Maybe half of us also come after seeing a 3-letter red CodeForces allow everyone to be Santa xD

sorry im so tired and i got something wrong. thx for the nice problems!

As every year: 2022 2023 2024 and 2025 : [image]

Let $$$x$$$ be the smallest real number s.t. there are at least two building intersect with the line, then the range of real number to be able to see all building is $$$(x, \infty)$$$. When $$$x \ge 0$$$, output $$$x$$$ is consider correct just because it have arbitrarily small relative error to the correct answer rather than it's visible on $$$x$$$, and when $$$x < 0$$$, $$$0$$$ is visible and the problem ask you to output $$$-1$$$ in such case.

wow I am legendary grandmaster now with the help of magik ✨

Taking the expert rank because everyone else is GM now won't make you special.

As a -50 rated, I can confidently say I deserve the LGM.

I did try that, but i just ended up getting a bunch of air conditioners :/

I am the lowest rated LGM rated in the last 6 months!

The first one is correct behavior according to the definition in the problem From a point P with coordinate x and height h, building i is considered visible if there exists a point Q on building i such that the line segment PQ does not intersect with any other building. For the second one, I guess it would work as long as $$$L = 0, R = 2^k$$$ for big enough $$$k$$$? Not sure how to estimate the error but it feel reasonable to have less error when $$$R$$$ is power of $$$2$$$ since computer

shit I remember seeing this one last year fun fact: this meme is half of his comments

Happy New Year! The gifts presented by Codeforces are truly awesome.

Not everbody [image]

MikeMirzayanov , How do I request the handle of an inactive user Mosaab ? as it says "This handle is currently in use".

Long double isn't precise enough, probably __float128 will work.

being tourist has aura that we couldnt feel even with magic

Let's define $$$f(x)$$$ as the frequency of $$$x$$$ in $$$[l, r]$$$, and $$$g(x)$$$ as the number of times frequency $$$x$$$ appeared in $$$[l, r]$$$. Mo's works by "moving" the range for a total of at most $$$\sqrt{n}$$$ times, and each move only takes $$$\mathcal{O}(1)$$$, so we can do this in a total of $$$\mathcal{O}(n\sqrt{n})$$$

Me too. I tried to change to the handle " Mosaab ", and it is inactive, but it failed.

Sorry for tagging MikeMirzayanov , I can't change the name Raven even through that user has inactived for 12 years (~2011,2012)

This is the magic of this magical community.

they did publish the numbers, read the comment above

Now, I can comment anything and my comment will gets upvotes.

Thank you I can finally be my true colour! Happy new year

instead, why don't you pray for you to get a lot of ACs?

Errichto

Excuse me, how to request the handle of an inactive participant.

being tourist has aura that we couldnt feel even with magic

suggest any hazardous threatening handle name (innovative)

more practise and participant contest (if you can participant on every contest).

[image]

Please don't set such high precision requirements! Is there a difference between 1e-6 and 1e-9? A lot of people find the right solution but can't break the limits of floating-point accuracy. Such a high precision requirement is difficult to meet and does not improve the "mental difficulty" of the problem, so it is meaningless. （Sorry for the bad translation software.）

finally, the only thing that makes Holidays feels like HOlidays has come!

I hope at the end of this year i reach expert

lmao I saw the background and refreshed 5-6 times and thought it was a bug until I looked closely and saw the snowflakes.

submission

Yep. But what I meant could be this test example: 2 1 99999999 100000000 100000000 For me I got WA with double , but AC with long double . The reason is that the precision of F was set to 10^-9.

Thank you so much for your answers guys ! The thing is, I really want to improve in Python rather than C++ since it's the go-to language for machine learning and AI, which I want to focus on. So this brings me to another question : - Do competitions like Codeforces or in-person events like ICPC have the same time limit issues with Python as CSES does?

The difficulty of F is centred on the consideration of the boundary case, and the precision.

Change r to 1e18 and it works. This is for the case when there's $$$(999999999, 10^9)$$$ and $$$(10^9, 1)$$$.

You can also solve it using a DSU to simulate the removal of an ticket after selling it to a customer(point parent to the previous ticket in the sorted order). here's my solution in python: Spoiler Anyway, you're better off using C++ as it has set/multiset data structure and most of the problemset has tight limits like this problem.

Well that certainly is an interesting way to look at it. I don't think it should become meaningless if AI becomes better than tourist. I think that, no matter how good AI becomes at math or chess, it will never have the type of awareness that would put it on the same level as people. So like, maybe it will always be worse at pure ad hoc contests.

For problem F, why do i set the upper bound of binary search 2e18 WA but 1LL<<60 accept??? code 2e18 code 1LL<<60

Wait wait wait, if my interpretation is correct, there is no need to be so complex. Each time you just need to find the first occurrence of each string between the '*'s, so you can just run KMP upon each of them (taking O(n) time), and scan thru the document, it should only take O(n+m) time. I'm now convinced I've misunderstood. Please give some more explanation :(

The (*) part is edited, it's supposed to be a marker for the equation

I'm not sure if I would still be passionate about CP if this is true... But I think I will even though I don't want to accept someone use it to get red (Master is my dream..

I'm not acquainted with regex expressions, so the following might be a misinterpretation of your problem, but it seems that what you are asking is essentially to find a series of non-intersecting occurrences of several strings. We should now seek a way to find all occurrences of a certain string, and find the first one past a certain threshold (determined by the last string's position). This can be accomplished by the use of a Suffix Automaton, which turns the problem into online queries abo

If there is a real AGI out there, I believe it's a good thing because we will achieve communism faster

Merry Christmas! Last AtCoder contest before Christmas :)

thanks for the code... I understand your approach, i must've done something wrong in one of my transitions

Bangali can relate the name of this contest LOL

The A.I being good, then it most likely the same situations with a student and his mentor? No. Over half of the people here (and much more in the whole society) are not genuine CP lovers. They will use AI for malicious purposes and we cannot stop them.

Nice contest! I had fun participating, and it was especially nice to see that the quality of translation has gone up by a lot. Some minor questions (not directly related to the contest, sorry if this is the wrong place to ask!): Is there a way to filter for only rated participants in the ranking page? It's not that important, but as someone used to atcoder and codeforces, I was somewhat surprised to see 2000+ rated participants in the ranking and no immediately visible way to concea

Never had it crossed my mind to put the absolute full path in the include directive. That's cursed af.

yes. [image]

I think the announcement has a mistake: It is Luogu Round 210 but not Luogu Round 10.

I hope to get rid of this color in this contest. Best of luck to yall too \(^~^)/