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Hello! Codeforces Round #634 (Div. 3) will start at Apr/13/2020 17:35 (Moscow time). You will be offered 6 or 7 problems (or 8) with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that **the penalty** for the wrong submission in this round (and the following Div. 3 rounds) is **10 minutes**.

Remember that only the *trusted participants of the third division* will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a *trusted participants of the third division*, you must:

- take part in at least two rated rounds (and solve at least one problem in each of them),
- do not have a point of 1900 or higher in the rating.

**Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.**

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of my work. Thanks to my good friends Daria nooinenoojno Stepanova, Mikhail awoo Piklyaev, Maksim Neon Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round. Also thanks to Artem Rox Plotkin and Dmitrii _overrated_ Umnov for the discussion of ideas and testing the round!

Good luck!

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**UPD**: Also thanks to ma_da_fa_ka and infinitepro for testing the round!

**UPD2**: Editorial is published!

vovuh logic

vovuh hello , I guess my rating will down to less than 1600,but I've registered before,will I be rated in this case?

May b nope...

I think the you won't be rated in this situation.

I think after rating is updated, you can deregister and register back

I guess you are lucky (?) :D

Maybe

Wow wow, another Div.3 but this time I will not get rate :D

I good opportunity to get my rating back from today's Div2 lol

What does "open hacking" mean, sorry I am new? How can you hack a solution?

After each contest, there is a time of 12 hrs where if we find someone's solution incorrect, then we can give input the case at which the submission fails, if hacked you are rewarded with some points sometimes.

Not after each contest LOL

is there a way to sort solutions by programming languages? Now, I have to go through all the submissions to find Java submissions.

On the right hand side of status page,you can find language option under status filter and select language Java.

status > status filter > languages

1)After this contest (the same for following div.3 rounds and educational rounds), there's a period of 12-hour hacking phase.

That means, you can hack anyone you want(while in normal rounds you can only hack your "roommate" during the contest).

2)If you find something wrong in other's solutions(but passed the pretests), you can give an input which will make the solution fail (wa,tl,...)

In div.2 you get 100 points from each successful hack and lost 50 points from each unsuccessful one

In div.3 there's no prize or penalty for hacks.

Simply That Problem Will be Marked as Unsolved

What's "roommate"? Sorry, newbie.

In a normal round, you are assigned to a specific room, with ~50 people if I'm not mistaken. During the round, you can attempt to hack their solutions, but only if you lock your own solution first (this is because you will be seeing others' codes, it wouldn't be fair if you could change your own code).

A successful hacking attempt will give you a 100pt boost, while an unsuccessful hacking attempt will take away 50pts from you.

and what does locking the solution mean? If hacking starts after the round then how can one change his code?

Hope it won't go too mathematical ;/ questions about string is more interesting;)

50iq is not bad actually idk

**BACK TO BACK CONTEST **

But comparison of other websites code forces server is good

It's not just good it is pretty awesome when compared to others. Some of them had to cancel their contest just because the servers weren't able to handle the load and the submission queue was getting overloaded.Did you ever participate in codechef cook off or luch time...they can't handle even 5000 participants properly...whereas codeforces handles nearly 20K participants...comparitively codeforces servers >>>>> any other cp server

Can't you guys take it as a meme...

Exactly!!!

These guys become really too serious.

Their serious and sensible reply let me know that CF's servers are not bad, very good instead. Next time I'll be more patient.

The girl is the middle would rather be saying that "Give the lengthy problem A and engage participants."

Did you ever participate in atcoder contests It servers cannot hold even 5000 participants. Codeforces servers when it is +5000 submissions at time it will get some time but when it is less it works good!

it was just a meme bro... I know CF is really good.. I apologize for posting this...fine?

After every contest i felt disappointed.But believes that practice makes a man perfect. Best wishes for all. <3

Hoping to see "Yet Another Problems".

finally a div 3 contest :(

Can "Hacking A Soluton" also be included in Div 3 ?? Just an idea.

When div3 was first proposed then it was said that a beginner should waste time in hacking instead of solving problem. So contest time hacking is not available in div3 rounds. And there is 12 hours hacking phase for improving hacking skill. But i think 12 hour is too long. It can be reduced.

Is Hacking Possible in Virtual Contest??

yes if contest is still in progress hacking

I think 12 hour just for global, maybe someone finish the problems then have to go to bed.When they weak up, they can hack.

Actually after 6 to 7 hours nothing for hacking. Almost all the hackable solution are already hacked in this time.

Thank you, as always I am excited for another div.3 :)))

I should try to increase my rating in this contest....

Hello, is this rated?

rated for div 3(below 1600 rating)

I'll try to skew my propagating wave at least this time. Pray for strong pre-tests

lol

Problem set in recent codeforces contests seems tough for me. Looking to get some easier problem this time.

Div 3 means higher participation. I hope the testing is fast!

hopefully some short problem sentence

Looking forward for the contest .✌️

Why setter is yelling at tester?

Because of my invaluable suggestions :)

reveal unto us, are the problems sexy?

I think you should explain about your suspicious behaviour in the past contest. Because it looks like you've

cheatedbefore.your suspicious behaviour

Apologize in advance if I mistakenly understand it.

I am not being rude, but I don't think so I need to explain you anything :)

The thing is, your code is totally same with another code in the past contest.

I don't want to be an annoying guy who repeat one thing everywhere. I just want to know what happen. If it isn't a cheat, I will sincerely apologize. If you've already be punished for that, I won't mentioned it anymore.

I will keep annoyying you if and only if you really cheated and haven't been punished so far.

Well You can see my rank in this contest :)

ok, got it.

Hoping for stronger pretests in A. Got hacked (unfortunately) in the educational rounds and it had been a while since I had solved 3 T_T and I missed it thanks to the A.

You should be more careful with some corner cases

That's more important and valuable than strong pretests.

I hope it will be like this for second time!!good luck and high rating bro

thanks bro

00:01 nice network

Yes , it worked good this time (^_^)

You just risked there by some basic observation, I don't think that on minute 2 you were able to prove your solution. So don't be happy by that.

I hope to see Math problem~~

you make me a green i dare you

good luck

Hope the system test won’t be too long.

It's good to see vovuh back but I expect that codeforces will do something in order to handle such large number of submissions during the contests and I guess this time it won't be queueforces :)

I hope that the gap between the div.3 rounds will be decreased , its the best roundHope my rating will go up.

I'm so vegetable......

Hope the statements will be shorter -_-

Thanks vovuh again for bringing such contests in this pandemic situation we highly thanked to codeforces and its community for bringing such contest in this period to make us somehow busy and involved.Thank you codeforces community !

The color spectra doesn't quite match Codeforces band XD

I am becoming

~~Expert~~Blueafter this round.I don't think so

Is it rated?

For you, yes. For expert or more, no.

Isn't it more logical to not allow experts or above from submitting solution for first 20 minutes rather than making problem A hard/confusing for div3 rounds atleast. Many participants don't even submit solution if they find problem A confusing or tough for it's bracket.

My rating is just 1600, so helpless

lol similar is case for guys having rating just 1900

Good Luck Have Fun!

what happens when i hack my own solution ?

pts — your points

So, if you hack task A, then:

new_pts = pts — points_from_task(A) + hack_points

i guess both points are equal.

If successful:

You will lose a problem solved

If unsuccessful:

Nothing changes.

So it's obviously meaningless to hack yourself unless you've found some mistakes in your code after the contest.

oh i got that ,thank you.

Good luck to you all

What you see:

`Codeforces Round #XYZ (Div. 3)`

What I see:

`Another unrated contest :'(`

architb_12 why the hell are u giving this contest ,if u have such a mentality,and posting this nonsense doesn't show that u are cool

pizza_hut what mentality? I am just saying that I am sad that the contest is unrated. It is much more fun when it is rated. What is your problem with that?

architb_12 whats the problem when contest is rated for 70% of users on codeforces

I think you have completely misinterpreted me. I meant that it is more fun to give a contest when it is rated for you.

hey, first of all, learn to give respect to others (doesn't matter he is highly rated or low) and don't talk rudely. He was just saying that this contest is unrated for him and of course this was unrated for him because he has done efforts to be at that position where Div3 even Div2 is unrated for him. So, please don't post such non-senses. Even today tourist gave the Div3, can you stop him...No nah?

Can the frequency of Div.3 contests increase ?

let's hope that i get one question correct on this site tutututu

This is what happens when Tourist attends Div 3.

How can one read the problem, figure out the solution and write down the code in 1 minute.....

Think about reading 3 problems, figuring out solutions, coding all 3 of them and getting them accepted in 3 minutes.

He doesn't read the problem, nor do any other red coders. They go straight to test case, find a pattern, and implement.

Unbelievable!

If problems see tourist comming they give up and solve themselfs.

You are very humorous

lol

Actually tourist has solved the problems before they solve themselves:)

technically, it was 36 seconds

answered here

he submits all the answers after that he reads the questions

What's the point of having 50000 testcases, if giving verdict still takes you > 10 minutes?

I'm sorry, what about you? I don't see any of your submissions were judged for more than 1.5 minutes.

F took > 10 minutes. It really didn't affect me. I understand its because these rounds don't have pretests, I guess.

Wow the servers have REALLY IMPROVED. MikeMirzayanov orz

tourist solved 7 problems in 22 minutes and still atheists exist. orz

Tests today was rlly fast, thx)

A nice and balanced round !!

How to solve E2 and F?

Easy +rating, nice contest)

I was logged out suddenly while submitting please solve this problem it occurs most often during contests with me!

How do you do E2? I spent like 45 minutes binary searching for the second endpoint(and I think it was probably the right way since you have blocks of xyx or just one large block of length x), so you can use binary search to find the optimal value for the second endpoint, but it wouldn't work.

EDIT: Just realized that I thought a_i<=26 for E2 (as it was in E_1) so that's why I was getting wrong answer

Observation:

The contribution of a number is limited by number of occurrences of prefix and number of occurrences of suffix, In other words, if you have a prefix of 7 occurrences and suffix of 2 occurrences, Then you're only contributing to the final answer with just 4(2 from the left side which has 7 occurrences, and 2 from the suffix), So contribution = min(prefix,suffix) * 2.

Knowing the above, We can just try for each number x, A possible prefix that has 1 occurrence.. 2 occurrences .. 3 and so on, you can fill the middle with an element that occurs the most, you have only 200 distinct element, just go over them all.

Oh my this is neat. I mean the idea about checking only prefix and suffix. Thanks!!

Video Editorials + Facecam

How to solve E2. Any idea??

store position of every element in a vector v[203].for each number between 1 to 200 run two pointer from start and end of its position vector and between them find maximum time occuring integer frequency using prefix sum 2-d array.

can you elaborate on the

prefix sum 2-d array? how are you storing in them?my approach for E1 was to check for the next max frequency of a number between the extreme positions of all the numbers having max frequency ...what's wrong in this??

more than 60k accepted submission and aprox 26k participants.wow amazing.

Why i can't hack someone? It's write me "Illegal contest ID" or "Неверный идентификатор соревнования"

Hack via status page,not standing page

Thank you very much :)

You're wellCUM! ^_^

Fast servers but speedforces :(

How did you all solve E2? I used Mo's and I wonder if that was overkill.

Oh man the alphabet is only size 200, Mo's was definitely overkill.

Yeah, I used Mo's too, don't know how others got that right

complexity- (2*10^5)*200 I will tell very briefly... for each index i let x=arr[i] and till now(including i) let it(x) has appeared fth time so if (Frequency of x in whole array)/2 >= f then i found index j of x from the end of array such that in segment arr[j...n] occurrence of x is also f and hence I iterate through all the number 1<=n<=200 an found the no of occurrence O= of such n between (i,j) . so length of

three blocks palindromewill be f+f+O.Do this and update the result if you got a larger length.

Here's my approach:

prefix and suffix(`aaa...aaa`

)middle part(`...bb...`

) between $$$i+1$$$ and $$$j-1$$$ using a precomputed table of counts for each prefix and each symbol.`middle part score + 2 * symbol count`

.$$$O(n\cdot k^2)$$$ in total, but the operation #3 is rare in practice, so it passes the pretests ¯\_(ツ)_/¯. Got MLE on the first attempt though, lol. Don't define int long long where it's not necessary.

Excellent solution. I thought O(N * K^2) would fail, but yours is super simple, so I guess it has a very LOW constant factor.

That is $$$O(n \cdot k)$$$ in total if you have position pre-calculation, not $$$O(n \cdot k^2)$$$. That is because $$$\left(\sum \text{occurrences of symbols}\right) = n, \text{not } n \cdot k$$$.

but for each prefix and suffix of given size and letter, he is going over all 200 letters to find biggest centre block. So I think its really $$$nk^2$$$ with small constants.

I think prefix table can be calculated for

`vector<int>(200)`

, then the max_element(for "b") can be found in`O(k)`

time. Since we traverse over`O(n)`

two pointer pairs`(i,j)`

, this gives total time`O(nk)`

.FYI this approach is almost optimal; but at step 2 you should initialize $$$i$$$ and $$$j$$$ to the two middle occurrences of the symbol (if the amount of symbols is odd, just ignore the middle one), then move them away from each other, towards the end.

The benefit of this is that you can start with an empty set of frequences $$$f[200]$$$, then initialize that to the counts of each symbol between $$$i$$$ and $$$j$$$. Keep track of the most frequent symbol whenever you update ++$$$f[\cdot]$$$. And then do --$$$i$$$, ++$$$j$$$, and update $$$f[]$$$ as you go. You can always keep track of the most frequent letter in the middle part this way, and complexity is $$$O(n)$$$ because you do ++$$$f[\cdot]$$$ at most $$$n$$$ times.

For E2, I coded a

`O(nklog(n))`

time algorithm using the same idea as yours except using binary search instead of prefix sums. I realised that the complexity is too large. However, for testcase 9, which has 134 repeated 2e5 times, my solution should simplify down to`O(nlogn)`

. However, when I ran it on my system, it took more than 30 seconds and gave me TLE on codeforces. Can you please help? Here is my submission https://codeforces.com/contest/1335/submission/76624993I also tried the same thing but got tle on test 15 .76625017

Later i think to remove binary search and precompute freq table but memory limit exceed on test 9 .76625886

can anyone help ?

Replace ll from int in your program

wtf , It worked .

how they can set such a memory limit.

E2 complexity analysis:$$$\displaystyle \sum_{c=1}^k freq_c \cdot k = \displaystyle k \cdot \sum_{c=1}^k freq_c = O(k \cdot n)$$$

Sum over $$$c$$$ for choosing the number in $$$1^{st}$$$ block. $$$freq_c$$$ for iterating over all possible lengths of $$$1^{st}$$$(and $$$3^{rd}$$$) block. Multiplied by $$$k$$$ for choosing the number in $$$2^{nd}$$$ block.

$$$\displaystyle \sum_{c=1}^k freq_c \neq k \cdot (freq_c)_{max} \neq k \cdot n $$$

$$$\displaystyle \sum_{c=1}^k freq_c = n $$$

What is Mo. I thought of doing binary search but failed to do so.

maybe Mo's algorithm?

https://cp-algorithms.com/data_structures/sqrt_decomposition.html#toc-tgt-4

Could you please explain how you applied Mo's?

The problem can be reduced to answering a bunch of queries of the form: What is the most frequently any element appears in the range [L,R]? If the alphabet is large, you could attack this with Mo's Algorithm.

Unfortunately, it slipped my mind that since the alphabet is so small (only up to $$$200$$$ symbols), you could more simply create 200 prefix sum tables instead.

I didn't quite understand from your code how you're generating these queries. Could you please elaborate how you're generating these ranges?

I did, but one must be careful to not use long long, since it used to much memory. We have to use a $$$32$$$ bit type.

76620592

My solution is similar to yours but it is giving TLE on test 2. I am not able to figure out the reason.

My Submission

Is Mo's algorithm the only way to do it?

Really good round! I loved problems D and F. Had an idea for F but gave up after 20-25 minutes of implementation after I realised I didn't account for something mentioned in the question. Had about 20 minutes for E1 and E2 (I regret not having attempted E1 and E2 before F). I believe I'd have been able to solve E if I made a wiser decision earlier but shit happens, lol. Also, ideas for E, anyone?

You can enumerate a and the number of a, and then enumerate what's b, it seems 200*200*n, but you can find the really time complexity is 200*n

Keven Can you explain your solution? Is it similar to this: https://codeforces.com/blog/entry/75908?#comment-602884

200*200*n, first 200 enumerate a, n is the number of a, so first 200 times n is n, time complexity is 200*n

can you explain me why it is not 200*200*n?

I got AC on E1? but i assume that complexity was 200*200*n, now i sent same code in E2 and i got AC

In your solution you have

`y = v[a].size() - 1;`

which is actually not always n. In fact it would make the make the factor of 200*n to just n.yeap, I already got it, thanks!

can anyone point out which test case is giving wrong answer in this code for question B of this contest. https://codeforces.com/contest/1335/submission/76554481

i missed corner case when we have to take 26 unique characters.my bad

In problem F, if N and M's range can be confirm instead of N*M <= 1e6 will be better

How do you guys solve D?

I finally figure out that we must change 9 postions which are respectively distributed in the nine squres.

So I enumerate the column and the row (maybe like N-queen Problem) and get the answer....

Author's solution is to replace all 2 with 1.

Wow, this solution is very cool!

I think replacing 3 with 1 / 2 / ... is also ok.

Why 2?

You can take any 2 numbers.

wow that's innovative, instead of figuring out which cell to edit.

Yeah, I also changed every 9 to 8. It took just less than 4 minutes for me to solve this. Fastest D problem ever for me.

I place a number

`9 - currentNumber`

if its non zero else`1`

in every cell, such that whose`row`

,`col`

and`block`

was not visited before through any cell. To check that I use simple hashing.IdeaYou only need to make changes such that each row, col, block have at least two same elements. If you make changes at indices (0-based) (0,0), (1,3), (2,6), (3,1), (4,4), (5,7), (6,2), (7,5), (8,8), you end up covering all rows, cols and blocks. So, make the required changes to those indices => problem solved.

CodeI did the same xD

Shhh, don't post anything remotely related to the word "same" :p They'll think we cheated

lol should I delete it?

Nah, hehe. It's cool. Now, it'll look like: Yeah, they cheated, but if they're talking about it openly, they might not have....

Jokes apart, I'm just having fun. GL for future rounds

a[1][1] = a[2][1], a[2][4] = a[1][4], a[3][7] = a[2][7], a[4][2] = a[3][2], a[5][5] = a[4][5], a[6][8] = a[5][8], a[7][3] = a[8][3], a[8][6] = a[7][6], a[9][9] = a[8][9] This was my idea. Later, got that replacing all 2's with 1's was good enough.

Funny way to solve F without detecting cycles: notice that if two robots get to the same cell, then they will always move to the same cell from then on. This means that we can find for each cell where the robot starting in that cell will be after some $$$2^{22}$$$ moves using binary lifting.

This is author's solution :)

Hi, this is my first contest. Can you please share the link to the author's solutions? I am not able to locate it. Thank you!.

I'll post them with the editorial. Please, be patient.

The contest was not at all balanced, I must say. Even you can see the rating of people and compare the no. of questions they solved. Starting 4 questions were not of much use as almost half of the contestant solved the first 4. Also, question D looked like it was meant for April Fool's 202 but was posted by mistake?

Why do you posting this from fake account? :)

You ignored my comment... What am I saying wrong according to you? You should accept the mistakes. :)

Which mistakes? Prediction mistakes? Maybe, I need to clean my magic ball to see the future better.

The only (and very doubtful) "mistake" here is the "gap" in $$$5400$$$ accepted solutions among official participants (which means that E1 was solved by 1/4 of people who solved D). This is

nota big gap.I could show you examples of really big gaps but I don't think I need to prove you something anymore. If you don't realize that

thisprediction with difficulties is good enough, I have nothing to say to you.And, if you post your opinion from fake account, it seems like you don't want to anybody know who are you actually :)

The last point is quite obvious.. :) I just wanted to say that these weekly contests are supposed to test our algorithmic (and not puzzles) skills. Thank you for replying..

Yeah, you are right. As far as I remember all the Div.3s I have given were perfectly balanced and excellent problemsets and all of them were made by you:)

Well, I disagree with you :) Most times there are some mistakes that you couldn't notice, but this time the contest went fine. There also were some minor mistakes but not as fatal as usual. This is the rare case when almost everything is fine and I'm glad to see that.

Thank you for the round! No queue, no weak pretests.I found problem E2 very interesting and problem D a bit hard for me but interesting too.Please make more div 3 contests, we really appreciate you :)

That's actually a really cool approach. Have you also tested a cycle-based solution? I am trying to implement a fairly straightforward search and it keeps running out of memory.

Looking at 32 pages of MLE results in Status, I think the memory limits for E and F could have been a little bit higher.

The approach with extracting cycles and doing some dp is much harder to implement, so I didn't even tried to do that. I could but didn't do this.

I ended up flipping the dimensions if N > M, and surprisingly this helped (76628096). According to the test results, it uses up 254.5 out of 256 MB, lol.

Well, I'm sorry to hear that. I really thought to increase memory limit, but my solution uses ~200MB (the two-dimensional vector of size $$$nm \log nm$$$) so I thought that the 50MB will be enough (because this is the only array which should have such size).

Hello ; can you please explain me why does N*M*M solutions pass for problem E2 ?

tourist has written a code with that time complexity ; here is his submission — https://codeforces.com/contest/1335/submission/76520445

And I am not able to understand how does that happen ; curiously ; I even did try to check the test cases ; and I see test case 9 which should break such solutions.

So clearly ; I am not understanding something ; can you please help me with this.

Thanks !

The solution of tourist has the complexity $$$O(nk)$$$ where $$$k$$$ is the size of the alphabet. If you take a look at the two outer cycles, you can notice that they just iterate over all characters of the string and this sum is obviously $$$n$$$.

Maybe, when I post the editorial, it will be more obvious to understand why this is $$$O(nk)$$$. Just be patient.

Oh ; I see ; I apologize for being a bit impatient ; and I thank u for the reply.

I understand the complexity analysis of that solution now ; Thanks !

vovuh please look at my submission for E2 this it got MLE 252700 KB but now I am submitting it got accepted this with 252800 KB (more space than previos)... :'( and because of that I was not able to submit E1 also during contest...

very sad!!! :'(

Thanks for teaching me something. I learned binary lifting for RMQ/LCA but didn't consider reusing the table for k-th ancestor queries.

I got E 5 min late -_-

Nice round!

help me please which problem i have with this code on problem B? https://codeforces.com/contest/1335/submission/76592383

your code is giving wrong output in this test case :- 1 10 10 10

your output contain only 9 distinct character but reuqired distinct characters is 10

Idea is wrong.....

My idea(1) Construct a string of length B with all distinct characters.

(2) Construct a string of length A — B with character same as last distinct character in String (1).

(3) Concatenate the String (1) and (2).

(4) Print String (3) (which has length A) N / A times.

(5) Print the remaining length (N % A) with appropriate characters from String (3).

CodeYou have taken 'g' twice in your array

How to solve E?

How to think fast like tourist... :(

When he was solving each question in about 1 minute, I was drawing on mspaint.exe and looking for the solution :)

practise,practise and practise .smartwork +hardwork

In normal practice I could solve A and B with 1000 difficulty in less than 2 hours. :/

Probably my brain crashed because of taking too much practices on last night lol...

they are doing coding for many years now , if you continue doing this you will be master in upcoming years

Do you know this website? https://vjudge.net/ I think is a good sit for practice. There are all kinds of projects to practice on.

Well pen-paper seems more reasonable for me

I have my own drawing pad & stylus pen so I don't wanna waste them too haha.

the best contest ever for meProblem DCan somebody please help me in finding out the problem of this submission. Apparently, the inputs are not being properly taken, instead some absurd garbage values are being shown. I was stuck with this for the last 40 minutes of the contest and still could not figure it out.

You declared the array as integer, hence it will read whole row at once (not character by character as it is supposed to be).

Input is not n*n integers, it is n strings in n lines.

You can't take the input as int sudoku[10][10]. Make it char sudoku[10][10] and change your code accordingly.

That memset tho...

cin >> sudoku[i][j]; this takes takes whole row as input. Not a single integer

https://codeforces.com/contest/1335/submission/76590246

https://codeforces.com/contest/1335/submission/76509810

both solutions are identical and the person who submitted it from different accounts, intentionally did so to hack (and get points if hacking gets you extra points, not sure tho) !

Maybe that's the reason why he's still gray.

Successful Hacks don't change anything except the victim will lose a problem solved

Hacks with this method are......meaningless, honestly

people making such D should be banned from making future contests

Can you ban MikeMirzayanov?

People making such comments should be banned from making future comments

I did read the "you can change any..." as "you can swap any numbers in the grid". Still have no solution for that problem.

you can see my solution i also thought the same

https://codeforces.com/contest/1335/submission/76594021 my sol

Can you please make some better to and provide them to problem setters so that the people who works so hard for making this problem can take some rest.

Haha , why should i do that. I only want to solve problems not make them. But if some problem is bad and i say that it's bad dosen't mean i need to make problems.

Here is a Chinese proverb for you：“你行你上啊”. no can no BB!

It's your problem...

2! According to me it should be 6!``

1st and 3rd part must have the same length x

1 1 1 1 1 is the longest palindrome. What is your 6 length palindrome?

Damn! I missed x==x :/

I was also stuck at this point for half an hour

Number of a's on both sides of b's has to be equal as per the problem.

It should be 5. The longest 3 block palindrome here is 11111

1 1 1 1 1 is the answer and here both x=0,y=5,x=0

basically every answer should be a palindrome with max 2 types of elements, and the 2nd type of element should be sandwiched b/w 1st type of elements , hope it gives you a little idea ! PS: thanks ashkANOn for pointing out the mistake.

For 1 1 1 1 1 x=0,y=5,x=0 i think

remember that the 3 palindrome should be of form x y x. the length of first x and second x is the same

It's really bad for me. 1335B - Construct the String said that " to construct a string s of length n consisting of lowercase Latin letters such that each substring of length a has exactly b distinct letters. " e.g. For test case: n, a, b are 22 17 12 my code output: abcdefghijgkkkkkkabcde

is it wrong?

My construction is:

abcdefghijkllllllabcde

Oh, it's a really silly mistake. Thanks for helping

The first substring "abcdefghijgkkkkka" has only 11 distinct characters.

Yes it is wrong for the prefix of size 17. I think you are trying to print this "abcdefghijkllllllabcde".

Can someone show me your solution to C?

My solotion is to binary-search the answer and check whether it is legal.

But my solution runs nearly 900ms.

Is there some easier and faster solutions?

Here. Mine runs in 61ms. Idea could be figured from the code but hit me up in PM or here if you need help understanding something.

CodeCan you please briefly explain your solution?

I have a little difficulty in understanding it....

Sure.

I map every

skillto the number of times it appears in the input. I do this because I'd like to find the skill value which is present in max number of people. This is what the variableholds. The size of the map will give the total number of different skill levels. However, one of the skill levels must be used to compose the team with same skill levels. So, the number of distinct skills areSameMap.size() — 1. And then it's a matter of evaluating which team size is maximal.Btw, the only check you need to do is whether the biggest count equals the size of the map or if it is greater. If they’re equal, return min(biggest count, map size — 1). If not, return the min(biggest count, map size)

Thanks bro!

I have understood your solution and finished the problem in 62 ms.

I can explain mine if you want to :)

SolutionThere is a greedy aproach to this. We want to maximize the number of people in both teams. So we want to check the maximum number of students with the same skill (let's name it max_same) and also the skill (let's name it num). After that, we want to check the maximum number of students with different skills, different than num (there is the risk to put the same student in 2 groups). Both of these can be done with a simple sort on the initial vector and a linear scan. Now, if the absolute difference between the number of students between the 2 groups is at least 2, then we can transfer one student from the second group that we didn't take previously to the first to maximize the number of different skills. The answer is the minimum between the number of students between the 2 groups.

Code: 76607507

Thank you, but I have finished the problem in 62 ms(^v^)

76583554 Can you please tell me why my answer is wrong in the second testcase. I have used the same logic as you. Please help.

Can you please tell me why my answer is wrong in the second testcase. I have used the same logic as you. Please help.76583554

Hi, can I be unrated for Codeforces Round #634 (Div. 3). Reason is with Problem D with Java. I submitted the code https://codeforces.com/contest/1335/submission/76566765 during the contest without PrintWriter but TLE'd on test case 2, although my solution is O(81*10^4) which should be under the time limit. After the contest I submitted the exact same code with PrintWriter https://codeforces.com/contest/1335/submission/76614841 and got AC. @vovuh vovuh

Best contest ever for div 3. Thank you god vovuh very much and hope everyone can become expert !

I didn't read D correctly and thought you needed to keep the "sudoku structure" ( just swap numbers around, no replacing ), so I came up with a backtracking solution and I just kept wondering how was it a div3D and how did so many people manage to solve it lol.

Strong Pretests and Intersting Problems, Fast Test and Good Network.

Thanks to vovuh & MikeMirzayanov

Plz change your time of contests in india.

What is the intended complexity for E2?

O($$$N$$$ * $$$200$$$)

The complexity of your code is

. Many other solutions with this complexity have passed. I think the best we can doO(N*200*200)?O(200*N*log(N))I believe my code is O(200N):

https://codeforces.com/contest/1335/submission/76611393

See my reply below. It can be done in O(N*200)

O(200*200*nlogn)

Not for E2

I can be done in O(N*200) analyzing from the "center" for each distinct value. Centers for value V means two distinct pointers (a, b) where the amount of V before a is the same as the amount if V after b. Then keep jumping a backward and b forward, both at the same time, but the jump is actually to the previous/next V value. While you do this, count the frequency of values in the middle of a and b, and keep the one with maximum occurrences, and add that to (amount of V before a)*2 keeping maximum for each time you move the pointers. This is linear for each value V

Nice solution!, initially I had thought of this method, but I tried to move the pointers closer instead of farther and it led to some complications in maintaining the maximum

Ah! I realise now that what I did was itself O(200*N). I did the complexity analysis wrong. Thanks!

Where can I report this self-hack? https://codeforces.com/contest/1335/submission/76616926

And then there's this one: https://codeforces.com/contest/1335/submission/76618522

And then this guy is hacking his own solutions by deliberately putting wrong tests: https://codeforces.com/contest/1335/submission/76620243

This one too: https://codeforces.com/contest/1335/submission/76616722

no need to report.

There's no prize for such successful hacks

Also there's no Best Hacker Rank in div.3, only in Educational Rounds.

In F,Why did you keep 1 as white and 0 as black ?

If your intention was to delay submission time by 2 minutes, congrats :D

Could you please help me understand why my code for

Problem Cdoesn't work? It seems pretty straightforward and I have seen other solve it my way, however my code fails on test 4. I have no idea why. Here is my submission.I don't understand your code, but it seems to fail on the case "1 1 1 1 1 1 1...". The answer should be taking the whole array (n)

My code fails on case 4, do you mean when all the numbers are the same? In this case the answer is 1. How my code works: 1. Counts all the different numbers 2. Finds the biggest amount of repeating numbers. 3. Takes the smallest of the two (+- 1) 'difs' — different numbers 'maxN' — biggest amount of repeating numbers 'curN' — current amount of repeating numbers

Don't compare two Integer(object type) using ==. It compares reference not value.

Thank yo soo much for your help. I used the intValue() method when comparing and my code is now working. Thank you so much for your help. Here is the working version.

I am getting automatically logged out when i press submit button for the first time in a contest.This is happening regularly with me from past 5 to 6 contests. I am still a specialist and those extra 15-20 seconds don't matter to me right now, but this might be a bug in the server that may need a correction. Please have a look MikeMirzayanov. Thanks!

Maybe your cookies are not working.

but,this is not happening with me on other coding platforms! If it were cookies problem, then it would have occured every time i hit submit, or at other platforms too! It does not happen on the light codeforces website even!

I have faced the same issue when clicking from the problem itself instead of going to the submit code tab

I'm also facing some problem while submitting my first solution of previous 2-3 contests. I get some html 403 forbidden error with some token value!

In E1 test 2 set 107 is "4 1 4 1" and the expected answer is 4?

Am I missing something?

I think that the counter in the evaluator that displays the WA test number is off by 1

No, I'm pretty sure that's the line. I made a submission that outputs -1 at 107th item, and it gets caught as -1.

https://codeforces.com/contest/1335/submission/76621065 wrong answer 107th numbers differ — expected: '4', found: '-1'

Then I submitted an app that prints only the 107th line's input. And it is "4 1 4 1".

https://codeforces.com/contest/1335/submission/76621266

It can't be 4.

It prints all the inputs starting from the first; 4 1 4 1 is actually the first test case.

On 107th iteration you are taking input for 1st time. It is taking 1st case as input not 107th case.

My bad, thanks. It's actually 1 1 1 4 1.

Is that input visible from the test case?

Why are some of the submissions containing this part?

if (t == 'something') cout << "OK" << endl;

Does it give any benefit?

They do that so they can get easy hacks from another account