Hello Codeforces!

Asymmetry and I are glad to invite you to Codeforces Round #743 (Div. 1) and Codeforces Round #743 (Div. 2), which will be held on Sep/18/2021 17:35 (Moscow time).

Each division will have **6 problems** and **2 hours** to solve them. All problems were written and prepared by Asymmetry and me. The round will be rated for both divisions.

We would like to thank:

- Monogon for coordinating the round and helping us with the problems.
- mnbvmar, gamegame, dorijanlendvaj, tnowak, Jasiekstrz, ijxjdjd, polarity-, lcaonmst, -_-INV4D3R-_-, ToxicPie9, Shinchan01 and Krzaq for testing the round and providing useful feedback.
- ToxicPie9 for being a VIP tester.
- finally, MikeMirzayanov for great platforms Codeforces and Polygon!

We've put great efforts into preparing this round and we hope that you will enjoy it.

Good luck!

**UPD:** We would also like to thank Sexpert for testing the round and KAN for translating the statements into Russian.

**UPD2:** Here are the scoring distributions:

Div. 1: $$$500$$$ — $$$1250$$$ — $$$1750$$$ — $$$2500$$$ — $$$2500$$$ — $$$3250$$$

Div. 2: $$$500$$$ — $$$1000$$$ — $$$1500$$$ — $$$2250$$$ — $$$2750$$$ — $$$3500$$$

**UPD3:** We are sorry that the round became unrated due to the long queue. We hope that you enjoyed the problems anyway.

## Winners

Congratulations to the winners!

**Div1.**

**Div2.**

**UPD4:** Editorial is out!

Since Monogon had ascended into the coordinator realm, I became the new VIP tester.

btw, as a VIP tester, I tested

That's okay, but were you tested too?

As a tester, I wish you sunny weather ^^

As a resident of a warm country I downvoted.

as a lover of your pp,I still downvoted you

as a -ummm ughh uhhh yes

As coordinator I want to thank the VIP authors Asymmetry and Markadiusz

.

But still takes contibution ;)

Ah, you misplaced dogs

As participant I want to thank coordinator Monogon and authors Asymmetry and Markadiusz

finally

When will next div 3 round happen??

Looking forward to participate

Challenge:hit 200 downvotes under this comment (bye bye contribution)

As a coauthor, I wish everyone best of luck and a positive skill delta after the contest.

As a tester, I wish everyone happy rating :)

Wish me happy rating for the next two-three contests so that I can have your current color

Because thats what heroes do:))

Why the huge difference between C and D problems !!!! ** it was the first time i see such a difference ** Is it okay >>

Don't wanna miss the contest but suffering from anxiety and depression for the past few days.Hopefully I will overcome anxiety and depression soon and start playing in contests

Mental health is much more important than competitive programming, so take it easy, hoping for some well wishes from a fellow stranger.

Yeah,I am going to meet my psychologist today for that reason.Being mentally strong is important in cp too.Because you can have all the skills in the world but if your mind is unstable during 2 hour contest I Don't think you can deliver your best.Anyway,I am going to consult my psychologist soon and come back strongly.And thanks for your kind words.Cf has really been like a family to me

I hope you become well in the coming days and I wish the best of luck for you in the contest

Thanks for your support.Unfortunately I am not stable enough to give today's contest but hopefully I will be giving every contest in cf from 23rd September.I wish you all the best in today's contest

Take good rest bro ,and come back strong

yeah sure buddy, pray for me.Good luck for today's contest.Wish you high rating!!!

Thanks Noelle-s-Stupidsta , I'm still working on a strategy to reach a better level but I think it will take some effort and time to reach a higher rank

solve 1/2 1200 rated problems

solve 1/2 1300 rated problems

solve 1/2 1400 rated problems

Do these three things everyday

Do this in every week

Do all these in the next

2 monthsand be confident and always believe you can be as good as anyone.Hopefully your rating will be between

1200-1300after thatFrom my experience, I think you just need learn the basic technique to get rating >1600. Do too much easy problem doesn't make your skill better.

I agree but my learning was always extremely problem-solving oriented.Even I never studied binary search.I felt the importance to learn it only when I had to do some binary searching in a problem. Well, every human-being is different.So, strategies may vary

I'm trying to do this recently and I'm learning new topics as well but it's not always easy to use new topics in solving problems it takes some time

Thanks for the advice I'm going to try it and maybe I will try solving problems on other websites as well

yeah, atcoder beginner contests are very helpful, you may check it out

i will thanks

kickstart round is also scheduled from 17:00 UTC on the same date, can u guyz plz prepone the round?

That way it will clash with Atcoder ABC

Whats your Atcoder Handle ??

As someone who wants to hit -69 contribution mark, I request y'all to downwote for high ratings. thank you!

As an old, almost retired contestant who grew up in the same school as the authors, I'm thrilled to see what my younger comrades prepared :)

Interesting division 2 scoring distribution. I expect (and hope for) a sizeable jump between C and D.

Early scoring distributions! :D

I am a newbie and started giving contests on this platform from this month itself. Excited for this another upcoming contest!! All the best everyone!

I am a specialist and started giving contests on this platform from last year itself. Excited for this another upcoming contest!! All the best everyone!

I am a pupil and started giving contests on this platform from this year itself. Excited for this another upcoming contest!! All the best everyone!

I am a future expert and started giving contests on this platform 20 yrs ago. I was excited for this then upcoming contest!! All the best everyone!

Lmao

Target to solve A, B & C asap to get a good rank, D seems to be much harder based on rating distribution.

Noice looking contest good luck :> HEHEHE HA

I hope a good result for everyone !

Hope all get positive points :)

`many tks for your efforttt hope to up rate uhuhu`

Very good stuff worked to prepare this contest I guess, thanks

your welcome

As your father,I want some contribution

I just want to say that 743 is a prime number.

why are you a foolishgoat ? Btw I just want to say that number of characters in foolishgoat is also prime (11)

Well, I'm foolishgoat because I'm just a goat that comes to Codeforces so that I can train and become a smart one.

Suiiiiiiiiiiiiiiiii!

all the best to all

Wish Me Cyan !!!

"An Unexpected error" is showing up whenever i am submitting my solution!!! Anyone else facing the same issue????

Me also

Check the selected problem either it's selected or not every first time submission.

Queue :(

too long queue :(

A contest after ages and what we see is a Queue

unrated please

QueueForces T_T.

such a long queue , it has to be unrated .

15 minutes since A is in queue.....

Thinking for 10min shall I get TLE in problem B or not :)

too long queue:(

while true: dequeue()

please extend the contest by 15 mins at least due to the long queue.

don't make it unrated because the problems are really good.

just cause you solved , this is unfare to wait for 20 min to get wa verdict

i never said it is fair to not make it unrated but an alternative is to extend it since a contest is much awaited by everyone.

imagine waiting for B and then getting wa after 30 min for some silly stuff , meanwhile some guy solved after 30 min and submitted in first go , I dont think extending contest will justify now

So good that you submitted A twice

it was a mistake, i accidentally submitted my B solution in A.

yeah, I thought so ...

Its not possible google kickstart is on the way.

for me at least any CF contest >>>> any other contest except(ICPC).

Lol, I guess you are only thinking about you. Possibly everyone here will participate in kickstart.

This should definitely be unrated!

The contest must be unrated . Such a long queue

I think there should be a clear guideline in Codeforces when the round should be unrated. Because queue in first few minutes can severely affect the rating in case there is a difficulty gap in the problems.

Why so long queue??

Too long queue, please, make this round unrated

The queue is too long today? i submitted a solution 5 mins back and still it's not judged yet.

20 minutes here bro!! I feel the same

30 min here

lol

Queueforces ;-;MikeMirzayanov what type of queue is used for submissions, my submission at 7 min is not yet judged but some user's 12 min sol are judged. Please help!!!!

That might be a stack, not a queue then!

might be a stack of queues lol

If unrated, my possibility of becoming CM will be ruined for this time.

If unrated, my possibility of becoming newbie will be ruined for this time.

Sed lyf, but the chances of being unrated is 99.99%.

Thanks codeforces. I will be back again.

Done :( Hope you break into CM next edu!

Done :( Hope you break into CM next edu!

Same :sob:

I feel you.

If unrated, my possibility of becoming pupil will be ruined for this time.

Same here . 2nd person to Solve C . Also A and B were pretty easy but here I'm crying

probably because of queue issues and many ppl left, just saying

Unrated please!!

Why queue is so long?

Is it unrated?

20 mins just to get wrong answer so sad

The contest should be unrated cause of long queue!

(https://c.tenor.com/SB66UNkGc0gAAAAd/sloth-slow.gif)

@mike @contest admins please give the info whether it will be rated or not

so that we can do some other things.

I am waiting for more than 30 minutes, still my solution is not judged .

Next contest shall be to program decent queue

Disgusting queues

To be honest round should be unrated as 30 minutes are too much just to get a wa.

Nothing to take from Authors you did a phenomenal job

but it should be unrated.

30 minutes are too much guys/

Such a bad queue :(

Me: submits

Codeforces: Click this

It is also gonna cost penalty and pain if it comes out to be wrong.

Imagine submitting before 30 min and getting WA. :))

It'll be unrated, there's no point participating. There's no better way to compensate for never getting results on pretests.

queueForces

Finally it's unrated now !!

it's unrated no doubt just have fun solving problems

Good bye , tata, Allah Hafiz

Nice round. It's a pity that this round is unrated(

F

Great questions, sad the round will be unrated.

Second person to solve C only for round to get unrated xD

Best round but very sad that this became unrated

I thought the problems were really nice, so I'm disappointed that the round went unrated.

will it be unrated for div 1 also?

still 0 announcement

yeah! it was announced that it will be unratted for div1 too

Bye Bye, See you at Google kickstart

When I solved A and B under 15 minutes, the round gets unrated. Of course, why not :(

Read first three problem . The questions were interesting . Sad that the round in unrated :(

same, pal. same

Thousands of submissions are in queue and only one of them is running. why?

thanks for hosting I enjoyed the problems I attempted. sad the queue times were too long.

B and C were so beautiful, very sad it's unrated.

this is not expected from such big organization please fix this for once I did so good in the contest (a and b) this fast (i'm newbie) :(((

Results of other OJ：

Result of Codeforces:

lol

3ms TLE kinda sus tho..

lol

where i can see it? im about first image

It's a pity that the round was unrated. But, the tasks were interesting. Thanks!

After so much time as a beginner i am finally able to solve a few and it is unrated now. Tbh i am pissed off :(

I have a theory that most problem-setters make tough problems just to avoid queue issues. LOL! Nice contest, btw.

problems were really good and don't deserve unrated(;_;)

Feels sad man , when first contest you give goes unrated :(

I_Hate_Swaps you shouldn't feel sad

problem B.Swaps

nowdays Long queue is common problem in cf. please do something MikeMirzayanov :(

Custom invocation is also unusable during contest.

Imagine waiting for a Codeforces round for 1 week... and this.

Unlucky comeback after ~2 years of inactive. F.

yup for me ~2 months F.

Please fix this!

unraited...

How to get rid of queues on CF?

The answer is...Make div. 1 rounds only

Unrated？

comeback failed, folks. :(

Could have been one of my best, really sad. But a lovely problem set. Cheers to the setters!

It was the first contest I liked in probably the last 12 months and seeing it getting unrated this way, I feel sad for the authors very much. Markadiusz and Asymmetry I liked the problems a lot and hope to see you guys with another contest soon!

Thank you for your support :)

Problem 1C seems to be the same with Problem C in 2020-2021 ACM-ICPC, Asia Kunming Regional Contest.

cucked my comeback

Problems was really nice. i think i will become specialist again after this contest :) but Unfortunately unrated :(

The problems are nice! Because it is unrated now, I can sleep early today LOL :)

do kickstart that is in ~80 mins from now

That's too late :(

This is still a good problem collection and worth trying out. I will still try to do my best even if it's unrated.

Can there be a announcement that they will rate the round? If not please tell me so that I can stop worrying. :(

Don't Worry Codeforces always keep their words... They said its unrated means its unrated...

1-gon's first round got unrated due to the long queue. NowAsymmetryandMarkadiuszset their first round coordinated by1-gon, and the round got unrated due to the long queue. RIP:(

And Both contests had nice problems :(

logic behind b?

Since a[0]!=b[0] we need to make those two elements so that a[0]<b[0]

That is, we iterate all a[i], and foreach one we find the smallest possible j so that a[i]<b[j] That pair of i,j need i+j operations, we search for the min such pair.

We can find the smallest j for a given i by binary search on the prefix arrays of a[] and b[]. make preA[i] the smallest value in a[0..i], and preB[j] the biggest value in b[0..j]. Binary search the smallest j where preA[i]<preB[j].

We can avoid binary search since we know the numbers in a[] are 1, 3, ..., 2n-1. Create an array c[] such that c[k] gives the index j for number k i.e. if a[i]=k, we get j by c[k]. To fill c[], we can iterate over array b and store index j for all numbers smaller than the current no in b.

Doesn't this require sorting, which yields linearithmic time complexity either way?

No, the numbers are relatively small compared to the length of the input (<=n), so a simple loop, described above, is enough.

Oh I completely misunderstood that comment. I got it now, thanks!

give editorial ASAP.

I have to do a lot of homework.

Spoilerdidn't got any AC.

Spoilerproblem A

how to change 2020 with 6 operation. help me please!!

well I think: 2020 -> 0022 -> 0020 -> 0002 -> 0000

swap two digits (you can choose which digits to swap, and they don't have to be

adjacent). how the following operation is feasible with one swapwe must have to do two consecutive swap.

0020 -> 2000 -> 0002

please clarify

swap 0 and 2?

i don't understand what you wonder...

p/s: they don't have to be adjacent, but adjacent is accepted

logic was sum up all non zero numbers and also increase by 1 if the position of nonzero element is not last...

i also have the same doubt becaz this---

"you can choose which digits to swap, and they don't have to be adjacent" has been clearly mentioned in the problem

2020 -> 2002 (swap 3rd and 4th element) -> 2001 -> 2000 -> 0002 (swap 1st and 4th element) -> 0001 -> 0000

The problems were good. (especially Problem B...able to do in O(N)) PS: All the best for Kickstart buddies. :)

First time my code was accepted in contest but fine i enjoyed the problem....

Any thoughts of problem E? It seems to be a dp problem, but I didn't figure out the recursion formula.

it looks more like a divide and conquer problem for me, i tried dp and it didn't work.

NVM. Thought it was D.

I dont understand why people in comment section are filled with sarcasm and down votes

For problem C, we first figure out if it is possible to finish the whole book by SCC (strongly connected component). If size of some (possibly 1) SCC are greater than 1, then it is impossible to complete the book. After that, we use 2 min heap to store pages with indegree 0. First heap stores current traverse. Second heap stores next traverse.

Link for my submission

Using SCC is a bit overkill and definitely a waste of time for this problem.

The goal (when checking if understanding the book is possible) is to check if there is a cycle. We can do that with a DFS (or multiple, in case of a DAG, which we are dealing with here) which, instead of marking a vertex visited, can set two different kinds of flags.

The DFS works as follows:

We can count the result with a DFS too, using dynamic programming on a DAG. Every vertex with no outgoing edges receives value 1, because that chapter will be understood during the first reading of the book. Every other chapter will be understood after all its prerequisite chapters are understood, so the result for the chapter is the maximum of the results for the prerequisite chapters... but slightly modified to account for the order of the chapters in the book.

I'm not the best at explaining so maybe my code will help.

topological sort :|

Yes, and?

Idk :D

I actually used a different approach. I made two heaps : the first one is the "main" heap, the second one is the "pending" heap

The "main" heap's purpose is to store the progress chapters you can learn right now given that you're already learn all required chapters. Meanwhile, the "pending" heap's purpose is to store the chapters which you have learned all the required chapters but the chapter number is less than the chapter we currently on

So let's process the chapters one by one starting by those which have no requirements. (E.g. the ones which have 0 in-degree)

Each time we're processing of the current chapter (let's say we name it U) we do this :

check all other chapters V which has U as requirements

subtract V's degree by 1

if V's degree become 0 then : A. If V > U, then this chapter comes after U, then put it on the "main" heap OR B. If V < U, then this chapter lies before U, since we can't read it backwards, put V to "pending" heap

on the end of each processing, if our "main" heap becomes empty, put all chapters in "pending" heap to "main" heap and increase the number of read by 1 (because this means we will start re-reading the book)

To check if it's impossible, just check if after we're done with all the processing, there exists at least a chapter we haven't processed before

My implementation

This work is O(nlog(n)) though due to the heap! But it's still a fine approach.

Video solution for C using topological sort

How to solve div2 D?

First of all, xor of all the elements of the array should be $$$0$$$ (xor of all the elements of the array isn't changed after any operation).

From here, I am assuming that xor of all the elements of the array is $$$0$$$.

If n is odd, it's always possible to change all the elements to $$$0$$$.

Note that if $$$a_{i}=0$$$ and $$$a_{i+1}=a_{i+2}$$$, $$$a_{i}\oplus a_{i+1}\oplus a_{i+2}=0$$$. This is the main idea of my approach.

Now select indices $$$n-2,n-4,...,1$$$. It's easy to observe that now the first element of the array will be $$$0$$$ and $$$a_{i}=a_{i+1}$$$ for all even $$$i$$$ less than $$$n$$$. First element will always be $$$0$$$ because it is the xor of all the elements of the array Now select $$$1,3,...n-2$$$. After these operations, all the elements of the array will be $$$0$$$ and total number of operations performed is $$$n-1$$$.

If $$$n$$$ is even, you can split it into two subarrays of odd length. So if you can find an $$$i$$$ such that $$$i$$$ is odd and xor of first $$$i$$$ elements is $$$0$$$, you can change all the elements to $$$0$$$, otherwise you cannot.

Oh holy crap this is brilliant. And here I got stuck with some greedy ifological strategy which... I'm reasonably confident would work but I didn't manage to implement it.

Yeah, such approach can work too but I had to consider many different cases.

Basically: we iterate from left to right, either A keeping all elements behind 0 or B leaving all elements behind 1 (for simplicity ensuring that the current element is 1 too after previous operations).

A: if applying the operation to the current $$$i$$$ will result in zero xor, we do it. Else if the $$$i$$$th element is 1, we ensure that $$$i+1$$$ is too; we know that $$$i-1$$$, if it exists, is zero, and we can zero $$$i-1..i+1$$$, and if it doesn't, we switch to B.

B: if applying the operation to the current $$$i$$$ will result in zero xor AND there's an even number of 1s behind us, we zero $$$i..i+2$$$ and 1s behind us two at a time. Otherwise we ensure that the next element is 1 (working through a few cases it can be shown that if it's 0, the current xor must be 1).

codeFirst you must have an even number of 1s in the array, otherwise it's impossible to change all elements to 0s.

Then group all the 1s in pairs: (1st, 2nd), (3rd, 4th), ... we will solve the problem by changing each pair of 1s to 0s repeatedly:

`101`

, then do the similar thing on case 1 but from right to left. Note that there are no extra requirement to this case.So the strategy is firstly find one pair that can be solved then solve the remaining pairs from that pair in both direction. If none of the pairs can be solved, it's impossible to change all 1s to 0s. Submission: 129221560

https://codeforces.com/contest/1573/submission/129205633 Can someone please help me in figuring out why this submission gets TLE?

How to solve div2 B ?

We need to find the minimum no of swaps so that no at the first index of odd array is less than the no at the first index of the even array . One way to do this is brute force. For each even no we try to get the nearest odd no which is less than that even no in the odd array. But this will be a O(n*n) algorithm and may give TLE for the tight constraints. So a better way is to store the indices in some ds like a map and for each no 2 4 6 8 ... We store the minimum of the indices of the current no — 1 and the index of previous no .And finally itterate through all the possibilities which will be only n checks and get the global minimal answer out of all the possible.

My Accepted code :https://codeforces.com/contest/1573/submission/129208921

Although round got unrated but I liked task B!

My segment tree solution for Div1 A

can you just briefly explain what you are storing/updating in the nodes?

Can anyone help me to find a counter case? 129205285

Correct answer is 3 yours gives 2.

I have used a similar approach with the topological sort, except instead of storing the chapters to be read in the next iteration in

`vec`

, i just added n to the chapter and pushed it into`pq`

. 129229753Thanks a lot, but i was so dumb that i forget priority queue stores greater elemnet first. I thought it keeps smaller element first. You just saved a lot of time mine.

I have tried a video editorial for problem C :book https://youtu.be/KWkkZEGffZw . Have a look at it..and if have any doubts let me know in comments

Can I change the statement that a[i] is (0/1) to [0,100000] on Problem 'Xor of 3'.Is there a solution?

Maybe you can solve the task in each bit, but the operation times may exceed $$$n$$$.

I think We can do that as well in n-1 operations.

I think my solution doesn’t use the 0/1 condition so probably yes.

Video Editorial for Problem Div 2 C / Div 1 A

editorial is also still ``in queue'' LOL

is the editorial rated ?

I'm newbie :(((

For Question B, Can Someone please tell me why my approach is wrong? https://codeforces.com/contest/1573/submission/129231370

My approach is that our task is to make a[0] < b[0], this can be done by two ways, either bring the closest number which is greater than a[0] in 'b' to the 0th position, let's say it is at j-index, then the answer will be 'j', or bring the closest element from 0 in a which is less than b[0], and the final answer is the minimum of both!

Any counter test-case which fails this approach would be highly appreciated!

answer is 2 not 3

Thanks, I got it!

You are only considering making the swaps on either a or b. In some cases it is beneficial to do the swaps on both arrays.

You want the clock to show 0 with as few operations as possible. In an operation, you can do one of the following:

1 decrease the number on the clock by 1, or

2 swap two digits (you can choose which digits to swap, and they don't have to be adjacent).

here in 2nd operation we cannot swap adjecent digits but no one is talking about it ! plzz reply am not getting the exact thing or it is something that was some mistake from creators and now no one is talking about it since they got ac?

Is div1 B supposed to make us feel pain?

Yes!.. No? Maybe.

Check this solution out!

Easy solution for B [Sparse Table]: 129189027

waiting for editorial ...

orz Molewus for the round being unrated.

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