Hello!

Welcome to the Codeforces Round 844 (Div. 1 + Div. 2, based on VK Cup 2022 - Elimination Round) that will start on Jan/15/2023 15:05 (Moscow time). It will be a combined rated round for both divisions and open to everyone.

This round is a mirror of VK Cup 2022 Elimination — annual programming championship for Russian-speaking competitors organized by VK. VK Cup started in 2012 and has grown to be a five-track competition in competitive programming, Mobile, ML, Go, and JavaScript.

All the problems are authored and prepared by me. Thanks to KAN, errorgorn, lperovskaya, dario2994, Monogon, Arpa for making this round better.

You will be given 8 problems and 3 hours to solve them.

**UPD**: Editorial

Congratulations to the winners:

and to maroonrk for getting the only accepted solution to problem H2.

omg tourist round

omg tourist round

omg tourist round

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omg tourist round

我的天啊我打破了你愚蠢的锁链!

omg tourist round

omg tourist round

omg tourist round

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omg tourist round

haha broke your silly chain

haha restarted the silly chain.

Omg tourist round.

You do realise I can just break it once it reaches a significant length again?

You do realize that I can just restart it once you break it again?

if(no_of_chain_makers > chain_breakers) cout<<"Relax"<<endl;

Now for everyone's sake,Im restarting the chain again. Thank's for the support guys.

Omg turist round.

You do realise that I can just break it once again?

I'll start again.

Omg turist round.

Omg turist round.

omg tourist round

omg tourist round

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Omg tourist round.

Omg tourist round

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Omg tourist round.

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omg round tourist

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turist lol

omg tourist round

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OMG tourist round

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Omg tourist round

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omg tourist round

HaHa! I broke your silly chain.

Ha Ha! and i started it again.

omg tourist round.

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We need more!omg tourist round

omg tourist round

.

omg tourist round

Notice the unusual time

omg tourist round !!

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OMG tourist round

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no

If it's possible can you please explain how did you get Specialist Rank?

just be good at programming lol

omg tourist round

omg tourist round~my love

Damn, 3 hours for only 8 problems. Scary

omg tourist round

omg tourist round

Damn, 3 hours for only 8 problems. Scary

omg tourist round

omg tourist round

I am so interested in your graph, how did you maintain a slope of 45 degrees ?

وانتا مالك؟؟

What language is this, looks Arabic I don't know it. Speak in English. Also, I expressed my interest in speedy_boy graph

NOT yours. Actually, your graph isn't interesting at all mr.newbie! The comment directed to you is below : Newbie detected, opinion rejected XDRespect to newbie

yes sir. all due respect! but he seems talking in a rude manner to me by a foreign language tho I didnt talk to him and replied to ur comment only :-( However, your graph with slope 45 is so cool ✪ω✪ speedy_boy

For your first 6 contests, you actually start at 1400, and as you do contests your hidden rating is updated as normal and your shown rating rises rapidly-ish to reach it.

He didn't maintain it after this contest though, if he did he should be orange by now.

newbie detected.. opinion rejected xD

Shortest announcement ever *_*

Yes, hope the problem statements will be short as well.

:DAlso the most upvoted one :)

Not even close...

(although this might still not be the most upvoted one)

damn!

Tourist can't compete against benq in this round, but he can make some crazy geometry problems to force benq to lose rating and get back to rank #1.

What an wonderful Opportunity. Haha to be crowned again.

This contest is only for Russians ig.

Yes VK Cup is for Russians, but this is a mirror of this contest and everyone can participate.

What a great idea

good

Benq is a fair contestant, he's like:" Amma beat him while he's participating"

3 hours? How difficult the problems will be

now you know

Maybe my "master experience card" will be expired in this contest.

talent, and nice avatar

I don't care about the difficulty lvl of the problems if it's Tourist Round. All i know that it is going to be fun.

Note the unusual timing

omg tourist round ...^,^

haha，that’s right

yes,its correct.

absolutely right,haha

Guess will have to wait for

tourist vs Benq:-(what you guys think is Benq gonna or not gonna participate this contest?

Benq will not participate, because tourist has intentionally made problems with topics where Benq is weak. This is to make Benq fall to second place in top rating chart, so that tourist can become #1 again.

xD

Oh,man !! Is there any topic on earth in which these legendary guys are weak?

I don't know, maybe some optimization of matrix multiplication algorithm from $$$O(n^{2.43})$$$ to $$$O(n^{2.42})$$$.

To make benq lose rating you have to specifically make it hard for him. If it’s hard for everyone he won’t lose rating.

But Benq is better...

I think Benq only wanna compete with tourist who is well-known as the

No.1in competitive programming. In my opinion, he will not participate contests without tourist such as this contestTomorrow i will miss Benq vs tourist but no worries because later they will be facing each other soon

Benq only compete with tourist in the contest.

Benq only participate in the contest in which tourist participate.

Omg 3hr Round ><

I think, Benq will not perticipate in this contest. But I will participate and face to the world number one Programme's problems...☝️

Score distribution.......???

tourist is a great personBut I really hate most of his problemsBut I can't skip his contestI'm crazy.

Who said you are not Crazy?

omg tourist round

omg tourist round

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Omg clash with ABC.

I am a pro memer.

.

My IQ has decreased by 20 points while trying to read this. Thanks.

This is the shortest announcement for a round that I have ever seen :))

Legend don't need to say much. Their presence is enough alone sir!

omg tourist round . Probme 1 will be 1200+ Rated now

Nope. It was 800 Geometry. Haha

orz

It is rated ？(⊙﹏⊙)

well, obviously, sorry bother……

TFW you expect tourist to claim back first place then see that tourist himself is the author

.

It was good one . Don't know what's wrong ? Maybe too rude or offensive for some people ?? Atleast you should have blurred the authors above. It's disrespectful for them.

Contest Clashing with atcoder Beginner Contest 285.

Tourist round >>> Any Other round on this planet. As Simple as that .

I might just end up top1 because problems are being designed by tourist.

tourist revealed.

when I saw by tourist, inner me: Don't worry, Better luck next time!

Orz tourist round.

score distribution when?

omg tourist round

omg tourist round

omg tourist round

while(1) { OMG Tourist Round }

There will be very Interesting questions. Very Excited for the round!!

GOAT

omg tourist round

ok

omg tourist round

Should be Interesting.

omg tourist round

~~Thanks to MikeMirzayanov for amazing platforms Codeforces and Polygon.~~~~Thnx to the community who made codeforce this good~~omg tourist round

Contest Collision (CF & ABC):TimingsAtCoder->17:30 ISTCF->17:35 ISTCF is based off VK cup. It can't ve postponed

omg tourist round

i do bad in tourist rounds, hopefully this will change tomorrow

why problem in this year contests are didn't given any rating till now?

conflict with ABC285... what a pity that i can't participate in both contests...

That's right...

Your profile picture looks great

Note the

unusualtiming.It's such a piece of cake for tourist to send out a splendid contest. & Just BY HIMSELF !!!!!!!!

tourist gang

The One Piece is real!!!

Originally I am not planned to participate this contest because I have a date then. But when I see the author, I just postponed the date and register for this contest. Wish I can turn cyan once again.

You guys are having girlfriends!

Reference

omg tourist

I wish the next round will be written by BenqObviously tourist round will be full of fantastic problems. Good luck, yeah! ^=^

I will always be a fan of tourist!

OMT Gods round

It is briefly before the beginning of the round now, and score distribution hasnt been announced yet...

is this rated for everyone?

yes

omg tourist round

Where is

score distribution?I think I may not be able to participate because I feel like going to the toilet right now. What a wrong timing. :(

glhf

OMG tourist round

I regret I couldnt participate in it as I got stuck with some personal work

omg tourist round

why the rating of H2 is higher than the rating of H1

it said that H2 is the hard version,doesn't it?

its not H1 vs H2 but rather H1 vs H1+H2

A solution for H2 always works for H1. So if you solve H2 you get the points for both.

oh

F is an amazing problem! couldn't code it in time though, but mindsolved it

Can somebody please tell the solution?

$$$dp[i][j]$$$ = $$$P$$$(Minimum Prefix $$$\geq$$$ $$$ -j $$$ )

In E, finding maximum total area is not that hard but finding the exact subrectangles seems quite difficult. Stuck on it's implementation. Any approach regarding the same?

Maintain a stack for the rightmost intervals for u, d, ud rectangles. It was very painful implementing this, especially when the contest started at 4am for my local time.

Hey, thanks. I just implemented the logic using your stack idea. Yeah, felt like handling a lot of cases!

My solution

Pretty hard contest.

No idea for D. Have idea for E but got WA. Now I'll return to CM.

how's C done ....?

It's pretty annoying. You need to consider all j that n is multiple of j, calculate how many char will be changed if you make the string to j kinds of different chars. You need consider 2 cases: j<j0 , j>=j0, where j0=the number of different chars in the initial string. If j<j0 you need change some chars with low frequency.

After you decided the optimal j, you need to add all chars you'll put to the final string into a "char pool" (implemented as int pool[26]), first try to make every chars remain the same(if(pool[s[i]]>0) pool[s[i]]--; flag[i]=true;) the assign i where flag[i]=false any char in the pool.

Well consider two indices i and j. now calculate all such x that make both a[i] + x and a[j] + x a perfect square. the rest should be easy

In E we can first shrink rectangles with same type (row1, row2, row1+row2) and then consider for each 2-height rectangle, shrink it if it's penetrated by any 1-height rectangle, and shrink every 1-height rectangles who doesn't penetrate it but cross with it.

However I got WA at pretest 15. Also C was very annoying. I spent about an hour for it.

What I did is calculate the answer firstly for two numbers, let's say a and b. We want to transform a into x^2 and b into (x+k)^2 (because b > a), and that means that $$$b-a = (x+k)^2 - x^2$$$, so $$$x = {(b-a-k^2)}/{(2k)}$$$. You can just try every k from 1 to sqrt(10^9) to see whether that k satisfy this condition, then, for each valid k, calculate the number of perfect squares you'd also obtain from the other numbers in the array.

Let's choose some x. If ai & aj becomes perfect square after adding x, then: $$$a_{i}$$$+x = $$$u^{2}$$$ & $$$a_{j}$$$+x = $$$v^{2}$$$ Substracting , we get $$$a_{i}$$$ — $$$a_{j}$$$ = $$$u^{2}$$$ — $$$v^{2}$$$

=> $$$a_{i}$$$ — $$$a_{j}$$$ = (u-v)*(u+v)

Now split ($$$a_{i}$$$ — $$$a_{j}$$$) into 2 factors f1 & f2 such that f1*f2 = $$$a_{i}$$$ — $$$a_{j}$$$.

So, f1 = u-v & f2=u+v.

Find, u from here. Substitute in $$$a_{i}$$$+x = $$$u^{2}$$$ & you will get x.

Find the count over all possible x in this way & take the minimum one.

I thought trying for every factor of a 10^9 number for 50*(50*49/2) times would cause TLE…

my thinking was similar to you though instead of tle i got wa

Now I upsolved D. My submission:189361440

The proof of that the submission is available:

In the algorithm we consider for all divisor k of (aj-ai) where j>i. The maximun number of divisors we check in one case is:

sum(1<=i<j<=n, sqrt(aj-ai)/2)

The "/2" is because if (aj-ai)%4==0 k must be even, if (aj-ai)%4==1 or 3 then k must be odd.

Notice that sum(a[i+1]-a[i])<=A (A=max(a[i])=10^9). We can see for each d, sum(1<=i<=i+d<=n, sqrt(a[i+d]-a[i])/2) is maximized when all a[i+d]-a[i] are same (can be proved by the mean value inequality) and the maximun value is about (n-d)*sqrt(A*d/n)/2=n*(1-t)*sqrt(A*t), where t=d/n. We sum up it for d=1...n-1 and the sum is about O(n^2*sqrt(A)), which fits the time limit.

Thanks my mate

It doesn't TLE, because you're factoring the differences of $$$a_i$$$ and $$$a_j$$$, all of which can't be around $$$10^9$$$ simultaneously.

it's around 4e7 operations (which should be doable even with 1s time limit), plus the time limit is 4s.

But for case a[i]=1+20408160*(i-1), it runs for 10622587 operations.

My testing codeOHHHHH I forgoted there's this line in the statement:

I missed in the contest.... I've thought sum(n) could be 2500 and missed the intended solution.

I just upsolved D with a similar solution. I iterated over all n^2 pairs of numbers. For each number, i iterated over all values of f1 (so sqrt(max(a)) time) and did some math to see if it worked. Then I took all the possible values of x I got, and I simulated the process for each value of x.

have idea for D but got wa

Problem A was kind of ugly. C also required (at least in my case) a quite hefty implementation.

problem C is tough :((( Can anybody show me some hint? :((

Definitely, not an easier one!!

I spend almost 3 hours but still couldn't solve it :((( But i see so many people could solve it :((( Pardon me for my bad english

Just brute every i from 1 to 26(i is how many letters will be left in string -> n % i == 0) and then you should calculate how many ops(call it cnti) you needed to make string balance(i used greedy in this part). So your first answer is min among cnti.

Implementation is nightmarish.

Can't agree more

Suppose you want to modify the string to have exactly $$$x$$$ distinct characters. How many times will each character appear? Which $$$x$$$ characters should you pick? Can you calculate the minimum number of changes to reach $$$x$$$ distinct characters?

ofc the first question immediately got hit by annoying geometry question lol(don't worry imo it's also interesting)

how's problem C solved ...... nothing struck my brain for like 1.5 hrs i took multiple examples but i couldn't generalize it .

You can keep only 1 <= i <= 26 characters with n/i frequency for each one

Try them all and minimize the answer

Can't believe I clutched C like that.

Seeing the drawing for problem A before reading the problem statement was intimidating lol

omg tourist round

Anyone know what pretest 11 was about in problem E ?

I got WA11 and 2 1 3 2 3 2 3 2 4 helped me find the mistake. When I was using only 1 row out of 2-row rectangle I forgot to clear the other row in the segment tree

How to solve D? :(

See here

The answer can always be 1. Now let's see what will happen if the answer is $$$>1$$$.

So suppose the answer is 2. This means that there are any two indexes(let's assume i and j) such $$$a[i]+x=c*c$$$ and $$$a[j]+x=d*d$$$. Here c and d can be any valid integer. So now $$$d^2-c^2=(d-c)*(d+c) $$$.This is equal to $$$(a[j]-a[i])=(d-c)*(d+c).$$$

Now if we factorize $$$a[j]-a[i]$$$. We can find the value of $$$d. and .c$$$. Using those values we can find the value of $$$x$$$. Now we will store all the eligible values of x and use the best possible option.

SpoilerHow's the factorization is happening. Suppose $$$a[j]-a[i]$$$ has a factor($$$A$$$).Then there will be also be another factor $$$(a[j]-a[i])/A (assume B)$$$. Now $$$A*B=(d-c)*(d+c)$$$. So $$$d=(A+B)/2$$$ and $$$c=B-d$$$.

I didn't get how you solved to get, d=(A+B)/2 and c=B−d

since $$$d+c=B$$$ and $$$d-c=A$$$. If we add both the equations we get $$$2*d=A+B$$$ which is equal to $$$d=(A+B)/2$$$. For the value of $$$c$$$ it comes from equation 1 by shifting $$$d$$$ from left side to right side.

I swear F is easier than E...

any hint on d?

D is simple but the constraints are way to big. Any idea how to solve it ?

Suppose that $$$b_i^2 = a_i + x, b_j^2 = a_j + x$$$, then we have $$$(b_j - b_i)(b_j + b_i) = a_j - a_i$$$. By enumerating $$$b_j - b_i$$$ we know what $$$b_i$$$ and $$$b_j$$$ are, and also the $$$x$$$.

So ... it is not simple, then?

its not. just realized :(

D was very good

Yeah, it was nice. I initially thought it was some sort of meet-in-the-middle trick, but was pleasantly surprised to see it was a much nicer brute-force.

How did u solve D ?

Spoilerlets consider i<j and a[i]<a[j] then if both a[i]+x and a[j]+x turn out to be perfect square then

$$$a[i]+x=n_{i}^{2}$$$

$$$a[j]+x=n_{j}^{2}$$$

then from $$$ a[j]-a[i]= (n_{j} - n_{i})*(n_{j} + n_{i}) $$$ you can find all possible x's in $$$O( \sqrt{n} )$$$

then for any x if total number of pair's it makes perfect square simultaneously is y then this $$$ ^{ans} C _{2} = y$$$ should hold