Hello, Codeforces!

Zlobober and I are glad to invite you to compete in Nebius Welcome Round (Div. 1 + Div. 2) that will start on Mar/12/2023 17:35 (Moscow time). The round will be rated for everyone and will feature 8 problems that you will have 2 hours to solve.

I feel really thrilled and excited about this round as this is the first time I put so much effort in a Codeforces contest since I quit being a round coordinator in December 2016 (wow, that was so long ago)!

We conduct this round to have some fun and we also hope to find some great candidates to join Nebius team. Solving 5+ problems in our round will be a good result and will count as one of the coding interviews. Apart from that top 25 contestants and 25 random contestants placed 26-200 will receive a branded Nebius t-shirt!

Some information about Nebius. I have joined this new start-up in cloud technologies in November as a head of Compute & Network Services. It's an international spin-off of Yandex cloud business with offices in Amsterdam, Belgrade and Tel Aviv. We offer strategic partnerships to leading companies around the world, empowering them to create their own local cloud hyperscaler platforms and become trustworthy providers of cloud services and technologies in their own regions.

We know that competitive programming makes great engineers. It boosts your algorithms and coding skills, teaches you to write a working and efficient code. We aim to hire a lot of backend software engineers for all parts of our cloud technology stack that ranges from hypervisor and network to data warehouse and machine learning.

Here you can check Nebius website and open positions. If you feel interested in joining Nebius as a full-time employee, go on and fill the form.

Special thanks go to:

MikeMirzayanov for his outstanding, glorious, terrific (you can't say enough, really) platforms Codeforces and Polygon

74TrAkToR and KAN for this round coordination (and for rounds coordination in general, good job lads!)

ch_egor for his help with one of the problem that has a peculiar setting which we will discuss a lot in the editorial

**UPD**Now that the list is complete we can say many thanks to all the testers! dario2994, errorgorn, magga, golikovnik, Hyperbolic, Farhan132, IgorI, vintage_Vlad_Makeev, aymanriadsolh, Vladik, Junco_dh, dshindov, csegura, Eddard, Alexdat2000, kLOLik, wuruzhao, Aleks5d, Ryougi-Shiki, qrort, ak2006, AsGreyWolf, farmerboy, vladukha, usernameson, ZiXeL, DzumDzum, Java, samvel, mister, Tensei8869, thank you for all your help on improving this round's quality!

Fingers crossed all will be well, you will enjoy the round and we will be back with a new one in several months!

**UPD2** The scoring distribution is 500 — 1000 — 1000 — 1500 — 2000 — 2750 — 3500 — 3500.

**UPD3** The editorial is here!

**UPD4** Congratulations to winners!

1-st place: jiangly

2-nd place: tourist

3-rd place: Um_nik

4-th place: isaf27

5-th place: Ormlis

**UPD5** T-shirts winners!

Auto comment: topic has been updated by GlebsHP (previous revision, new revision, compare).I have received a mail that my solution of C problem in Nebius Welcome Round is coinciding with Somebody else . In mail it was written , to post about the evidence .

C Problem was not too difficult for a person with average coding skills . The only trick in C was to get a control over the loop . I have seen many tutorials of past contests and in C problem , I used that for controlling the loop as the most common approach in these types get solved in nlog(n) or n root(n) . So I did it for n root(n) and then the code was simple a Brute force logic . It took me a number of contest and practice to reach specialist . Please reply me if i need to submit any supporting proof and what else can i upload . Please help GlebsHP , MikeMirzayanov . please do reply.

Make sure to have strong pretests this time ;)

Make sure to submit strong solutions. Good Luck Omar.Gawdat and everyone else.

Yeah sure, but both is still better, isn't it?

Last round tourist got FST, so I believe if tourist can't make sure 100% from his submitions then none of us will :D

Edit: Don't know why would anyone disagree with that :)

LoL I got FST :D

Woah, Glebs and Zlobober. I haven't seen these two names in a long time.

How long have you been looking for these names?

ooh ,Errichto, i haven't seen you on youtube for long time , we want you to continue on youtube :)

Cool! Will this contest be rated for everyone?

Yes!

How to solve Div2E/Div1C ? my idea was to sort the vector of vectors based on the max value of every vector and in case of equality i sort them by the one with the less number of inversions ,, i felt it is correct but getting WA on test2

Actually you did it correct but you did not check for the case where n==1

Can you specify the case more clearly ? because my solution just works very well with small test cases

Wrong blog

No testers this time?

Number of testers >= 1

Can you please share how will you randomly select 25 participants for the t-shirt? Like the seed value can be the sum of the total points of the top 10 participants, or the total number of solves of all the problems, etc. It was done in one of the past rounds.

There appears to be a conflict with Atcoder Regular Contest 158. Pushing back the start of the round by 30 minutes will avoid the conflict.

Link to round: https://atcoder.jp/contests/arc158

There's no conflict because we have 35 minutes break

Atcoder is starting an hour later than the usual time this Sunday.

This is the start time of Atcoder (9PM Japan time, 1PM UTC): https://www.timeanddate.com/worldclock/fixedtime.html?iso=20230312T2100&p1=248

Atcoder start time is regular, and there is a 35 mins gap between these contests. The attached link opens at 1200 UTC instead of 1300 UTC. You might have DST issues.

Hoping to become specialist in this round, have been trying for specialist for quite a long time now, every time I was close there was one contest that would bring me down, this time for the past 5 contests, I have been gaining +ve delta, I am at my all time highest rating, need +23 to get to specialist. Sometimes, it demotivates me that I am not able to hit specialist despite all my other friends are already specialist but at the end of day, it's all about learning new topics and loving problem solving....

Why is this round (and some others) not numbered?

I guess it's because the dates of some of them may change

Why do these people always do job offers like any one would like to take a job at there company.

How I think these people areI don't even wan't to waste my time on these contests they are very hard.

Only one problem can be done hardly ever.

I advise you to not waste your time in this and help me with my " I wan't Div.5 " campaign.

https://rb.gy/rshslt

Make sure to have strong pretests this time)))

Good luck and have fun!

Make sure to make the first question easy so that number of participants who solve the first question increases.

agree

As long as it is a good problem, Doesn't' care about the difficulty to be honest , i will solve it anyways.

Nebius is an israeli company, so i will boycott the contest. I hope all arab do this as well

Indecisive dummy makes 5 edits and ends up at the same statement.

:skull:

good luck everyone:)

Wish i can do well in this div

Auto comment: topic has been updated by GlebsHP (previous revision, new revision, compare).Hi. Is it a requirement to move to one of Belgrade, Amsterdam or Tel Aviv in case of job offer?

At my first glance, I read the round as

Newbie Welcome Round— simple and friendly problems to welcome new people join Codeforces.Turns out to be opposite :)

Will there be some classical problems?

Let's gooooooooooooo

When can we expect to see the score distribution?

now)

8 problems in 2 hours!?

As a first-time tester, I'm sad I can't participate:(

u will be much happier, when you will get salary.... xD

Testers don't get paid

as a new account you know that... and even after participating in around 100 contests, I don't know that.

My whole life is a lie...

magga orz

magga orz

No Score Distribution ?

No Score Distribution?

Scores distribution???

Pretest passed but wrong in test case 99....

Auto comment: topic has been updated by GlebsHP (previous revision, new revision, compare).Hope everyone gets a positive delta :)

As a tester, I'm really happy to test this contest. Zlobober's codes are super straightforward and easy to understand so I studied his codes when I was a candidate master. And Finally, I could become a master and His codes were really helpful to be a master.

Zlobober is my eternal idol, and I am really proud to test his contest.

Please, Let me become GrandMaster!

I'm going newbie again!

D is the Worst Problem I Have Ever Seen!..

Yeah just concatenation of two different trivial problems :p

Which problems?

Find minimum, and find maximum. Both are trivial but asking for printing them separately with space is obviously a new problem.

This problem teaches to write stress test

D is literal ad hoc nightmare

is problem C literally just luck?

I guessed on the upper bound of the triangle number to search and passed pretests, so I'd say yes it was possible to be lucky. Although I feel like there was some closed form solution to find the minimum triangle number in O(1), I couldn't find it

Sum $$$1 + 2 + 3 + ... + 2n = ((2n+1) \cdot 2n) / 2 = n \cdot (2n+1)$$$ thus $$$mod$$$ $$$n$$$ you did a cycle.

I was lost making maths equations in first 45 minutes than realized there is a cycle :)

do brute force for every 1 <= k <= min(p,2*n) for the value f=k since if no f in this interval satisfy region 0 as result, it is guaranteed that there exist no k value such that with f = k <= p, it'll land on region 0 (guaranteed to work since sum of n from all test cases only reaches 200000) (also this works since the roulette is basically a mod n complete residue system)

I was not sure about the upper bound of i, where 1 <= i <= p. Submitted with bound to n, but then realised it should rather be 2 * n, as the formula (x + (I*(I+1))/2) has a division of 2, so the bound will be multiplied by 2 instead.

How to solve C ?? :((

Basically the problem asks for if there exists a value $$$1\leq i\leq p$$$ such that $$$x + \frac{i (i+1)}{2}$$$ is divisible by $$$n$$$.

Since For $$$i>n$$$ we can break the fraction into something like $$$\frac{(an+b)(an+b+1)}{2}$$$ you know testing $$$i$$$ until $$$n$$$ will be enough because you think it's just a repeating pattern afterwards — no, don't forget there is a 2 in the denominator, the pattern length can be as long as $$$2n$$$.

Just check that when f is 2*n then it will come back to its original position.

So run a loop from 1 to 2n and check

My approach:

Given $$$f$$$, the displacement is $$$\frac{f(f + 1)}{2}$$$. We want $$$x + \frac{f(f + 1)}{2} \equiv 0 \pmod n$$$. But this is equivalent to the existence of $$$q$$$ for which $$$x + \frac{f(f + 1)}{2} = qn \iff 2x + f(f + 1) = 2qn$$$. So we can simply try each value of $$$f$$$ from $$$1$$$ to $$$\min (2n, p)$$$ to see if $$$2x + f(f + 1) \equiv 0 \pmod {2n}$$$. We don't need to check beyond $$$2n$$$ because the LHS only contains multiplications and additions, which are invariant under modulo, i.e., trying $$$f$$$ yields the same result as trying $$$f \bmod (2n)$$$.

(Note that division is not modulo-invariant, so trying $$$x + \frac{f(f + 1)}{2}$$$ for $$$f$$$ from $$$1$$$ to $$$n$$$ is not sufficient; you have to try until $$$2n$$$)

ImplementationForces all the way

Personal take: ImplementationForces sometimes is actually good! Many real-life problem only require brute-force, yet hard solutions. ICPC-style contests also features many implementation-heavy problems. You do not need crazy algorithms or math knowledge to solve C,D, yet the amount of solves are perfect for problems of that level. Stress testing for D to find the exact solution is also a skill we need to have in real-life programming... The problems are also not misleading, so the only person that you can blame if you WA is you, right?

What was the logic for C ??

Math knowledge: Sum from 1 to N is N (N + 1) / 2.

Modulo knowledge: T (T + 1) mod N = (T + N) (T + N + 1) mod N

Therefore, just check I from 1 to N for I * (I + 1) / 2 + X mod N == 0.

That's what I thought, but WA on case something... change for from 1 to N to 1 to 2 * N works, but I don't understand how :D

Because if N=2 for example, the sum from 1 to 2 is not 0 mod 2(due to the over 2 at the bottom)

Just check that when f is 2*n then it will come back to its original position. So run a loop from 1 to 2n.

Because in congruent equations, from I * (I + 1) / 2 + X == 0 mod N, you can obtain the equivalent equation by multiplying everything by 2 (including the mod). Thus it will be equivalent with I * (I + 1) + 2*X == 0 mod 2*N. But now you have to check all possible remainder after division by 2*N.

My intuition behind it is because of something like this :

If $$$\frac{d(d+1)}{2} mod N = T$$$, then at some point, it will produce some modulo cycle (e.g. adding movement with the value of $$$d_1$$$ will results in the same finish point for another value of $$$d_2$$$ for some value $$$d_1 < d_2$$$)

And the cycle will repeats every $$$2N$$$ times. Because when $$$d = 2N$$$, then the value of $$$\frac{d(d+1)}{2}$$$ will be $$$\frac{2N(2N+1)}{2} mod N$$$ or $$$N(2N+1) mod N = 0$$$ because $$$N(2N+1)$$$ is a multiple of $$$N$$$

Hope my annealing simulation will pass systests in E.

Upd: it did!

How to solve B? Anyone. Failing in Pretest 2 & 3.

You could do binary search or greedy since the contraints allowed it.

Here is my greedy solution: 197134552

Binary Search (Upper Bound methold in C++ STL). Here is my Solution. https://codeforces.com/contest/1804/submission/197100967

How to solve E faster than (2^n)*(n^2)?

I think it's the fastest solution

so for A my idea was to alternate with the longest side so it'll double up and reduce one and if both inputs share same pairty it's double

why didn't this pass ? ~~~~~ Your code here...

int a = sc.nextInt(); int b = sc.nextInt();

~~~~~

It will always be x+(x-1) if x > y I think, because in the last step you reached the destination

And for x == y it is just x+x

i messed that up thanks dude

Oh no! I have been writing question D, but it continues to WA on pretest 5. Can anyone explain that? thanks!

My idea is to find two consecutive $$$1$$$ in a row and divide them into a group for the minimum value. For the maximum value, divide $$$01$$$ and $$$10$$$ into one group, in addition, divide $$$00$$$ into one group, and group the rest separately. If the number is not enough, divide the two into a group.

my code: https://codeforces.com/contest/1804/submission/197124751

Why was the rating change predictor( carrot or CF-Predictor ) not working for this contest ??

Deleted. :(

Even for me it is not working..

could someone please tell me ,why am i getting WA on this submission 197128613

rainboy finally rainboyed a contest, by being the only one to solve the hardest problem. Congrats!

How to solve D?

Auto comment: topic has been updated by GlebsHP (previous revision, new revision, compare).One of the nicest contests in a while. Very cool problems!

D was looking doable but i couldn't finish it :(

same here. Just finished writing a working code with example pretest. Waiting for the judge to finish to submit it to see if my idea is correct.

The worst D problem I've ever seen

why ??

this user BFU-marve is submiting in assembly, is this allowed?

Solved A-D. I passed pretest of E but I think my solution will be very likely getting FST. If it passed system test I'll get a large positive delta.

A: First, WLOG we can assume a>=0, b>=0 (by setting a=abs(a), b=abs(b)), and by looking for small cases we can get ans=a+b+max(0,abs(a-b)-1).

B: The problem is equvalent to "each pack of vaccines can be used on some patients with arrive time {t1, t2, ..., t_m}, where m<=k and t_m-t1<=w+d". Thus we can solve by greedy by finding maximal blocks of patients where they can get a same pack of vaccines.

C: Force f is good iff f*(f+1)/2 % n = (n-x) % n. Because (f+2*n)*(f+2*n+1)/2=f*(f+1)/2+n*(2*f+2*n+1), ( f*(f+1)/2 % n ) has a period of 2*n, so we only need to check for 1<=f<=min(2*n,p).

D: Consider for each floor separately. We need to put m/4 double-rooms at some places of the floor. Let the total count of bright windows of this floor is L, and the count of "11" double-rooms is t, then the number of occupied rooms is L-t. Therefore we need to minimize and maximize t. For maximization, we need to check every maximal block of consecutive 1s, and for minimization, we need to check every maximal block without "11". Be careful that we need to adjust t into range [0, m/4].

E: If there's a solution, we need to find a simple cycle in the graph, where the distance of every node and the cycle is at most 1. I used brute DFS to find the cycle, and it passed the pretests, but I think it could get TLE on some other tests.

Update: Now my submission of E has passed system test.

For problem E, i don't think you solution is correct.

Take care of this testcase:

The node $$$4$$$ can directs the calls only on one side. Am i missing something?

I mean "find a simple cycle".

for problem B ... my idea was to let the patients wait as long as possible by add arrival time with w and make it the start of first session and add d to start time and make it the end of session ...... so every patient within that time slot get vaccinated ...... here's my submission can anyone tell me where i went wrong ?

https://codeforces.com/contest/1804/submission/197143526

You shouldn't do

`ans+=cnt/k`

because you will waste some vaccines which you could open them later. In fact you need to break the loop when`cnt==k`

.ohhhhhh i get it now ....... thanks

A VERY intresting problem C. Thanks!

Problem D : The Prince Of Ad-Hocs

Why my problem C skipped? I write it myself.It is easy for me.May be it is too short so it easily looks similar with others code?

I think you have submitted twice that's why

oh,thank you.

Please, for the love of God, do not call an undirected edge a direct connection in the future :D

Great contest overall, btw!

C has shitty pretests

because your alt failed on them?

wish me luck to become specailist after rating change.

finally become specailist. Dreams comes true.

Why did my code fail for problem D? Did I fail to consider an edge case or was my logic simply incorrect?

My idea for calculating min : If there are two lit up windows next to each other and we haven't hit the max amount of double apartments, assume it is a double apartment. Otherwise, treat it as a single apartment. And then, once I hit the max value for single apartments I treated every next pair of windows as a double apartment, and the same thing for single apartments.

My idea for calculating max : If there are two dim windows next to each other and we haven't hit the max amount of double apartments, assume it is a double apartment. If there is one dim window and a lit up window next to each other, treat it as a double apartment. Otherwise, treat it as a single apartment. And then, once I hit the max value for single apartments I treated every next pair of windows as a double apartment, and the same thing for single apartments.

My code : https://codeforces.com/contest/1804/submission/197105009

Thanks in advance for help.

Try this:

output:

explaination:

Thank you very much I understand now.

Int overflow issue with code, should have just accepted input as a string.

Appreciate the help.

Edit: still getting WA on test case 3 even after fixing input method. What else is wrong? code now works for your example case.

Edit: new code https://codeforces.com/contest/1804/submission/197135065

Try this counter example:

Answer:

And these:

Answer:

Answer:

try this:

1 8

00010110

o/p:

2 3

Didn't like B, but C was cool ;)

Why I get wa on pretest 2 in that code in problem A:

https://codeforces.com/contest/1804/submission/197096441

if a == b then I think it's 2*mx

yeah if a == b, you essentially add -1 with that code

Tnx.. I got it..

You didn't considered the situation when a=-b.

Tnx... I got it

Please put your code in spoiler next time or give a link to your submission

ok..

I very much liked E, F and H!

This is the best contest I have ever seen, I guess. Here are my approaches:

Solution for AProblem A is just greedy; if you have solved some grid problems like this, you would know this. It is good and optimal if we go only in 2 directions; for example, it can be up, left, up, left... or up, right, up, right.... or etc. But this is useful when the difference between absolutes of coordinates is at most 2. If you want the proof, just take two coordinates with the difference of absolutes at most 1 and check. If it is like this, then solution is abs(a) + abs(b). Else again, you go like this till when we reach the same line with the maximum coordinate, and we go remaining part twice to reach that coordinate.

Solution for BFor B, solved with binary search. Let's take the opening of one packet with a[i]. Now we need to vaccinate all patients. For this, we need to open a packet as late as possible, because we need to find the minimum number of packets. So, we try to open each packet as late as possible. Through this, we can take more people. Now we want to open a packet as late as possible, then we will push limits, we will open the packet at a[i] + w(the limit of the patient), and then the packets limit will be a[i] + w + d(the packet limit). So we just go to the upper bound here and calculate the answer.

Solution for CSo this is a nice math problem. In short, we need to check there is a value y which is 1 <= y <= p and x + y * (y + 1) / 2 mod n = 0 Here, if we loop till p, it will give TLE. So to avoid this: y * (y + 1) / 2 = (y^2 + y) / 2 (y^2 + y) / 2 = ((y + 1/2)^2 — 1/4) x + ((y + 1/2)^2 — 1/4) mod n == 0 => 2x + (2y + 1)^2 mod 4n == n Now we see that we can iterate from 1 to 2 * n and check if there is an answer. That is it.

I think this is a really good contest. Thanks for all the efforts. Upvoted :D

for B i had a similar idea but my submission failed can you check it out .... https://codeforces.com/contest/1804/submission/197147147

Thank you for the great round! Good luck everyone in a future:)

Nice E!And C,D are also interesting.

I have no idea why my C (197116314) got accepted. Pure luck I think.

You'll find that the task is just find whether there's an $$$y$$$ that $$$\frac{y(y+1)}{2}+x\equiv 0\pmod n$$$.

If n is odd,it's obvious that there's no influence to calculate with $$$y$$$ or $$$y\bmod n$$$,using $$$\frac{y(y+1)}{2}$$$ or $$$2\times(n-x)$$$ because 2 has its inverse.In that case since you find one,then it is correct and ends with

`f=0`

.So it's $$$O(n)$$$.If n is even,it goes a bit complex.Set y as $$$kn+b$$$,where $$$1\le b\le n$$$,then recalculate the left and you'll get $$$\frac{k}{2}n+\frac{b(b+1)}{2}+x$$$,and in the first check both part were doubled so $$$+\frac n2$$$ has no influence.Then it has only 2 situations depending on $$$k\bmod 2$$$,so

`f`

could only be $$$0$$$ or $$$1$$$ and there's a result.Also it's $$$O(n)$$$Am I the only one, who thought that Question C was easier than Question B?

You managed to dodge the wrath of test case 4 on problem D. Congo

Link:- https://codeforces.com/contest/1804/submission/197135076 my solution for B question. It's giving TLE for test case 37. I think the time complexity of above code is 2*n which is O(N) but why I am still getting TLE. Please help me out.

your solutions is O(n*n)

ImplementationForces, ObservationForces, LuckForces all satisfy today's contest. What is going on with CF? No need to force for contest number instead I want quality over quantity. We want quality like older contests, now it's like we r just lucky to get through a ques or not at all. Not learning much stuff from contests nowadays. Pls just decrease the contests numbers and make some pretty questions. :/ pls

I wouldn't join you. Today's contest was actually impressive. If you say for A, A was greedy, for B, B was binary search, and for C, C was math. If you think that you guessed or smth else then you didn't learn bro unfortunately :)

I know concepts were there, but no one can deny that older contests were fun, they were less but had good concepts, but i dont see much now. A was simple maths for most of us, B was binary search?, I did sliding window only :/, and regarding C huh, I cant comment bcz I was mad after getting to the logic there. But all have perspectives and I wish they just add some different problems.

Yeah, sliding window is also one of the solutions; just to be quick and straightforward, I used binary search. I think for us, A B C don't focus on harder topics or other different topics(If I misunderstood you, sorry).

No worries buddy, It's great that everyone liked the contest, I was just Expressing and sharing my thoughts here, but now from the down votes I got that I shouldn't be sharing my views..

Btw, great template bro

Can someone tell me the mistake in my logic of C ? https://codeforces.com/contest/1804/submission/197111331 Thanks in advance !!

I am checking for every cycle of n and updating the starting point after each operation. The moment a possibility arises inside p, or n goes beyond p, or a repetition in starting point arises(hence it is under O(n)), I am returning the answer. ( I know the simpler solution now, just want to understand mistake in my one)

Well I couldn't understand your solution there. But I will try to explain an easier solution.

Basically we have to go from sector x to 0. By selecting a force

`f`

the arrow moves`(f*(f+1)/2)%n`

sectors ahead. Or it moves from`x to x + (f*(f+1)/2) % n`

. So we need to find a particular`f`

such that the resultant sector is 0. We can find this`f`

by iterating over`min(p,2*n)`

and checking for every number`i`

whether it satisfies our condition`(i*(i+1)/2)%n = (n-x)%n`

If the condition is satisfied we can be sure

`i`

is our answer.Why

`min(p,2*n)`

?Because the force

`f`

to be applied must be smaller than`p`

. Also the remainders for the operation`(i(i+1)/2 )%n`

repeats after every`2*n`

numbers for even`n`

and it repeats after every n numbers for odd`n`

. Hence if there is no number from`1 to 2*n`

which will satisfy our condition then there is guarantee that there will not be any number`> 2*n`

which will satisfy the condition.My submission: https://codeforces.com/contest/1804/submission/197106694

Can you explain how did you know that the pattern will repeat each 2 * N times

((2*n)*(2*n+1))/2 % n == 0

Oh got it thanks <3

Fun fact: 173 submissions passed pretest and got FST(30.8% among all submissions which passed pretests) at 1804E - Routing. Maybe they didn't see $$$O(N^3*2^N)$$$ solutions could pass pretests :(.

orz

The whole contest was fine, until I realized that this round is for (a 404 not found country) an Israeli company.

Can anyone help me find out what's wrong with my code? It fails for the maximum part.

I first find all the '01' and '10'. Then I find '00' until there is no double size room remained. If after this process, we still have

remaindouble size rooms. Then these rooms should be occuppied by '11'. It passes the visible part of pretest5.Python Code

for the maximum, let's pick the double size rooms, the only bad pick is 11 because we lose a 1, so we count how many good picks we can do:

then the numbers double size rooms left is max(0, m / 4 — cnt) each of these rooms will be 11 which means we will lose a 1. and to get the answer we just subtract this number from the number of ones in the string.

Thank you. You are right, considering not picking '11' is enough.

for your code, what would be the cnt and answer for input like 1 16 0010011111111111

Is it right that cnt = 5 rm = max(0, 16/4 — 5) = 0 answer = 12 — 0 = 12?

But I can get only 11: 00|10|01|11|11111111 +1 +1 +1 +8 = 11

Finally, I find a failed case: 001001111111

PLEASE HELP ME, I dont know why my solution for C goes TLE if its o(N) only https://codeforces.com/contest/1804/submission/197134868

The sum of $$$n$$$ over all test cases doesn't exceed $$$2\cdot 10^5$$$, but the sum of $$$p$$$ ($$$m$$$ in your code) can be very large.

YES, but the loop is running till s which has max value 2e5. Oh you wanna say P * testcase will make it to TLE???

Yes, but there are $$$10^4$$$ test cases, so the loop will run $$$2 \cdot 10^5 \cdot 10^4$$$ times in total in the worst case.

I got your point, I fucked up I think so. It can easily be accepted I chose 2*n in place of 2e5.

How many CPU days do rating changes take? (I'm guessing they are not rolling out because of cheaters)

I don't think it's a good contest. Just this?

in problem C (f+n)(f+n+1)/2=f(f+1)+f(n)+n(f+n+1)/2 now f(f+1)+n(f+f+n+1)/2 when divided by mod n answer is f(f+1)/2%n why cant we check only till n-1?

why problem A is the same problem with the same solution from another contest!!

1452A - Robot Program

This happens on really easy problems quite a bit.

I have received a mail that my solution of C problem in Nebius Welcome Round is coinciding with Somebody else . In mail it was written , to post about the evidence .

C Problem was not too difficult for a person with average coding skills . The only trick in C was to get a control over the loop . I have seen many tutorials of past contests and in C problem , I used that for controlling the loop as the most common approach in these types get solved in nlog(n) or n root(n) . So I did it for n root(n) and then the code was simple a Brute force logic . It took me a number of contest and practice to reach specialist . Please reply me if i need to submit any supporting proof and what else can i upload . please do reply.

Why has my rating been decreased suddenly without any email or any error from 915?? And it's not showing rating of last two contest?? please reply

ratings have been rolled back for a while, they will be reverted back soon.

Why for last 2 contest??

they are checking plagiarism.

Just be notified that I'm being skipped for this contest due to the significantly coincides with jiangly's solutions 197093293 from mine 197126608 for problem F. Yeah the code is exactly the same to my surprise.

I suppose once figuring out the problem F, the solution is very limited: building graph with query stamp on the edge, writing a BFS lamdba and do several binary searches. These steps are independent, very standard which is easy to write the same code.

Finally, I don't think I have the ability to ask the round winner jiangly for the code sharing. Hope there will be a fair reply and judge, thanks.

I have received a mail that my solution of C problem in Nebius Welcome Round is coinciding with Somebody else . In mail it was written , to post about the evidence .

C Problem was not too difficult for a person with average coding skills . The only trick in C was to get a control over the loop . I have seen many tutorials of past contests and in C problem , I used that for controlling the loop as the most common approach in these types get solved in nlog(n) or n root(n) . So I did it for n root(n) and then the code was simple a Brute force logic . It took me a number of contest and practice to reach specialist . Please reply me if i need to submit any supporting proof and what else can i upload .

Please upvote.

Please help MikeMirzayanov . please do reply.

Same happened here:

https://codeforces.com/blog/entry/8790?#comment-926270

Lol, it's time to join Jiangly Fan Club!

I just went through both submissions and I think they are similar. But I don't think they should be treated as plagiarization.

Here are the reasons:

For this problem F, there are 3 parts, graph building from input, binary search to get the ans, BFS to get the max dis.

From submission 181795376, we can see the graph building and BFS are exactly the same,

From submission 122792860, we can see the binary search part is exactly the same.

So I think the similarity is just a coincidence just because sd0061 and jiangly having similar coding styles.

Besides that, I also don't think Codeforces is doing the right thing when 2 similar codes are found, Codeforces skipped sd0061 but not jiangly. I think if such a case is found, both side should be skipped.(in this particular case, I think both of them are innocent and should not be skipped)

Problems are similar?

Mike: It's very common and normal.

Solutions are similar?

Mike: Cheater!

P.S. As both of jiangly and tourist's coding style are very beautiful, lots of coders learn their coding style, this thing will happen again and again in the future.

If we just read the author's submission history that the coding style is consistent. And I think one seemingly possible reason (apart from code logic) for sd0061's code being detected as duplicate code of jiangly's code is that both:

`std::`

prefixBasically, if two coders use mordern c++ features and have a good coding style, once the room for algorithm implementation is limited, it's quite possible that the codes reads similar. We should allow human re-evaluation for such cases, not blindly skip the submissions.

Rating have rolled back for this round. What is the reason???

They rolled back like 5 contests yesterday and they are runnung plagiarism checks which take a pretty long time, the ratings for others graduallly came back so i guess this one will too

I submitted my code once after the game, and they all passed the evaluation.

But the game showed that I was wrong

Because they are runnung plagiarism checks which take a pretty long time?

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