Hello Codeforces!

I am glad to invite you to Codeforces Round 958 (Div. 2) which will start on Jul/15/2024 17:35 (Moscow time).

The contest will last for 2 hours with 6 tasks for you to solve. The contest will only be rated for those with a rating not higher than 2099, but high-rated competitive programmers are also more than welcome to participate out of competition.

The score distribution is: 500-1000-1000-2000-2500-3500

The contest will be impossible without the help from:

- 244mhq for wonderful coordination.
- gyh20, Dominater069, Um_nik, A_G, omeganot, tkl2006, harsh__h, oursaco, Hori, kostia244, ishaandas1, maomao90, waidenf, Stefan2417, MisterReaper for testing.
- MikeMirzayanov for Codeforces and Polygon.

Good luck and have fun!

**Update**: editorial

**Update**: video editorial

wtf a div2 with 3500 problem? Why not add more problems and extend it to a 1+2?

That's the score, not the problem rating.

I know, but very often they are almost the same.

I was also surprised but it appears to be more common than I thought: three of the last 10 div2 rounds had a problem valued at 3500 points, including Codeforces Round 949 (Div. 2) and Codeforces Round 940 (Div. 2) and CodeCraft-23. Additionally, I'd say that a good 1+2 needs at least 3 problems of this level (judging by Codeforces Round 942 (Div. 2) and Codeforces Round 942 (Div. 1)), so it's not as easy to extend as it may appear.

This is the first time I've seen a grandmaster's comment receive so many downvotes. People generally tend to upvote whatever a grandmaster says.

I think you are right.Maybe it is hard to add more problem.Or it will have too many div1 and div1+2 if every div.2 like this add more problem to div1+2.

A huge difficulty gap b/w C and D O_o!!

Unrated only for Legendary Grandmasters:-O

Wow feecIe6418 round! I missed you a lot!

speedforces!!

"> ...

Maybe, they call speedforces because the difficulty level in between two consecutive problem is too high. The person having more skills/practice got beaten by the speed of some other participants just because he was unlucky in starting problems in some of the contests. And anyway both candidates won't be able to solve the next problem because it's too difficult that results in negative rating change. <- My perspective :)

this. in some contest solving first 3 problem FAST is sufficient to be expert of even CM but solving first 3 problem SLOW can result in gray performance < 10000 rank.

this high variance of distribution of problem difficulty scares many div3 people including me.

Right. Solving first 3 problem FAST (in 15 minutes total) is much much better than solving first 3 problem SLOW (in 2 hours). What's your point?

yes but having extra room to upsolve (problem D) may help those who failed early e.g) multiple WA on pretest, already gray but problem D is too high: (problem_rating — current_rating) > 500 -> game over. However if difficulty gap between B and C or C and D is reasonable this cannot happen.

so the fair contest should form problem rating of arithmetic sequence, imo

Nothing is 100% fair, just solve faster. I don't see your point.

lets make contest with 1 gray *800 problem with 4 *3500 impossible problem.

just solve faster.. okay.

You know I didn't mean it like that, but ok enjoy your upvotes.

Yeah, so, about that...

As I tester, I enjoyed the problems and encourage you to participate!

I, Tester

Goal: ABC in 30 mins, D before the 2 hour mark.

as a tester

as a contestant

as a

as

404

817

Hopefully, the problem statement will be short and precise just like the announcement!

Goal solving ABC by the end of contest

us

As a tester, I am sure about tolbi is Nutella.

Also participate this round, problems are very cool.

why isn't this on home page?

"I am glad to invite you to Codeforces Round (Div. 2)" . you haven't said div2's id XD

A + B + C = E.

B=C

B+C=D

A+B+C=E

A+B+D=F

A+C+D=F

etc.....

Thank you to all

Great! I hope to reach +1750 in this round

if you want 1750 rate then solve 2000 problems

are you asking for downvotes?

maybe

hoping to solve 3 problems

Same

problem statements are the main reason as we read a big story and makers make more confusing some good coders ignores the stories and go read testcases. some problem have confusing story and does not have testcase exxplained. took more time! good wishes ;)

@tkl2006 and @ishaandas1

I Want some tips on how to be eligible for system testing of these contests...

@feecIe6418 can also answer

1000?2000!

Every round should be unique so it's fine to have bit different score distribution

i have a strong conviction that this will going to be BIG speedforce for div3, fast solve ABC or ff.

you were literally correct lol

OH! gyh I miss you a lot

omeganot orz

As a participant ,i love the testers.

Let's see if I can avoid bricking this contest.

hopefully i reach 1434 rating this round

I hope to reach 1100 in this contest (ノ^_^)ノ

I think this round is going to be mathforces

I am good at math. (•-•)

He's from China :(

I knew it. why some codeforces rounds are so maths heavy

This is my first contest. Thanks to CodeForces.com. I will do my best for this contest.

How did it go?

I blew it, due to my poor skill, I will study harder and harder. For now, I satisfied with my codechef result with 4 stars. https://www.codechef.com/users/enthusia0209

But now, I do think that I must learn from CodeForces.com contest.

Thank you, dear.

Gl next time, everyone starts in 5 digits here XD

I ve solved A,B and D. thank you for encouraging me.

What happend to C lmao

tried for 30mins but gave up and skipped D.

You had Entire Saturday and Sunday. But no, we will host the round on Monday. wow.

There is a Div. 2 round scheduled for this Saturday :)

i hope to be expert today

this time your goal is realistic :)

enjoy in contest.

oh this is the good chance that i can improve my rate :) HOPE CONTEST IS EZ!!!

C is 1000 points. Speedforces incoming.

Hope I can be Expert in this contest!!!

GLHF

You meant to say that the contest

wouldbe impossible without the help from..., right?Now it seems that we won't be able to solve any tasks unless we contact the creators :D

Mistake

2 accounts are forbidden, so you will be banned.

What's your idea for e?

Dont change the topic mate you are cooked. lol.

Sorry bros

What are you doing? Rofl. You can edit your comment a thousand times but people are not blind.

And I keep falling! Am I getting more stupid or are CF contests getting more difficult?

number of people who cheat are insane now. When ABC is hard cheaters will have the advantage because honest people wont be able to solve it but they can copy paste it from somewhere doesnt matter what the difficulty is.

The first one to pass problem E~

Also my first time to be the first to pass a problem on Codeforces! Happy~

smurf account detected.

Question D Looked so innocent to me, I thought it was only about Bipartiteness of the graph, Later I understood it's way more complex than that.

++

Fooled by score distribution! all eyes were on B and C, A surprised many.

I can't believe that more than 12k participants solved problem A all by themselves.

Less than past contests, maybe cheaters needed some rest

A is easy, am I missing something?

Personally, I didn't come up with a formula for problem A. I just used multiset to simulate without thinking too much, with an extra optimization which is to not insert all the 1's inside the set again

(My first solution got TLE because I underestimated the problem and didn't add the optimization lol)

I wasn't able to solve A. I wasted too much time on it and because of that low rating in B and C.

lol

At first it looks like the most innocent-looking bipartite graph problem lol

haha

Ther is a problem in a persian contest that is really similar to problem D.

Basically this problem there is no $$$a_i$$$ but the problem gives you a tree and asks you to set a positive integer value to all vertices in such a way that no to connected vertices have the same value and the sum of the values is minimum.

It is basically problem D but with $$$a_i = 1$$$ for all $$$i$$$.

It is not so far from problem D and i know two solutions for the problem in persian contest that can be changed to AC solutions for D by changing a few lines.

This problem from 2015 Facebook Hacker Cup is also similar: https://www.facebook.com/codingcompetitions/hacker-cup/2015/round-1/problems/D

Wasted 15 minutes reading OR as XOR :(

man, that was TOUGH :)

don't know why but I smell brute force in D.

bro please clean your nose you are cooking wrong

probably the hardest div2 A I have ever seen

I solved B & C but failed in A :<

i solved A but failed in B idk why i missed some cases :( i need to work more

Thanks for the contest. especially for task D. the first idea that comes to mind — two days is enough (painting tree in two colours) passes the sample but is wrong :(

I think it requires O(logn) rounds to kill all monsters?

I guessed that with a bit of intuition, but not sure how to approach a formal proof of it.

.

me too

A >>> (B + C)

OMFG got accepted E in the last 2 minutes. My first E in div2 for nearly 2 years.

I was thinking of using segment tree and binary searching through the upto which range our ai is the current minimum. Is this approach correct. In theory this will work in O(N log^2N). What do you think

It is still ambiguous. Can you say it more clearly?

Let's say our original sum is F

when A[i] is removed we calculate from 0 to i. The minimum idx which have min(idx, i) is A[i], lets say X . and maximum idx which have min(i,idx) is A[i], lets say Y. Then count the subarrays between [X,Y] in which i is included. Subtract this*A[i] from the F for this index ans

instead of segtree + binary search, you can just use a stack...

I haven't really solved problems using Stacks. And yea this is one of my weakness that my mind focus on Seg trees whenever something related to range comes. gotta work on that. And I will try to solve the problem with both approaches. Thanks for the info

Damn, I got too many penalties on B for concatenating to a new string instead of just using normal variables.

A is harder than C, Nice.

In what sense bro, i couldn't come up with any idea how to approach C :(

lets say we have bits in positions 4,3,2,1

what's the optimal here ? first lets take all bits then we take 4,3,2 then 4,3,1 then 4,2,1 then 3,2,1

now its just in reversed order :) that's why I said what I said

Cool!

A is easy too....just think like : Break the current number(>1) in max possible 1 you can break, leave rest for next step.

EX: S = 5, k = 2

1

41 1

31 1 1

21 1 1 1 1

EX: S = 6, k = 3

1 1

41 1 1 1

21 1 1 1 1 1

if n would have been large this brute force would have failed, but within these constraints this solution seems okay

Yes, after getting this logic i checked the constraints.

since it's problem A i was sure that constraints will be low.

But tell me the proof why every time we convert it to maximum 1..... We need to minimize the operations as well.... How would you conclude that this will take minimum operations??

If it comes in your mind, then it is easy else it is very hard to think with proof for Div2A....

This is the Brute force solution which one can think (obviously by taking some test case by themselves and proving whether this is a correct approach ensuring minimum operations).

I think most of them denied this solution because they might have thought it would give TLE (and the pressure to solve A in minimum time too). So they chose to think in some optimal way and messed up (can you believe some used DP to solve it).

After thinking this approach you can see i implemented it in the brute force way too, but you can see from editorial after getting this logic one can also write answer as (n + k-3) / (k-1) which didn't striked in my mind.

Donyou mean by taking some testcase...if it is minimum then it is minimum.... You shoudl have concrete proof for it.

How do you prove that Brute force will produce minimum operation

Your possible approach becomes a solution once it is correct for every test case given along with the question(at least). And also if your are not satisfied you take some test case by your own and check if by any other method (like how it was done in qn) you can get a answer less that what you get by your approach.

I think you didn't understand what i was trying to explain with the solution.

Breaking the current element into max 1 will itself ensure you are choosing a minimum way(but not optimal, of course). Let's take another example:

S = 4 , k = 4

Break it into max 1 we can: 1 1 1 1

No. of ops = 1.

Suppose you're developing an approach so you might also think how about breaking it into:

1 3 (

insert no more than k positive integers, and we have chosen only two i.e <=k)but this will not lead you to minimum ops.

Just take a big n and small k within the constraints and simulate it yourself, you will get approach.

Both the observation and implementation of A are simpler than C

for the first time i couldn't solve A, but i was able to solve B and C.

Did A in 9 mins, but was not able to do C even after throwing my whole mind at it for 1.5 hours.

It's interesting that the problems time limit is in ascending order.

ok i swear if the submission for D that im about to send ACs im gonna be so pissed (i was 3s late)

also here's my approach for D

observation 1: you only need to delete at most 3 times

first deletion will split everything into trees of depth 0 or 1 (if it's depth 2 then the non-leaf can be deleted in the first move and it's guranteed to be better)

2nd and 3rd deletion just "cleans" the trees of depth 0,1 i let dp[a][i] = min price to delete subtree i WHILE deleting node i at t = a

then for all child v1,v2,...,vk of dp[1][i] = val[i] + min(dp[2][v1], dp[3][v1]) + min(dp[2][v2], dp3[v2]) + ... + min(dp[2][vk], dp[3][vk])

dp[2][i] = 2*val[i] + min(dp[1][v1], dp[3][v1]) + ... (you get the idea)

then the final result would be min(dp[1][0], dp[2][0], dp[3][0])

however this gets wrong answer on pretest 2 (either it's wrong or it's my implementation that's wrong)

just go with log2(n) instead of 3

Can you tell what's the intution behind log(N) rounds

i tried firstly with 3 got wrong ans then thought max it can be 100 looking at the constraint then got tle then i just felt since my logic and implementation is correct it must be logn

8 1000 1 1 1000 10000 10000 10000 10000 1 2 2 3 3 4 1 5 2 6 3 7 4 8

wow...

so turns out i threw top 80 because of that

still positive delta tho so yay..?

at least it was better than last div2 where i solved a wrong version of E and spent forever debugging it (only to realize the mistake AFTER the contest)

I tried the exact same thing ;-; , but using this logic idk why it didn't even pass the given tc , will have to upsolve later:).

so hard :( but nice problem. I think E is pretty good.

resolved

$$$k$$$ needs to be atleast $$$2$$$

oops sorry, my bad

The minimum value of K has to be 2

can anyone tell Whats the logic behind problem -d

my logic — just doing dfs and marking all alternate nodes in a tree then the answer will sum of all nodes + min(sum of marked node, sum of unmarked nodes)

Think of a case like this one:

1

4

100 5 5 100

1 2

2 3

3 4

it fails this test case

1

4

1000 1 1 1000

1 2 2 3 3 4

oh thanks a lot bro got it

i think its because lets say u have 4 straight connecting nodes and u can imagine like an array and counter test case will be something like 100000 1 1 10000 since best option is picking 1 and 4 instead of 1 and 3 or 2 and 4

if the tree is 10 — 1 — 1 — 9 the answer is not 31, but 24. the optimal in this case is with 3 turns instead of 2 (I'm also thinking about the logic, but this was my counterexample of the dfs idea)

this shouldn't work, counter scenario: think of an array where you need to maximize the total value you pick without picking adjacent elements! this can be solved by a simple dp; and so this each path in this tree can act as an individual array if you think

is there a direct bitwise formula for C I solved it but nevertheless found the bit manipulation I did quite complicated.

Can you explain your logic in brief Please ?

You can take a look at my solution here. You basically iterate over the bits from highest to lowest significance, and if the bit is on, you turn it off and print the rest of the bits as they were. Also don't print zeros and that's it.

You can look at my solution if it is of any help

take n in binary for example 23 is 10111

the answer will look like :

(the number of ones + 1)

00111 10011 10101 10110 10111

look at my submission

The number of submissions for problem C is beyond my expectations and nowadays looking at no of submission always demotivates me(i feel how this fast submission crossed 1000)

true, It feels like I am too dumb for not solving A in 5 minutes.

Auto comment: topic has been updated by feecIe6418 (previous revision, new revision, compare).Not able to solve B after lot of attempts can't find the case where the code goes wrong.

let me help you bro

think of this test case 00000011100000000

here in this test case in first move you will combine all prefix 0 with 1 in the second move combine all suffix 0 and in the third move apply operation on the remaining array so answer will be 1

the corrrect approach is to combine all consecutive 0 into one '0' so that no one gets into waste and then checking is count of 1 is greater than 0 or not

Just replace all consecutive zeros with a single single zero(make a new string) and then count c0 and c1 and apply the conditions given in the question.

Can someone explain in short the idea for C ?

For each number, just unset one set bit starting from msb to lsb

My solution : keep searching the bit which is 1 from the smallest bit. When it is 1 , replace it with 0 and make the bits after it to be valid for A | B = N (So we can keep finding smaller number)

(Sorry, My English is bad, I am not native speaker.)

Therefore, the answer of 14 => 4

Great. But How can you be so sure that in every number only one '1' is made zero ?

like can't we make multiple '1' to '0' ?

Because there are two rules we need to follow.

To achieve the rules mentioned above, make only one '1' to '0' is a good method. ( If we make multiple '1' to '0', since we need to make sure A | B = N , then the total amount will be smaller than make only one '1' to '0' each time. )

thanks bro, for now my intuition is built up but If I will spend some more time I will definitely understand.

XEI XEI ( sie sie) :))

Your welcome : )

Lets build $$$a$$$ from the end.

$$$a_k$$$ cant be larger than $$$n$$$ because that would mean $$$a_{k-1} | a_k>n$$$.

It turns out that $$$n$$$ can always be put as $$$a_k$$$. (dont know how to prove that sometimes $$$a_k$$$ smaller than $$$n$$$ can result in bigger $$$k$$$)

Now for every next number $$$a_i$$$ we construct (actually previous in $$$a$$$) we will try to decrease it as little as possible compared to $$$a_{i+1}$$$ while the $$$a_i | a_{i+1}$$$ is equal to $$$n$$$.

in C, is the len of the longest sequence equal to => (no.of set bits in n)+1, except for the cases where the no.of set bits is less than equal to 2..

when no of set bits > 1, the answer is just no of set bits plus 1; otherwise, just print 1

Probably the best contest I ever gave on Codeforces. Got wrong answer on A then understood my mistake and then solved C with 1 minute 29 seconds remaining!!!!! Very happy rn.

Hello,

These problems were trash.

Yours faithfully.

Comment from your main.

Have you considered that it could be a skill issue?

solved a, b in 14 mins but could not solve C. Can some one throw some light. Am i missing somethng obvs?

take n in binary for example 23 is 10111

the answer will look like :

(the number of ones + 1)

00111 10011 10101 10110 10111

look at my submission

print number in binary format

N = 13 {1101}

unset the msb

0101 = 5

set the previous unset bit and unset the next bit

1001 = 9

do this again

1100 = 12

1101 = 13

the length of this seq will be equal to == No of set bit + 1

Special Cases(2^N)

Length = 1

sequence is 2^N only

started 1hr late solved B and C only stuck at a cause A, doubt was whether there exist an easy sol in constant time but would have implement in log(n) complexity rather if no time pressure. Will try best to wakeup early next time

For problem A: imagine you have a block of length $$$N$$$, and you can cut it at $$$N-1$$$ different positions. During one move you can make $$$K-1$$$ cuts (this splits a piece into $$$K$$$ pieces). You want to make cuts at all $$$N-1$$$ positions, since this will result in $$$N$$$ pieces of size 1. This means the required moves will be $$$\lceil \frac{N-1}{K-1} \rceil$$$.

i had same approach but mine was little mistake i took ⌈N/K−1⌉

i think D was too hard

I turned my head away from that problem and went to E as soon as i realise its prolly DP on trees LOL

Hello everyone, this contest is good. I solved 2 questions in 13 minutes, but I'm stuck on the 3rd question. Even though the 3rd question is challenging. There was a live stream running on YouTube called "Sharabi Lal". If someone official is watching, please block this. However, even if they do, they might create another channel. What can we do? Maybe you should increase the quality of the plagiarism checker or something similar. If someone is found cheating, you could mark their ID differently or publish the name of the cheater after the contest. Or maybe not. I'm just frustrated.

Yes, you could also have a "Report Suspicious" button where someone could report if they suspect someone else of copying code.

Please edit out the channel name else more cheaters would get to know about it and follow him.

Is D dp, trees or both?

Basically Dp...try to traverse from one leaf node...

DP, the move in which you kill a subtree's root must be different from the time in which you kill its children. You can have $$$dp[i][j]$$$ = the minimum cost to kill the subtree rooted at $$$i$$$, if you kill the root during move $$$j$$$. For $$$dp[i][j]$$$, we take $$$A[i]*j$$$ damage from the root, and then we then have to pick $$$j' \ne j$$$ for all the child subtrees.

Thanks

Rainboy round , I want to know which country he belongs to.

he is asian but respresents USA at IOI

How to not choke in questions like A :<<

What I try to do is Think simple and don't overcomplicated anything and be greedy(Although it doesn't work everytime)

This contest let me become purple and >1900 for the first time, i love the problems!!!(O.o)

Congratulations bro

What's the idea behind problem E?

Why was my D-question never system tested?

Haven't you finished the test yet?

Hi everyony. It's written in the description that "The contest will only be rated for those with a rating not higher than 2099". But my rating hasn't changed yet. Should I just wait for some time or there is some kind of mistake?

Wait for some time

It'll change in a few hours

I saw A tutorial it's really easy I think I couldn't solve it because I restricted my self that in each operation n should be divided by number <= k

I got WA in 270731172 during the contest. I submited the same code after contest and it accepted. 270754276. Why ?

I don't think both codes are same, you must have made some changes

Oh yes. I change local vector to global vector.

ur code fails here.

It checks whether the n is power of 2 or not. Also this lines present in the AC code. I think I got WA because I use local vector and when I use global vector I got AC. I don't why.

no, it got lucky AC

`pow(2, bits-1)`

returns double, and double precision is not that good.try to use

`(1 << (bits-1))`

instead, it will get accepted whatever vector u useOkay I'll try it. thanks.

Problem D got me good. Oh well, We will get 'em next time.

good contest!

Auto comment: topic has been updated by feecIe6418 (previous revision, new revision, compare).But tell me the proof why every time we convert it to maximum 1..... We need to minimize the operations as well.... How would you conclude that this will take minimum operations??

If it comes in your mind, then it is easy else it is very hard to think with proof for Div2A

I Got 42 plus if not that skipped it would have increased by 60 or more

Hello

Hi

same dp and name

Bruhhhh

I have been practicing DP and Greedy problems for the last week. Solved A quickly, realized it was just a single formula.

Solved B, using the another solution mentioned in the editorial.

Looked at C, thought it was gonna be easy and then saw the constraints. I have 0 experience in good bitwise and tree based questions and got cooked. I knew it had to do something with 2^64 and the loop running 64 times but not the logic lmfao. I was so disappointed with my performance, I wanted to solve atleast 3 questions but none of the topics I practiced came.

Finally became a CM, thanks for the contest!

bro again became Expert

Do you guys know why some questions are skipped in judging?

due to more than one submission of same question or cheating...

What does it mean more than sending?

i mean like submitting the already solved question's answer again during the contest.

I'm sure none of these things are possible, how can I communicate this with Codeforce?

Recently i got message from codeforces about same solution with Rudransh58 but i even don't know who is Rudransh58. It is totally conicidence that core idea of problem is same of both. The method of doing question is totally different of me and Rudransh58.It is totally unfair.

Dear Codeforces Team,

I recently received a notification regarding a solution coincidence for problem 1988C with solutions Rudransh58/270724137 and 2022ucp1403/270738520. I would like to address this concern and provide some context.

First and foremost, I want to assure you that I have always adhered to the rules and guidelines of Codeforces and have achieved my rating through my own efforts. I have not engaged in any activity that would violate the rules, including sharing or copying code.

I am unaware of how this coincidence occurred, but I can provide the following points to support my position:

Independent Work: My solution was developed independently based on my own understanding and problem-solving approach.Code Sharing: I have never used public platforms like ideone.com with default settings or any other medium to share my code. I take the confidentiality of my work seriously.Common Sources: It is possible that the coincidence is due to the use of common techniques or standard algorithms widely known in competitive programming. If there are any common sources that could have led to this similarity, I am unaware of them. I am committed to maintaining the integrity of the competitive programming community and am willing to cooperate fully to resolve this issue. If further details or explanations are required, I am ready to provide them.Thank you for your attention to this matter.

Best regards, Rudransh58

Your coding style completely changed between problems A/B and problem C. Even the indentation and formatting are different.

Problem A and Problem B, which use <bits/stdc++.h>, include macros for

llandmod, and solve everything in the main function.Problem C, which uses different includes, no macros, uses a solve() helper function for testcases, and has different spacing/formatting.

It's because i've used diffrent moduled code which i've pre written on my vs code, so i did use diffrent includes, no macros, used a solve() function that's on my vscode!!, and changing the code writing style does'nt mean i've copied the code.

Respected MikeMirzayanov, feecle6418 and the system admin team of codeforces,

I recently received a notification regarding a solution coincidence for problem 1988C with solutions ksp.77/270714693, 07-priyanshu/270714907. I then got some more notifications regarding the same. I would like to address this concern and bring some facts to you.

In this regard, I would like to express that i always adhere to the rules and regulations of the amazing platform codeforces and its guidelines. I never intend to share my code with anyone or copy others code. Also, this is the first time my solution got skipped and i am experiencing something like this. Whereas, ksp.77's solution from last 5-6 contests are being skipped which is not the case with me.

I am unaware of how this coincidence occurred, but I can provide the following points to support my position:

Simple Code:My solution used variable names as 'v' for array, since i use 'v' in cpluscplus for vectors. Ans since the example test case was initally not passing when i was in initial phase of my coding my solution so i moved to Python. I usually use 'a' for variable name of array but since i just transalated my code myself so did not get the intiution to change the variable name. Also, the first loop was very standard method of bit shifting and must be available on internet beforehand and once the algorithm or idea of this question is understood , the second loop can be formed easily.Common way of approach & python language:I checked the submission history for this solution and i find that many solutions in C++, C or java have the exact same approach but they have not been flagged similar to me. This could happend because python or PyPy syntax is standard and can look more similar to other compared to other languages like C++, Java , Python. Also, my solution was already very optimised.Time of submission:For this problem I think i submitted before other who have been tagged similar to my solution.Independent Work:My solution was developed independently based on my own understanding and problem-solving approach.Code Sharing:I have never used public platforms like ideone.com with default settings or any other medium to share my code. I take the confidentiality of my work very seriously.Common Sources:It is possible that the coincidence is due to the use of common techniques or standard algorithms widely known in competitive programming. If there are any common sources that could have led to this similarity, I am unaware of them. I am committed to maintaining the integrity of the competitive programming community and am willing to cooperate fully to resolve this issue. If further details or explanations are required, I am ready to provide them.Thank & Regards, 07-priyanshu

(Thanks Rudransh58 for guiding me how to draft the above message as i faced similar situation as you)

Why many contests of july 2024 are skipped ?

what is the reason and how can i contact codeforces?

Hello. I have a question: I got 1700 this round and I did see that my rating increased about 261 or so, but after 2,3 days, this became unrated!! Is this a bug? Could somebody please tell me why this would happen, please?

yea same... they made it unrated or what? Without giving any reason

I have the same question...

Wow I got the rate back. I wonder what that was though...

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