Hello Codeforces!

On Dec/18/2023 17:35 (Moscow time) Educational Codeforces Round 160 (Rated for Div. 2) will start.

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be **rated for the participants with rating lower than 2100**. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest, you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given **6 or 7 problems** and **2 hours to solve them**.

The problems were invented and prepared by Adilbek adedalic Dalabaev, Mikhail awoo Piklyaev, Maksim Neon Mescheryakov, Roman Roms Glazov, Artem Ferume Ilikaev, Ruslan AcidWrongGod Kapralov and me. Also, huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

**The problemset partially intersects with the Open KFU Olympiad, so if you participated in it, please avoid taking part in the round.**

Good luck to all the participants!

Our friends at Harbour.Space also have a message for you:

**WORK & STUDY OPPORTUNITY IN BARCELONA @ HARBOUR.SPACE UNIVERSITY**

*Harbour.Space University has partnered with Giga (Unicef) to offer Master’s degree scholarships in the fields of Data Science, Computer Science and Front-end Development, as well as work experience.*

*We are looking for various junior to mid-level candidates:*

**Data Scientist:**

*Strong ML knowledge**Experience with Data Visualization Tools like matplotlib, ggplot, d3.js., Tableau that help to visually encode data**Excellent Communication Skills – describing findings to a technical and non-technical audience is essential**Strong Software Engineering Background**Hands-on experience with data science tools**Problem-solving aptitude**Analytical mind and great business sense**Degree in Computer Science, Engineering or relevant field is preferred*

**Data Analyst:**

*Cleansing and preparing data**Analyzing and exploring data**Expertise in statistics**Analyzing and visualizing data**Reports and dashboards**Communication and writing**Expertise in the domain**Solution-oriented*

**Front-end Developer:**

*Solid understanding of HTML, CSS, and JavaScript**Familiarity with front-end frameworks and tools such as React or Vue.js.**Strong problem-solving skills, attention to detail, and a passion for creating intuitive user interfaces are essential*

**Full-stack Developer:**

*Interest and experience in web application development, data products and OpenAPIs**Comfortable with on-cloud deployment services (preferably Azure), Git and CI/CD pipeline and deployment processes**Experience with open-source projects is highly preferred**Strong ML knowledge**Experience with data visualization tools like matplotlib, ggplot, d3.js, Tableau that help to visually encode data**Excellent communication skills, — it is incredibily important to describe findings to a technical and non-technical audience**Strong software engineering background**Hands-on experience with data science tools**Problem-solving aptitutde**Analytical mind and great business sense**Degree in computer science, engineering or relevant field is preferred*

*All successful applicants will be eligible for a 100% tuition fee scholarship (29,900 €/year) provided by the partner company.*

**CANDIDATE’S COMMITMENT**

**Study Commitment:** *3 hours/day*

*You will complete 15 modules (each three weeks long) in one year. Daily class workload is 3 hours, plus homework to complete in your own time.*

**Work Commitment:** *4+ hours/day*

*Immerse yourself in the professional world during your apprenticeship. You’ll learn from the best and get to apply your newly acquired knowledge in the field from day one.*

**REQUIREMENTS:**

*Industry experience**International exposure**Eager to learn**Sustainability is a key topic for you**You want to work for an NGO*

All rounds are created equally.

i wish, i can solved the C no problem in this round ...!

i want to know the end of the story how many problem did you solve ?

i had done problem A is after 1 times trying and problem b is after 3 time trying.... and didn't touch problem C ...cause i had maybe 7 to 10 minute left after solving A & B... and i got maybe -37 ratings after this contest (According my standing on this contest).

sad but i become grey from green in this round...although ratings is nothing i still very sad

no issue brother..please don't be sad...the last div 3 contest (in the past night)...i can't solved the problem A in 1 hours 10 minute and then i ditch the contest...it's hurts me a lot yesterday...but i am okay today...somewhere, i had read..."A master has failed much times that a newbie doesn't even try"...this line inspired me a lot...so don't be sad brother...i hope...one day we will became RED coder...just keep trying brother...!

in fact,i solve 6 problem last night(i never do it before) you do not solve A maybe because you just do not see this kinds of question before. Try to solve more question!

okay brother..!

Gl for everyone!

Hope the questions are not confusing.

Yes, only after the announcement for the problem C :(

I'm back with one more post-contest discussion stream

I will discuss the problems I solve in this contest.

and i'll be there to discuss it with you aryanc403

i've been watching your past streams too .

Thanks

get job

I have one :)

that wasn't for you. urta teer q lete ho bhai

UBER SDE2

Thanks for that , would love to participate in that ......hoping for a positive delta this time...

Educational rounds are always exciting, don't know what they surprise.

hope i get positive delta :D

same

For some reason I didn't get the notification that the round would be unrated for me (rating must be between 0 and 2,099), and there is no unrated marking on the registrants list. Is this intended (or am I misremembering)?

Sometimes the educational rounds being hard for me But I feel this one will be interesting ☺️

Hope to become specialist this round!!!

Is it rated?

Yes, it is.

Hoping to perform better than my previous contest.

What's the difference bwtween normal rounds and educational rounds? It really bothers me.

Different scoring type and usually educational rounds are supposed to be "education" so you might encounter more standards.

So educational rounds have less ad-hoc problems and require us to use standard algorithms more?

Not necessarily. It might have more standard but still needs ad hoc and similar. Tbh not too different from a normal . For example last round had ad hoc problems but problem E was standard problem.

USACO bronze annually is clashing with this contest :(

let's make it Cyannnnnn

Hope to become an expert after this round.

What is the difference between edu and general round？

goodluck everyone!

Speedforces

C <<<< D

i think there is huge gap between them

I couldn't able to come up with an solution for this problem tried using bit manipulation but got wa on testcase 3 do we have to solve this using any other approach.

you mean problem C?

yes

My solution for problem CTo count how many we have for each power use vector or just ordinary array with 30 elements for each power of 2 that could be entered (0 to 29).

For the first type of operation, just increment that value in the vector.

For the second type of operation, go through from 29 to 0 and subtract until the values are finished(refer to the first vector for how many you have with that power of two) or v which was input has less value than the power of 2 we want to subtract. At the end, check if v is 0 or not, if it is answer is "YES", otherwise it is "NO"

You can refer to my submission here: 237768705

Great approach,thanks for sharing your approach. according to you what will be the rating of this problem

Probably, ~1300, but i'm not sure

Thank you for sharing

Another approach....

So to calculate we can get w from out list or not , i just started traversing each bit of w , if at ith bit is 1 in w then we will look that we have ith bit or not in hashmap.If No then we print NO , If yes then decreament ith bit from our hashmap. if there is still some remain in ith bit then we can push it ahead , like we have m[2] = 3 and we have used it one from it so now we had 2 that is (2^2 + 2^2) we can write this as 2^3 so we can increament m[3] by 1. so in genral we have

countextra for ith bit then we can increament m[i+1] by floor(count/2.0)..

good contest

After doing ABCThis is uncomfortably true...

How to do E?

Using Flows

I guess, Flows with Demands should satisfy the problem. Just throwing in some virtual nodes for outflow from each node and inflow to each node should do the job. Here, a node is a row/column, and edges are delta-based flows on flipping cells. Note that we know the original sum across rows/columns, and only care about needed deltas through flipping. This delta determines the demands on inflow/outflow virtual nodes.

Agree. I'm about the same.

Use minimum cost maximum flow. Specifically, create a node x[i] for each row i and a node y[j] for each column j.

For each cell in the i-th row and j-th column, create an edge (x[i], y[j]) with a flow of 1. Flow represents filling the cell with 1. If the cell originally contains 1, calculate the cost in advance with a unit cost of -1 to offset the premature calculation cost. If the cell originally contains 0, the unit cost is 1.

Connect the source node to x[i] with an edge of flow a[i] and cost 0. Connect y[j] to the sink node with an edge of flow b[i] and cost 0. If you understand that flow is “1”, this step will be quite natural.

Finally, run the maximum flow algorithm. If the flow is equal to sum(a[i]) = sum(b[i]), it indicates that the assignment is valid, and the answer is the minimum cost; otherwise, it is invalid.

Pretty much the same problem

is rating change delta which shows during contest accurate?

Usually the actual rating change is higher than what is shown in by carrot extension, especially in edu rounds. Unless you get FST

Oh Ok thanks bro

someone will become master then :),congrats

Thanks

How to solve E? Is it related to graphs somehow?

Make a flow network with n+m+2 edges. One is sink, one is source and n vertices for rows and m for columns. Make edges with capacity r[i] to all rows from source and edges of capacity c[j] and cost 0 from columns to sink. Also make edges from row i to column j with capacity 1 and cost 1-a[i][j]. Your answer is minimum cost max flow plus the number of a[i][j] == 1 where the i -> j edge was not taken.

E reduces to this CSES problem: Grid Puzzle II

I couldn't able to come up with an solution for C problem tried using bit manipulation but got wa on testcase 3 do we have to solve this using any other approach.

i tried to solve using multiset itself but got tle for 8th testcase, tried using two vector after the contest got summited

https://mirror.codeforces.com/contest/1913/submission/237823558

Speedforces for someone

Unbalanced Contest,

A,B,C were too easy and D is too hard.

Wrong answer on test 24 in E — Matrix Problem ,, can any one give a test case sorry for my bad English

This test fails

Also this

And this

D is somewhat monotonic stack and dp, I thought so

I tried the formula dp[i] = Sigma(dp[j-1]) with a[j] >= a[i] and j <= i result is the product dp of 2 sequences a[0] ... . a[minimum_index] and a[n-1] ... a[minimum_index] but wrong, do you have any other solution

I just thought so, but didn't come up with a clear solution

.

Unfortunately it was easy :(

D fucking boggled my mind. I thought I had it like, 3 or 4 times, only to fail sample, find out that my method is wrong, come up with new method, rinse and repeat. I wrote everything from monotonic stack to segtree today 🤡

A to C is trash btw, but D made the contest worth giving I think (even though I still cannot get it)

Yup, C isn't that bad but A and B are just trash and the round is unbalanced AF

C is also kindaa repeated idea

knew D was div-conquer with min but couldn't implement 😭

How to solve D?

What's the idea behind D

DP on suffixes, maintained with a monotonic stack.

Compute for each index how many arrays you can reach that start with it, ignoring elements before it. You can delete any prefix starting from it that contains no smaller values (the monotonic stack is for this) or delete a prefix immediately after it to reach smaller elements further away. I handled this part with a set but upon further inspection it’s really just a copy of my stack ;)

Could you explain the part "or delete a prefix immediately after it to reach smaller elements further away", it's a bit ambiguous for me.

The first case, where we delete until some index before the next smaller value, does not cover cases where the element immediately after the first one is smaller than it. Consider for example the following case:

4 3 2 1

The array

`4 2 1`

can be obtained by operating on [2, 3], but since 2 comes after the next smaller value after 4, we would not have counted this case. Arrays where the second element is smaller than the first will be obtained by operating on [2, j], if we want the j-th element (smaller than the first) to become the second. And it's only possible if a_j = min(a_2, ..., a_j). So we maintain the set of possible indices j, and remove elements from the set over time.The problem is quite similar to Indonesian NOI contest here

I think the approach should be quite similar. Assume $$$dp_i$$$ is the number of ways that the array ends with $$$p_i$$$, then we can consider that for a range $$$(l, r)$$$ we can calculate number of ways when $$$p_l$$$ or $$$p_r$$$ is the minimum

For each $$$r$$$, there are two possibilities :

find conditions for all values $$$l < r$$$ where $$$p_l$$$ is the minimum from $$$(l, r)$$$ then $$$dp_l$$$ is a contributing factor to $$$dp_r$$$

find conditions for all values $$$l < r$$$ where $$$p_r$$$ is the minimum from $$$(l, r)$$$. To do this, find an index $$$i$$$ such that $$$i$$$ is the previous smaller element of $$$r$$$. Then $$$dp_{i+1}, dp_{i+2}, ..., dp_{r-1}$$$ are contributing factor of $$$dp_r$$$

For the first case, we can maintain a monotonic stack and maintain the sum of all dp within the stack

For the second case, we can use a monotonic stack to maintain previous smaller elements for each index and to find the sum of the dp we can use a prefix sum of dp

you can do it with DnC on the minimums of the array, split the array on the minima and calculate recursively for the left and the right part and merge those results

Which data structure for the splitting would be better than the Linked list or the segment tree?

Can you explain your approach, what would be the base case and the merging condition?

Thanks for such a neat explanation.

Thanks!

Thanks for explaining the approach.

There is just a correction that:

If both l≠1 and r≠n, like you said we are overcounting the case where we remove every element in [l,r],

therefore we must subtract 1 instead of adding it back.Fixed, thanks!

another speedforces round!

I think Problem D is same as this, Except the constraints are tighter.

The hardest step of problem D is to come up with a correct DP. Optimizing it from $$$\mathcal O(n^2)$$$ to $$$\mathcal O(n)$$$ is easy.

Luckily no one submitted this problem during the contest :(

Speedforces lol

Easiest D ever seen

bro you didnt even solved it XD

I did, just didn't submit

Why would you not submit lol

How to solve E?

mincost maxflow on a bipartite graph

With flow min cost, I didn't manage to create a net during the contest sadly ;d

any hints for D? also can someone try to hack my C?

about ci think my approach for c would give a complexity of number of elements in the multiset * number of queries of the second type * (some constant) but i am not able to think on a good test case for that

make both 1e5 / 2

I tried ,apparently it passes. No idea why.

Why C failed at test case 5?237765289 Could anyone point out what i missed ?

what do you do in the add?

Try this case

Test casespeedforces moment

lol was about to comment that

Today is my birthday, and I wrote this round.lol

hoping for +ve delta!!!

I have just upsolved problem C and I like it the most!

Thank you guys for the great contest!

I think problem F is quite similar to this problem.

why is C on C when it's quite literally just greedy knapsack and zero idea involved?

Well the proof is not obvious. It only works for set of numbers where all smaller numbers divide each element in the set, for example here it is powers of two. It wouldnt work in cases like {1,3,5}

I would say it's pretty obvious considering it's rated *1200~1300 on clist...

Do you really think people prove it before implementing it? Its just wishful thinking

agreed

I don't understand why there is such a large gap in difficulty between C and D.

speedforces

if i had an hour more, i can definitely solve D.knew it was some dp and related with monostack stuff.nice problem anyway

F has appeared before: coi15p2. And no, it wasn't like the idea coincided or something, this is the exact same problem. I mean, problem-setters could just avoided this by literally search their problem up, but they didn't, which I don't really get.

As a result, an unusual number of consistently low-rated contestants AC F, which are solved by like, what, 30 people, LGM included. A person whose skill are around Expert — Candidate Master level should not be able to pull that off in a 2 hours contest.

If enough people complain about this, the contest could probably get unrated, or... since F is solved by so few people, and these guys solved so few problems that it shouldn't matter. I don't know. Personally, I want the contest to get unrated to punish these MF

I'm getting +ve delta so :noo:

It doesn't effect larger group of people so it should not be unrated

It has been said before. Problems will repeat, but the round will not be unrated if not done intentionally.

lol cry about it

Very interesting contest, for me problem " C " was super easy even easier than " B " as I love math problems , Thanks for authors <3

Could you share ur intuition for C? C got a lot of ACs . Everybody figured it out so quickly except me

Sure, we know that pow(2,n) it's binary representation is 100.. ( 1 followed by zeroes )

now for any number X it's binary representation is ones and zeroes for example " 1011 "

we can write it as 1000 + 10 + 1 which all of them are in the form of pow(2,n) .. So, if we subtract the maximum pow(2,n) less that X form X we will reach to X = 0.

in this problem we just add one condition that pow(2,n) must be inserted, also we subtract this number from x until it's frequency in the vector or until X become less that it

Wish you got it

C got accepted 30 seconds after the contest :(

what is the brute force solution for problem-D?

Is it to check for all subarrays something.

That problem D was just so

hard!!!Hi! How long does it usually take for the editorial to be published? Thank you :)

Can someone please hack my C? It's O(n^2) in some cases and I think it should TLE. But I tried the case of 50000

`1 0`

queries followed by 50000`2 50000`

queries and it didn't TLE.UPD: I resubmit it to judge it on the new test cases and it passes in 1013 ms. So I may have gotten away with it this time. I'll take it as a learning experience.

Seems unhackable. C++ compiler optimization too OP

The complier understand what you are doing with

`while (sum >= p && stock[i]) sum -= p, stock[i]--;`

And convert it to a O(1) operation.

I don't think this is true reading the assembly on godbolt

You are right. I carefully checked the assembly on Compiler Explorer and I can see the loop even on -O2.

So maybe because everything is in CPU L1 cache so the loop runs super fast.

Yeah. The loop is also very short, the branch predictor is also probably never failing.

currently trying near $$$48000$$$ add queries, and I don't quite know why it isn't getting hacked (or why the setters even set limits too loose)

seriously what the fsame solution different complier

RT https://mirror.codeforces.com/contest/1913/submission/237820946

ACC https://mirror.codeforces.com/contest/1913/submission/237828694

This is because of the y = 0 case when you use __lg: https://mirror.codeforces.com/contest/1913/submission/237832586 This is because the implementation of __lg may be different on the different compilers. Functions starting with __ are not necessarily consistent across compilers.

why this get so much downvotes?? just for D was too hard?

Folks, why in the problem C the greedy approach for the queries of type 2 works? For example, in the normal coin change problem the greedy approach won't work afaik, why here it works? Thank you guys, appreciate!

2^n is greater than the summation of 1+2+4+8+...2^n-1 so you can be greedy about using your smaller powers because they can never contribute to be a bigger one

can anyone say why greedy 0-1 knapsack working for problem C, 0-1 greedy knapsack don't work for all denominations right. did there is any special properties of 2 powers that is making greedy knapsack work?if yes what is that prroperty

binary representation of a number?

how it is related, please elaborate. i saw your sol. you are taking maximum possible sum from coins of maximum denomination, like why maximum possible sum from coins of maximum denomination is related to binary representation

so practically when we make the binary representation in base 2 of a number we always take the biggest power of two that is smaller than our number and make that bit 1 and apply the same procedure but for a smaller number, this works because every number has an unique binary representation

Uh how does rating work? Will I get rating for this round regardless of my score? Someone explain pls

Yes, you always get a new rating for all contests that are rated for you. This contest is rated for anyone under 2100. Considering that you're currently unrated (that's 0 basically), it IS rated for you.

Rating works as follows — for any rated contest that you give, your rating is updated based on your performance. If you perform well, it increases, else, it decreases. Anyway, rating is just one factor to competing, in the long run, the experience results in a net positive anyhow. Have fun! :)

Ayo thanks for the reply. Okay so this is a pretty cool system. I did very badly, but eh it's just for fun. It's actually kinda refreshing to compete in programming just for the heck of it.

Can anyone suggest me a more optimal solution for C? My Submission

Problem C was much easier than i thought TT

solve()What's the worse time complexity of your approach? Isn't it O(m^2)?

It is O(m^2) without compiler optimization.

The compiler understand the loop

`while(sum+(1<<i)<=x and aux[i]>0 )sum+=(1<<i), aux[i]--;`

and generate a better O(1) code to achieve the same result.

That's cool. I had never known that.

Yes! It's o(m^2) and I've noticed a bit late so i received a hack XD. But I resubmitted with a binary search optimization and got ac too with O(mlogm).

I'd like to see that O(1) code.

Wasn't aware that your code was hacked. Curious what the test case was. I try using

and it does not work

if you find the test case i would appreciate if share it plz. Also I think of the most obvius test case like that one but no get hacked

What is your binary search approach?

In this line:

`while(sum+(1<<i)<=x and aux[i]>0 )sum+=(1<<i), aux[i]--;`

Instead of adding one by one, do a binary search over

`lo =0, hi =aux[i]`

and check if`sum+mi*(1<<i) <=x`

Alright got it now. Thanks

The binary search can probably be replaced with division.

$$$desirable$$$ = $$$(sum - x) / (1 « i)$$$

$$$available$$$ = $$$aux[i]$$$

$$$taken$$$ = $$$min(desirable, available)$$$

$$$sum$$$ -= $$$taken * (1 « i)$$$

my Divide and conquer solution to D:

https://mirror.codeforces.com/contest/1913/submission/237844397

good one !

can you explain your approach how the index of element is helping you to construct the ans

as you stored (element,index) in root , what is the contribution of left ans and right ans

correct me if Iam wrong

even I thought of the contribution of minimum index but couldn't proceed:D

for each segment l..r i am getting min value and it's index, let it be (mI). then making a call to its left and right excluding the index. left = l..(mI-1) right = (mI+1)..r

now, there can be two cases, case 1: when we include element with index mI in our required final array: if that is the case answer is simply leftAns*rightAns.

case 2: when we don't include element with index mI, this case is difficult in my implementation:

in this case we have to know whether is there a smaller element than mI on left side and/or right side. thus bools isL and isR

if isL, then we have to remove all left side including mI, then ans will be added is right side. similarly, left side might be added. note: left side and right side includes empty array also, then there can be case where both isL and isR, then empty would be added twice thus remove isL&isR.

Nice One!

Speedforces

Plese explain D

Scroll up

Can anyone understand the code for problem D of tourist? I can't understand what dp_nxt and dp_sub mean.

you need to be true sigma ioi enjoyer to understand this code

Honey, I love you too.

How to solve Problem C using bitset.

Won't we get solutions?

Will we get extra rating only by successful hacking or also by submitting correct solutions?

Where can I find editorial?

Probably with that rating change of this contest

okay

Is there a reliable rating predictor available as of today? This one doesn't seem to be working accurately any more.

carrot extension

https://mirror.codeforces.com/contest/1913/submission/237789418

https://mirror.codeforces.com/contest/1913/submission/237900686 The same!!! The first one was hacked The second one was accepted؟؟؟؟؟

The hacked case is not considered in the original test cases yet

bruh...you literally submitted it in practice

Hello.Why ratings hasn't come yet?

Rating changes for educational rounds usually take more time.

Thanks

How much time....it's already more than 20 hrs.

how to solve D ????

What about the editorials and rating changes?

They forgot about them

Can anyone tell when will the rating of this contest release. It's already more than 21hrs...hoping soon.

yaa i am just hoping it is not unrated

yeh div3 is about to start and no changes so far

did they forget to update the ratings? Div 3 contest will start soon. How will the ratings be given if they dont update it today lmao.

When will this contest rating update and where is the editorial

Problem C was terrible for its place.

not gonna lie, I couldnt solve it in the contest cuz i overcomplicated it but once i saw the correct idea, the code was easy af.

Please add the editorials.

Since no tutorials have been provided yet (as well as my rating change), here are my short hints of

problems D, E, F.Hint of problem DSuppose the minimun element of array is at position $$$p$$$, then any operation across this element will be equivalent to operating on both sides of $$$p$$$ separately.

Code: 237781872

Hint of problem EBipartite graph flow.

Code: 237869255

Hint of problem FJust think when we change the character at a certain position, what will happen to a longest palindrome substring centered at another position? Then calculate the contribution to each possible operation.

Code: 238134018

Anybody mind taking a look at my C and giving some insights as to why it is incorrect? After looking at the test case, it seems that there is an iterator position for a map that is before begin, but I do not know how to properly handle that case. Link to code 237787623

When will the rating of this round be updated? I was hoping to become specialist and then give today's div3 as a rated round.

40 min left for next contest to begin and no change in ratings so far. Uneasy situation.

Problem CHey can anyone tell the error in this solution 237781165. I am getting runtime error.

I managed to pin point the issue.

You can see these two submissions:

238078175 Accepted

238078130 Runtime error on test 3

The only difference is that the loop variable

`i`

being`int`

or`long long`

I think I figure this out.

The issue of your code is that, when you vector

`v`

has no element, i.e. for type`2`

query,`b`

is`0`

,`v.size()`

is`0`

and`v.size() - 1`

should be`-1`

.However,

`v.size()`

actually returns unsigned integer, so`v.size() - 1`

becomes really big (underflow) and it pass the for loop condition`i >= 0`

. Then accessing`v[i]`

will cause asegmentaion fault.I am not sure why this doesn't happen to test1 and test2 though,

To solve this bug, you can simply push back a

`0`

to the vector`v`

to make sure that underflow never happened. See 238078855 for added lines.Why are the ratings not out yet?

There is an announcement telling that ratings will be updated after the div.3 contest due to technical reasons. See div.3 post for reference

Thanks for bringing it to my notice.

Please upload the editorial and new ratings. Also, please try to elaborate on the concepts used instead of the method.

Is the contest unrated?

check up div3 announcement

Attention!

I received the following message from system: Your solution 237799004 for the problem 1913B significantly coincides with solutions mayank246/237799004, Naytive_Ritninja/237800400. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://mirror.codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

As that was the approach used by most of the candidates and the candidate is unknown to me. It's just a matter of coincidence that my code matches with other person. Kindly review it. and give me my rating back. BledDest

Your last div3 submissions are also skipped. Coincidence there too?

I got a codeforces system message that my solution for problem C coincides with other people, but I haven't done anything and it is a sheer coincidence that my code matches with others, please give my rating back.

I am sorry for my mistake, before the contest that day I had mistakenly logged into my old (SAZIBUR)account. I mistakenly submitted the solution to the contest without logout of that account before the contest. So later when I see it's my old account so I logout it then change some code and submit with my main account. And rest of the time I complete this contest with my main(Sazib204061) account. I apologize for my mistake. I promise I won't make this mistake again. So please don't block my account.

My solution to Problem A was rejected in the hacking round. Where was the problem?

Wait what.

I thought it was rated for Div.2The screenshot is taken at 12/21 2a.m.(GMT+8)

Edit : Seems like it's back to normal.

What the difference between "Edufcatinal" with nomal?

Is it easier or it have more rules?