Hello Codeforces!

The series of Educational Rounds continues thanks to the support of the Neapolis University Pafos.

On May/30/2024 17:35 (Moscow time) Educational Codeforces Round 166 (Rated for Div. 2) will start.

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, Ivan BledDest Androsov, Maksim Neon Mescheryakov, Roman Roms Glazov and me. Also, huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

**UPD:** Editorial is out

another BledDest round. Sheesh!!!

I got it.

Orz

Let's run this shit down boys

mewing cat ready 4 the contest

Hope the competition topic is more interesting！

Hope ABC!

Hmm Educational round "Mathforces" for a reason!

Mathforces round are the best

hope ABC

let's hope for delta > 0

Best wishes for everyone's A, B, C and D.

My post contest discussion stream for problems ABCDE

need +1

Hope everything goes well !!

Am I the only one for who the contest page is broken? I can't read problem statements or watch standings or submit.

Probably among a handful at best. Pretty sure the blog would be lambasted otherwise.

yes I am also the handful one

Nice problems, but solved only ABC. Sure D is simple counting with stack, but couldn't quite figure out till the end.

today first 10 min my codeforces didnt run well

Bad network

I've wasted about 30 minutes because of 502 and cloudflare

It was constantly showing an error on Brave browser. The issue was resolved when I switched to chrome for some reason.

I opened a B! :)

had 11 WA2 and took 1.5 hours on problem B until get AC

couldn't solve C

worst nightmare for me, expecting -150 delta

but.. you didn't even give the contest?

alt probably

Yes, it seem, this tickbird is made for leaving comments only. Here is his

altjinsuwhat is the idea for D?

Let $$$b$$$ be array of string balance. Then, if you do operation on $$$(l,r)$$$ you change all $$$b_i$$$ $$$(l \le i \le r)$$$ to $$$2 \cdot b_{l-1} - b_i$$$.

Hint — 0

We can only reverse EVEN length segments. We can only reverse segments, where number of '(' is equal to number of ')'.

Hint — 1.

Lets take one sample string ->`(((( ))))`

. Figure out, how many possible reversals are there.Hint — 2.

We are allowed to reverse till last two '('. We can not reverse last three '(' . Why ?

To get an answer to above question, try this on pen-paper...

Keep one counter. Initial value of the counter = 0;

Travel the string from left to right. When you encounter '(' , counter++, else counter--, Store above values into some array.

For each segment start l we want to count number of correct segments' ends r, such that sequence remains balanced, and that means:

Counting suitable ends can be done with grouping positions by balance and binary search

Here is my solution.

`diff_map[(diff_cnt[i]-1)/2] = 0;`

The different from`0`

to`diff_cnt[i]-1)/2`

is invalid. But other value will be set later.For each position the answer will be all position of current diff_cnt. (

`ans += diff_map[diff_cnt[i]];`

) Because the difference between these position is 0.map<int, int> diff_map; for(int i=1;i<=s.size();i++) { ans += diff_map[diff_cnt[i]]; diff_map[diff_cnt[i]]++; diff_map[(diff_cnt[i]-1)/2] = 0; // cout << "diff cnt: " << diff_cnt[i] << " map: " << diff_map[diff_cnt[i]] << endl; } cout << ans << endl;

263353060

Excellent mental exercise... Loved the problems.

thank you for the contest. :)

I really enjoyed solving C today.

Said no one ever!

I don't really understand C

For the last test case 3 1 4 3 3 4 1 5 5 4 5 2 and for the 3rd candidate (programming skill, and testing skills) would be (4, 5) (3, 5) (4, 5) (1, 2) we need 3 programmers and 1 tester so for the 1 tester we take the 2nd candidate (3, 5) and for the 3 programmers we take (4, 5) (4, 5) (1, 2) Doesn't that mean the result is 5 + 4 + 4 + 1 = 14? why is it 13?

since first candidate is having higher testing skil, he must be consider as the tester.

When you're skipping 3rd one,the tester shouldn't be the 2nd one rather 1st one because according to the statement until programmer or tester slot is not filled, you should always take the one with greater skill value.As the first one's testing skill is more than programming skill and tester slot is not filled yet(initially 0),you should take the first one as tester and rest as programmers.

you have to assign in order, so the first candidate will be assigned as a tester since his bi (5) > ai (4), then the second candidate will be assigned as programmer

ah I just thought since it was most suitable position, I thought I had to get the max possible value. Dam I actually did so much extra stuff when I could have solved it

I could not figure it out a case where the output is zero for Problem B , I guess this is where I got WA, could anyone provide one please? Thanks in advance!

you always need to do the copy operation once so the answer can never be zero

Exactly, however in the problem statement it mentions that the number of operations made can be possibly zero, this is the part I am not getting, how can it be zero if we need to do copy operation?

maybe a mistake

I just check the meaning of "aforementioned" that means change in value by increasing or decreasing so that possible that we wont change

So it's correct statement

It's not correct because copying counts as an aforementioned operation

The minimum answer for this problem is one because when we move any element of A to last it will cost us 1, So we won't get 0 as output

your code is wrong because if lastIncluded is false at the end of the iteration you just add one but more than that many operations might be needed

Thank you for your feedback, so I added lastIncluded to check if the last value of array b is included or not when transforming the other elements of the array a, if it is included then it means I do not have to perform an extra operation to convert it; as at some point it could be appended to array a and match last element of b. In the case is not included then I need to add an extra change to match last element of b.

Your point was right, just got it accepted, needed to check for more operations instead of adding just one; I thought during the contest that it had to be with possibility of zero, thanks for your help

Aah i missed updating the condition in A!! Cost me wayy to much

simple solution for A, if string is sorted YES else NO

Couldn't think of this in the contest so just brute forced it but damn, how did i miss this observation.

i solved D one minute after contest ended :(

May/31/2024 00:36UTC+8 263359883

If someone could explain why I have a Runtime error on this PyPy submission (problem C) I would really appreciate it.

https://mirror.codeforces.com/contest/1976/submission/263361104

Nevermind, after the contest ends you can see the error messages, it's a stack overflow.

Can Someone spot the mistake in my code or tell the test case ? I know the logic is correct but i am unable to find the error.

https://mirror.codeforces.com/contest/1976/submission/263360996

if someone does it, he will be blessed with high ratings :)

3 wa's for problem 1...i think its time to leave :(

Don't leave. Keep trying you will do better. I saw, you solved both A and B which is great! You shouldn't be demotivated by few WAs.

thanks :))

Solution for D with step-by-step optimizations:

263354202

First preprocess this thing to

`arr`

:For every starting position

`pos`

, the valid number of ranges is "count of`idx`

such that`arr[idx]==arr[pos]`

", where`idx`

must be between`pos+1`

and`last`

, where`last`

is "the first position after`pos`

such that`arr[last] > 2*arr[pos]`

" (because positive is subtraction after inverse, we need`2*arr[pos]`

to clear it to zero, and one more to make it invalid)Now we need two lookups:

`find_first_gt_from`

finds the position of the first value greater than`x`

starting from`pos`

.`count_equal_in_range`

counts the number of items equal to`x`

in range`[first, last)`

.`find_first_gt_from`

is optimized with`map<value, vector<pos>>`

and`lower_bound`

. While`count_equal_in_range`

is optimized with binary search plus`range_max`

, which can be optimized with a sparse table.How to solve C?

Calculate the answer for answer[n+m+1], call it sum. Store the job given to each indices in this arrangement(You can use a set for example). Also, store the first instance where someone's job flipped(i.e. they were more skilled in testing but only programming is left, or vice versa). Notice that after that, there are no more flips of the opposite manner, since after a flip, every job is the same. Finally, just run through from 1 to n+m, and pretend to remove this persons job. It would go to either a) the first person flipped infront of this index(in which case their job goes to the last person) or to the last person. For a more formal look, you can checkout my sol here 263338267

263401471 thanks for the logic bro

Simplest is bruteforce with memoization, because taking/skipping 1 element will affect very few elements.

But of course there much more clever solutions

is carrot working for anyone?

no

damn seems everyone made the same mistake in E

What mistake?

Can the 3rd test case of problem B Increase/Decrease/Copy be done in 7 operations?

how

Great problem D, the solution is quite simple but still not completely trivial. E is also nice, sad I couldn't implement it in time...

I don't want problems like today's C to get normalized

i don't want comments like this get normalized

nowadays not many people have sense of humour :(

Why? I enjoyed it.

good for you!

Why? Because it forces you to think in a different way?

it forces you to "if else if else if else if else if else if else..."

I don't have many if statements

Well there is an implementation which doesnt have such complications. One thing to notice is that there are only two types of distribution of workers. n + 1 programmers and m testers and n programmers m + 1 testers. For each i, if a[i] > b[i] remove it from the net sum of the n + 1 configuration, otherwise remove it from the net sum of the m + 1 configuration.

You are damn right

C — any idea why I get WA for this?

https://mirror.codeforces.com/contest/1976/submission/263362135

For example:

1

2 2

1 2 3 4 2

3 4 5 6 1

Your code outputs 14 13 13 12 14

For the second person, the answer should be: 3+5+4+2=14 (he would've been a tester, so the third guy becomes a tester instead of him, and the last guy becomes a programmer).

Came up with the following solution during contest for F but didn't implement it in time:

https://mirror.codeforces.com/contest/1976/submission/263362159

TLDR, always remove the two deepest branches, and make them into a cycle. How to prove that:

a) this is optimal

b) this runs fast enough

Edit: fixed spacing of comment

Edit 2: solution was hacked

Can anyone provide a testcase where this submission fails. 263337520

SpoilerCorrect answer: 3 Your answer: 4

Thanks

baler contest !!!

Today educational codeforces round is the best contest ever seen.

Any

hintfor 'C'?Remove last one and choose everyone to progers or testers by naive. What will happen if you remove $$$i$$$-th person? $$$(1 \le i \textbf{<} n+m+1)$$$?

D was nice, C was too much case work (for me) but anyways nice problem

can we do better than n * log(n) * log(n) in D.

yes, my submission is n log n

You can do binary search + sparse table or segment tree to solve in $$$O(n \log{n})$$$

i use bs over range maximum using segment tree it's

it's $$$ n\ log^2(n) $$$ can you please tell me how to reduce it to $$$n \ log(n)$$$

You are using segment tree to find maximum in $$$O(\log{n})$$$, but you can use data structure sparse table to find maximum on segment in $$$O(1)$$$.

Thanks!

In general, you don't really need to do binary search separately on the segment tree. CP Algorithm's blog on SegmentTrees is pretty good; I'd suggest going over the blog as a whole, but even the following section should give you an idea: Typical UseCase — $$$O(log^2(n)$$$ -> $$$O(log(n))$$$

Thanks! i learnt a new technique.

We don't really need dedicated RMQ data structures such as SparseTable/SegmentTree. Set/Map worked for me.

Reference Submission (C++ : Jiangly)

Reference Submission (Java : Me)

Complexity remains the same though, i.e., $$$O(n*log(n))$$$

*Edit: Added C++ reference submission (Jiangly's)

D can be solved in O(n): 263368704

Thanks! interesting soln.

Dang that's actually so clean

`max(a[i..j]) ≤ 2×a[i]`

can u please explain thisAfter inversion all +1 will become -1 and after subtraction will become < 0, invalid brackets

got it

Cool simplification!

Got accepted C and D after upsolving. Feeling pretty bad today for corner case (or maybe some stupid handling in my implementation) :(( .

I too was not able to find why I got WA for 30 minutes. I am Too slow at implementation and debugging.

Hello Master DuongForeverAlone! hru? do u remember me?

Yes

MasterDuongForeverAlone! You had a bad day, but you made a comeback just another day, and also today you did really well. Congrats! and yes i'll be focussing on >=1600 it just took a lot of work to figure out 1600 level problems (as of now) so i can't do more than 2, i have university exams for the whole month as well. Can you give me a tip on something? that how much time should i spend on problems >=1600 on my own without seeing the editorials?Thanx in advanceTry to direct message me instead of comment, as this is a personal problem.

I cant see you last reply LOL so i just replied myself XD

The A was really bad

I too thought the same. But the shortest way to solve the problem A is to just check whether given is sorted or not. it is so elegant.

Oh wow, didn't notice that. But still, i'm pretty sure that most participants brute forced the solution which was really annoying to code.

First time I solve C in a Div. 2 contest! A and B were pretty easy and C was such a nice problem, I took a bit too much time for it but I liked it.

how to solver B

consider each element as potential candidate which can be use to perform 3'rd operation

if you choose a_i for 3'rd operation

think about how you can minimize cost for i'th element because for all other j (j != i) cost is fixed that is abs(a_i — b_i)

Hi! Do you have any idea whey (possibly zero)?

_ the minimum number of operations (possibly zero)_

no. of ops will always be > 0

Dp, If don't understand then ask.

problem D the time limit is very strict. O (n * logn * logn) solution using Recursive segment tree gives TL, although I know nlogn solution exists.

I use sparse table + binary search to find "the index of the first item after pos that is greater than x"

263354202

I used map<int,vector>

My solution is the same and runs in O(n * logn * logn). But I just copied segtree's code from GFG and it runs in less than 1000ms.

It seems that my purple name want to stay with me for a longer time :)

Somehow I come up with an solution to D using segtree, and it worked! :)

compare my first submission in the contest and after contest in the D and cry with me T^T

How to solve Problem C using Binary Search? I solved without using it.

263329045

https://mirror.codeforces.com/contest/1976/hacks/1028125

Is this hack not weird? Why would they randomly hardcode wrong answers. Seems like a trick by the hacker to gain points by hacking solutions like this.

damn

I couldn't submit B for a long time cause cf wanted to verify if I am a human.

'C' was really mysterious!!!

God, I read B incorrectly and was thinking that the third option was to copy the number to the initially empty array

`b`

...Spent 30 extra minutes do to that =(

Another amazing performance for JinnyW For what i know he is only 8 years old Could he be the next big chinese competetitor?

Where did you get such information? I still find it hard to comprehend an 8yr old CM :(.

Could someone please help me by finding a test case where I am wrong for today's C https://mirror.codeforces.com/contest/1976/submission/263348497

input:

correct output:

Thanks. Could you also tell what strategy did you use to find this test? I tried to debug it for an hour but was unable to find a counter case.

i failed test 2 in the contest, so i wrote a test generator and brute force code to check my solution for task C, and it got accepted. here is my code

test generatori used the same strategy to find a counter case for your code

Thanks

Try to write cleaner code! it helps while debugging your solution.

C is really a bad problem. Just lengthy implementation. I just got scared and could not implemenet it properly

How is a problem bad just because it is lengthy? Most real-life problems are way longer than this tbh I enjoyed C it had so many factors to consider before arriving at the solution instead of one single crazy trick ¯_(ツ)_/¯

A lengthy problem in 2-hour contest is a bad problem.If it's in a 5-hour or 14-day contest, it's not really bad.

Personally I don't think it was that bad, it took me almost all the time to solve it but that's coz I'm still a noob, I can see a pretty significant amount solved C in sub 40 mins

I didn't find the implementation that lengthy, 263365416

The solution to problem D involves identifying a pair

`(i, j)`

that satisfies the following conditions:`A[i-1] == A[j], A[j] != 0`

, and for every k in the range`i <= k <= j, A[k] <= 2A[j]`

. Here, the array A is a prefix sum array derived by considering '(' as +1 and ')' as -1. To illustrate, consider the sequence "((((()))))".Interestingly, while many have approached this problem with a time complexity of

`O(n * log(n) * log(n))`

, there exists a more efficient solution. This problem can actually be solved in`O(n)`

time 263375129Any idea how could I improve my code for C, giving TLE in test 4

https://mirror.codeforces.com/contest/1976/submission/263348500

Assume all the candidates come to interview. Solve the problem twice, once for N+1 coders and M testers, and once for N coders and M+1 testers. In both solutions track whether each candidate is appointed as a coder or as a tester. This takes O(N+M+1) steps.

If the candidate i does not come to interview, then check whether i is appointed a coder or tester in the two solutions. If a coder in the first solution, then answer is (first solution score — candidate i score as a coder). If a tester in the second solution, then answer is (second solution score — candidate i score as a tester).

You will never have candidate i appointed as tester in first solution and as coder in second solution.

Oooh right, thanks for pointing that out!

Can anyone tell why this submission fails for C? https://mirror.codeforces.com/contest/1976/submission/263357375

line 124: cout<<ans1-t[i]; There is no space before the next output.

Dammit!That was the only mistake.Thanks for pointing out.

I would really appreciate it if anyone can point out why my submission for C is wrong: https://mirror.codeforces.com/contest/1976/submission/263355148

Nevermind, resolved.

whats wrong in my first solution to B ?

You need to check if $$$b[n]$$$ is between $$$min(a[i], b[i])$$$ and $$$max(a[i], b[i])$$$ for any $$$i$$$ in loop instead of just at the end using

bottomandtopvariables. If this condition holds, just add 1, otherwise handle the case as you're doing it right now.E is no easy, I can't AC T_T

codeforces.com is down in some time interval during the contest, and around 15 minutes later the mirror.codeforces.com recovered but the root domain didn't and wasn't accessible even at the end of contest. I don't see why this round should be rated even the terrible server issue occurs. By the way, C is really a bad problem.

The mirror was fine the whole time for me.

I wasted about 30min because of 502 and cloudflare.

m1,m2 and m3 didn't work well, too.

It seems that m1, m2, m3 haven't been working well for a few months. I don't know why Mike didn't fix it.

A lengthy problem in 2-hour contest is a bad problem.If it's in a 5-hour or 14-day contest, it's not really bad.

I solve problem C Using Recursive DP

DP[n][2]

this person take place as prog 0 else 1

and send remaining places of prog and test

this is recursive function

two states only person prog or tester

i think it's most easy solution to this problem

can you provide an AC link ?

can you please explain it a little bit? like in my mind I need at lease 3 states index,n,m

which is 1e15 I looked at your code but I couldn't get it because I'm not used to java can you explain the logic ?

if we solve it for first n people for example if we stand at index 7 and remaining programmers 4 and testers 3 if we remove index 3 this is programmer we go to index 7 with 3 prog and 3 testers yes? if we remove index 4 this is tester we go to index 7 with 4 prog and 2 testers yes?

if we remove index 5 this is programmer we go to index 7 with 3 prog and 3 testers yes?

which meaning we always go to any index with prog-1 or testers -1

so I don't care about n and m in first case I don't remove any one

thanks I think I got it dp[i][s] means that we are here at index I and we are missing person of type s right ? I will try to implement it to make sure that I got it

i also kinda solved it with DP(Non Recursive)

263356128

Please take care of chinese,We can't even get into codeforce.com

can try m1.codeforces.com?

didn't work well.

m1.codeforce.com is not work well,and a great number of times it can't load problems

exactly it is the case. or try use some vpns or build it via employ a server located not in china(shadowsocks). or you can ask the content of the problem to friends(in this contest only C is not loaded with m1.)

can someone give a hint on problem c first I sorted, but how to handle the case when you have to choose whether this person should be coder or debugger ?

never mind I thought we need to maximize our sum

what is the idea for C?

Hint : only 2 will change, first forced employee and last index one and calculate in reverse order

Thank you for the great idea. I get ac. 263424028

calculate in reverse order means?

mean first we will calculate n+m th doest come on interview then n+m-1,n+m-2....so on till 1,0th index dont come at work. or maybe forward can work but i have written in backwards manner

I have this test case for problem B 1 3 2 4 6 1 5 2 13

shouldn't the ans be 21 for this. ans=(2-1)+(5-4)+(13-6)+1+(13-2)=21. for making the copy of 13 we first have to go from 6 to 13 which is 13-6=7 operation and then after appending the copy(+1 operation) we have to come back to 2 (13-2=11 operations. And for the remaining numbers in a ans+=(2-1)+(5-4). what is the flaw in this logic ?

First operation is to copy the 6 to the end, giving 2 4 6 6. Then using +-1 operations convert to 1 5 2 13, taking (2-1) + (5-4) + (6-2) + (13-6) steps. Total 14.

You have to append 6 to the back of

`a`

and then make it 13. So the answer will be`ans = (2-1)+(5-4)+1+(6-2)+(13-6)=14`

I am puzzled at the statement of problem F for an hour and couldn't explain the sample output, until I realized that

lower vertexin condition below is by depth instead of vertex id! Somehow I couldn't realize it earlier despite noticed the input is a rooted tree.My friend had also misunderstood at this place before,though he solved the problem at last.When he was reading this part,he was confused why the problem repeats "root has only one son" again and again.So he asked questions and specifically said this to the organizers at question part.However,the response isn't satisfying and there are still a lot of people who neither realized the contradiction nor got the correct meaning.

I thought that I have to maximize the sum in problem C I just noticed that I don't have to

Results,when.. ??

When will the ratings change?? I am checking my ratings from yesterday Night!!!

It will take some time but soon will get updated.

The problems are quite suitable for Educational Round,I like this style.

can someone tell where is my code failing for problem c,i calculated skill if candidate n=m+1 th didnt come and then using it found if candidate i th didnt come(i>=1 and i<=n+m) 263427040

want a tutorial for C PLZ!!!

Forget that you have to pick n and m guys. Imagine that you want to pick n + 1 programmers and m testers.You can do it simply by checking a[i] > b[i]. Let us say the sum of a[i] (for prorgrammers and respectively b[i] for testers) is S. Now for all guys selected as programmer , the answer , when they are excluded is simply S — a[i] , as you can discard i_th guy from the chosen set. Similarly do it other way i.e. solve the problem for n programmers and m + 1 testers.

thanks!!!

how will i get to know if the i_th person which is absent now was contributing as a programmer or tester?

You can maintain that using some array. What I did was whenever a particular i was a programmer , I did was ans[i] = -a[i]. After getting the final sum S , I had to check if ans[i] < 0 and then do ans[i] += S.

but there will be 1 person that will contribute as both programmer and tester when we will take (n+1 programmer and m tester) and (n programmer and m+1 tester) right?

Yes , you are correct , and both gives same answer , so there is no issue.

Div-1 participants(rating > 2100) are rated for this round?

no. This round will be

rated for the participants with rating lower than 2100.Why are people above 2,100 given a rank in the common standings?

but rating more than 2100 are showing in official standing :)

Educational rounds has always been 'official' for all participants. However, 'official' != 'rated'. It is 'rated' only for participants with rating lower than 2100.

oh, thanks

I am new here. My rating is 1210 will this contest be considered rated for me or unrated.

You will be rated, if you made at least one submission during the contest.

Everytime there is a contest, the codeforces.com always asks me to prove I'm a real person. why?

This actually happen when there is lots of traffic at the codeforces. Just to avoid the denial of service attack during the contest this is done by the cloud-flare.

Will we get the rating updation before 949 Div-2 starts?

This is a nice question, if you became master in the EDU, I believe both contest will be rated for you.

Can anyone look at my C and help me in figure out what I'm missing? 263424813

Is this wierd if my all submissions got removed and my rating did not. It shows i did not attempt :/

when will the ratings be updated?

what is system testing? and why a 2 hour-long contest is taking more than 12 hours to conclude. I am new to cf.

“System testing” is a procedure that retests all the submitted code with successful hacks during the open-hacking phase, as well as extra tests (if available) provided by the contest managers. Edu rounds and Div.3 rounds may include a 12-hour open-hacking phase, during which everyone can try to hack every submitted code in the contest.

Okay. Thanks stephanez.

when will the rating be updated ?

Can someone explain why this code is getting accepted on C++17 but not on C++20..... 263515419

I have helped debugged your code.

In this section:

`it1--`

crashes. Looks like you are missing some boundary checks.input:

Debugging environment: MacOS ARM64 clang++

`-Wall -std=c++20 -Wextra -Og -g3`

with slightly modified`include`

section of your code.but how did it get accepted while submitting in c++17 it got WA on test 7 in c++20;

If it==s.begin(), then when you do auto it1(it); it1--; you are stepping off the start of the set s, and anything can happen.

If *it1 happens to give the same value as *s.begin(), or some unpredictable number greater than *s.begin(), then you get the right answer.

Thanks for the explanation!

Was it rated or unrated?

system testing got tle

awoo can you write a tutorial

why the rating updation is getting this much delayed unusually??

awoo , BledDest , MikeMirzayanov

Even hours after final standings, neither of today,yesterday's contest rating changes are published yet ?

is there any reason or just ratings TLE....

How much time will it take to display rating??? i cant see myself in 700s anymore. This is torture to a newbie.

Will be updated soon

.

look he left u on pupil and himself sitting at expert. Betrayal is unreal

Will the rating even be updated or not?

had my refresh button right now, been the submit button in contest yesterday

PUPIL confirmed :)

Great contest! Does the rating update usually take this long, lol?

They say sometimes it does. I've only been on a handful of rounds so far, but one took 12+ hours to update. There was no open hacking that round, so add that time to this round. And dunno about R949. Hope to see the updates by tomorrow morning.

Cheers, both of them got updated finally.

When rating will be updated?

They are updated, you can check now.

Why I'm still

unrated?the same problem

Why i am still blue even my max rating has been changed to candidate master?

It seems that all the people participating in this round but not the next round and getting level upgrade have encountered this problem.

can't even understand how this could happen...

It's quite fun seeing I am blue with the rating of 2007 :)

but I believe it will update soon.

I think they don't know about this bug As this bus is in that div only The least round has no bug

You have the best programmers in the world but takes you 2 days to update the rating of 2*10^4 users

Something went wrong with the new rank

give me my rating pls dawg

Rating not updated I have not gotten plag either

Why isn't the rating updated yet? what's happening??

It has been updated for my friend

Why am I still Newbie with a rating of 1200+ :(

MikeMirzayanov

I still didn't receive my new rating yet.

same with me

When will the rating change?

I don't know either because I am still unrated.

Gave the contest even had 1 correct submission but still not showing any change in rating and not being shown in my given contests section as well

Why isn't the rating updated yet??It has been so long...

Just got this e-mail that plague has been detected in my code and I am genuinely so confused. My solution for D problem is only slightly similar to one random guy's solution!? This contest was a huge rating boost for me. Please review this 263359271 This is my submission and the following is the other guy's submission: 263354669 MikeMirzayanov

Hey Mike Mirzayanov , I got a plagiarism check notification for this contest with another random guy . Here's my logic:- For each test case, it calculates two sums,

`edison`

and`ethan`

. For`edison`

, it iterates through the arrays, selecting the larger values from`a`

or`b`

based on limits`n`

and`m`

, and records these choices. Then, it recalculates for`ethan`

using a similar method but with a different selection order. Finally, it computes and prints values reflecting the contributions to the sums from the chosen elements. The process ensures that the sums are maximized according to the constraints. I am sure that this check is purely coincidental and I don't even know the guy who. I got the check with...the points were extremely important for me so please remove the check if possibleI tried running my code on ideone in default settings and i was not aware about that. It won't be repeated next time.

Hello Mike Mirzayanov, I have got a plagiarism check for problem C in this contest.

The provided solution of mine aims to maximize two different sums by selecting elements under constraints. For each test case, the solution initializes the arrays and two sums,

hiper_edandhiper_et. It then employs two selection strategies: the first strategy prioritizes elements from the first array if they are greater and within a limit n, and otherwise selects from the second array if within limit m. The second strategy inverts this logic, prioritizing elements from the second array if they are greater and within m. Both strategies ensure that the constraints on the number of selections are met. Finally, the solution calculates and outputs a value for each index based on the differences in sums, reflecting the contribution of each element to the overall sum. This approach uses standard problem-solving techniques for array manipulation and constrained selections, making any resemblance to other solutions a coincidence.The provided solution is an original work derived from applying well-known techniques to solve the given problem effectively. Any similarities with other solutions are due to the nature of the problem and the methods used to solve it, which are widely accepted and practiced in the programming community.

The approach used in this solution is a common strategy for problems involving array comparisons and constrained selections. The specific logic of iterating through arrays, comparing elements, and maintaining counters is a typical method used by many programmers to tackle such problems. Any resemblance to other solutions is purely coincidental and a result of applying standard problem-solving techniques.

This plagiarism will affect adversely to my graph in this placement season. Can you please look at this. Thank you for consideration

Students are welcome