Hello! Codeforces Round #834 (Div. 3) will start at Nov/18/2022 17:35 (Moscow time). You will be offered 6-8 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. However, all of you who wish to take part and have a rating of 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ICPC). Thus, solutions will be judged on preliminary tests during the round, and after the round, it will be a 12-hour phase of **open hacks**.

You will be given **6-8 problems** and **2 hours and 15 minutes** to solve them.

Note that the **penalty** for the wrong submission in this round is **10 minutes**.

Remember that only the trusted participants of the third division will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a *trusted participant of the third division*, you must:

- take part in at least five rated rounds (and solve at least one problem in each of them)
- do not have a point of 1900 or higher in the rating.

**Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.**

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of our work. Problems have been created and written by ITMO University team: MikeMirzayanov, myav, Gol_D, Aris, Gornak40, DmitriyOwlet and Vladosiya.

We would like to thank: mumumucoder, Enkognit, orloffm, TeaTime, ilyamzy, Olympia, oukeree, 74TrAkToR, molney, elseecay, bigDuck, Nickir, Be_dos, OAleksa for testing the contest and valuable feedback. List of testers will be updated.

Good luck!

**UPD**: Editorial

Div 3 after a long time...

YEAHHH div 3 is the best opportunity to become a specialist.. ;D

You will become newbie.

You will become unrated.

Your account will be deleted

Thats why y solve only 1 problem on this easiest contest

why is it so? I too solved one. But didn't get any rating?

y get ratinf after 12 hours of hacks of solutions

takes atleast a day for the rating to be updated.

I will be unrated but you are MC, BC, GC.

As a tester give me contribution

As a not tester, I gave you contribution

As a not tester, I gave you contribution too :)

As a participant, I gave you contribution too.

:))

As a contestant, I gave you contribution too.

:/

As a newbie, I have nothing but to see.(^///^)

as newbie too, I can understand you my friend :/

You are pro-newbie, I am just newbie. (●'◡'●)

Negative contribution given boy....

OMG DIV 3! HOPING FOR GLACEON COLOR!!! never, Leafeon. is the best :sunglasses:

Good luck!

try flareon color

Omg blue round

Yeah it's blue!often happens..

Div.3! Hope to Specialist!

You will become newbie.

Why?

because he is a miser

And you are BC.

You will become unrated.

Aleksamaster is the best tester , Oro najjaki

Among us

End term Exams and contest on same day :(... I guess I'll miss my becoming pupil chance ..

GL

Good luck to everybody who gives this contest. Cheers.

All the best everyone for the contest.

I swear, this time I am going to change color!

Delta will -500.

Delta of delta will -99999999

First put danda in your gand then my rating will be -infinity.

all the best!!!

OMG Blue Round

OMG Blue Round

OMG Blue Round

Time to change my rating.

Hope not by negative delta.

It will -500 for you.

We will see. It will be like "swap(anuj_1234.rating, MohamedEmara.rating);"

See, aand is always below laaaand.

It will -99999999 for you.

opportunity to get some plus after a long time..

I relate!

I can relate too

Really thats so true. But because of that university exam my mind was already exhausted and messed up in such easy problems :(

HOW CAN WE KEEP PHOTOS IN COMMENTS?

I will try my best to become expert.Only need 20 scores,best wishes.

Div 3 is the best opportunity to reach expert

You will become newbie.

Bro,why you say that?I just want to be stronger.Hope you can be expert too.

You will become unrated.

Good luck!

Best wishes

really be an expert.thank you all.

congrats, bro!

Hoping For 1300 :)

OMG div 3 , hoping for big +ve delta

someone please help me understand this statement. what is the true meaning of this statement? how it will work?

Note that the penalty for the wrong submission in this round is 10 minutes.Every time you submit a wrong submission 10 minutes of penalty gets added to you

after getting acceptedon that problem.the penalty is calculated as the time spent on all the problems you solved (for ex. if you solved A in 8 minutes and B in 35 minutes your penalty is 8 + 35).

the penalty is used to rank the contest's participants, if two participants have solved the same number of problems the one with

smallerpenalty ranks higher.thanks

****I am waiting.I will do my best.

OMG Div. 3 it is good opportunity to get GLACEON COLOR!

Never miss the rounds that written by ITMO University team

True. They always have interesting problems.

where I will get the contest link

Go to the contest section Codeforces page. Find round 834 and register before the contest

Div3!!

Hopping for green color(Pupil)

Good luck everyone, I hope everyone enjoys the round. Don't worry about rating. As a programmer we should enjoy contest more than having the tension of rating change.

Good luck for everyone . Do your best to be the best !!

I didn't compete with you a month ago, so i am in a hopeless situation, what should i do?

It would be nice to finally fix language selector issue and remove excessive use of flags.

Sources with supporting arguments:

Codeforces Language Picker -- chrome extension to see how fixed codeforces language picker would look like.

Please support the initiative and stop reinforcing poor UX practices.

Good luck everyone :)

Goof luck everyone!!!

Hope to become an Expert.

nice performance

Congratulations!

Thanks!

What if it is my 5th rated round? Will I become trusted participant if I will solve a problem in it?

great

omg blue round

I hope to become a specialist soon

This round seems to be a more interesting one:)

NO！！！！！ I lose again

How to solve G?

I used a greedy idea:

Starting from the reverse direction......for every a[i] place the max unused number from the range (1 ... a[i] — 1) before it, if no number is available in the range, then it's impossible.

However, I haven't proved this idea in the contest, so I'm not sure about the systests. If someone else who have proved it can mention it here, then it would be helpful.

Why max unused? We want the minimal permutation, so should not we place minimum in the range?

Notice that I am iterating from index N to 1, so using the maximum ones at higher indices is more optimal than using the minimum ones.

It seems logical...

Everytime you pick the maximum possible choice and you ignore smaller values which have higher chances of being used in another pair than the max value ($$$1$$$ is smaller than $$$n-1$$$ values, $$$2$$$ is smaller than $$$n-2$$$ values, and so forth...)

So even if it can be used somewhere else also, no problem because there is another smaller value you can put there (and if there isn't, it's impossible to construct the permutation).

Getting the max guarantees the smallest permutation also.

I used a Maximum Segment Tree.

`int[] answer`

of length`n`

which will store our final permutation. On every odd position(0 based indexing) fill the elements of`int[] b`

. Because we want minimum number to be in front position, so it makes sense to keep the maximum element on the second position.`answer`

array. Now we iterate integers from n to 1 and check if its not present in answer, find the right_most element in answers array which is less than this current number. So we can safely place this number on left hand side of the found number. Now delete this number. This type of operations can be done by segment trees.My solution is not clean as wrote in hurry in contest and its a newbie's code. My AC 181520850

If you are a braindead grey (like me) and never thought to solve the problem backward, you can also binary search on a lazy add/min segtree.

https://codeforces.com/contest/1759/submission/181510473

Solution for G:

From $$$i=n/2 \space to \space1$$$,for each $$$b[i]$$$,find $$$max(x)(x<b[i],x≠b[j])$$$ .

Proof:

Note $$$X=max(x)(x<b[i],x≠b[j])$$$.

Consider $$$"... Y\space b[j] ... X\space b[i] ..."$$$ and $$$"... X \space b[j] ... Y\space b[i] ..."$$$.Because $$$X$$$ is the largest number less than $$$b[i]$$$,we get $$$X>Y$$$,the former has a smaller lexicographical order.

Yeah It is so easy compare to other problems:(

What was the correct approach to solve problem C?

You can consider cases like: a == b then answer is 0 distance between a and b is >= x then answer is 1 distance between a and l or a and r is more then x and then distance from that end is more then x ( to the b ) then the answer is 2 and the last case is to make the deistance to one of the endpoints equal to at least x, and this just adds 1 to the previous case, so the answer is 3 in the last case, if you can't do that than the answer is -1

Check if ans==0

If possible use 1 step, else

if possible go to left or right end, then to b, else

if possible go to left end then right end, or to right end then left end, then to b.

else not possible.

Hi, Could you probably tell what may be wrong in my approach?

https://codeforces.com/contest/1759/submission/181515633

you only have limited startegies: go left boundry then right boundry thn target go left then target go right then target go right then left then target just check if you can do the transition in these cases and print the minmum number of steps you can do to reach target (b) using one of the above stretegies or print -1 if all of the above stregies are invalid ps : ofcourse if a == b you don't need to do anything so the answer is 0 so the answer will be 0, 1, 2, 3 or -1

Hey, could you probably tell what edge case I might be missing on?

https://codeforces.com/contest/1759/submission/181515633

hmm, honestly i don't understand your logic what i did was simply for each strategy to check if you can go from a point to another using abs(p1 — p2) >= x that was very simple you can also take alook at mycode for example i check if i can go from a to l and from l to r ad from r to b as follow: abs(a — l) >= x and abs(l — r) >= x and abs(r — b) >= x if this condition is true you can use this strategy

Thanks for responding. Honestly, I think I am doing the same thing.

These are each of the if conditions.

1) If starting temperature is the ending temperature, then print(0)

2) So, if I cannot go to the right extreme and I cannot go to the left extreme, then I am printing -1.

3) If the difference between starting point and ending point is greater than x, then printing 1.

4) If I can only go the left extreme, then I am checking for three conditions again :

5) If I can only go the right extreme,

Take a look at Ticket 16432 from

CF Stressfor a counter example.Thanks I was actually printing -1 in the else case but should've printed 2.

yeah you are right i got confused from that too many conditions in your code :) btw, ignore the message i sent too, i discovered that i misunderstood your replay about starting and ending point and couldn't sent another replay because i got limited messages

Np. Figured it out. Thanks.

The important thing to realize is that the answer is between $$$-1$$$ and $$$3$$$.

what's the silliest mistake a person can ever make? i got 3 penalties in E just because i was sorting array before taking the input.

like thismaybe i could have got my first 2 digit rank in Div 3.

I feel like F is about finding every corner cases (which I wasn't able to)

I felt problem D to be tougher than E & G :|

Me the same. I wasted too much time in D and then found out E was much easier to figure out. But it was too late... :(

Can someone tell me what's wrong in my submission of D

I am not sure what is your code doing. But i found a test case:

Thanks!

Problem C was interesting...

Can any one tell, what would be the difficulty level for the problem D?

My estimation would be $$$1300$$$.

can someone explain problem c and d

The answer of problem C lies Between 0 to 3.. Otherwise -1. Try to figure out how this logic is true!!

My solution for D:

k zeros at the end of a number means that this number can be represented as

`10^k*a`

, where a is some number.`10 = 2 * 5`

, therefore, in the answer we need to maximize the number of pairs`{2, 5}`

1) Decompose n into prime factors and count the number of 2 (let it be q2) and 5 (let it be q5). If

`q5 > q2`

, then we need to add`q5 - q2`

twos in the price increase coefficient. If`q2 > q5`

, then q2 — q5 fives. If`q2 == q5`

, it means that we cannot "collect" tens from two numbers and add a zero to the end of the answer.2) After that, I started picking up a number from [1; m]. We need as many 10 as possible in the answer multipliers, so I multiply the number (2 or 5) that I lacked to create pairs with multipliers of the number n,

`abs(q5 - q2)`

times by itself.`tmp = (q5 > q2 ? 2 : 5)^abs(q5 - q2)`

3) Then, I need to maximize the quantity of 10 at answer and, if there is solutions with the same number of zeros, need to get maximum. So we can increase

`tmp`

variable. To increase the number of tens, you need a number that can be represented as`10^k * a`

. This is easy to do if you take a digit of the highest digit and add to it`(the length of the number - 1)`

zeros. For example, when`m = 21345`

, you need to take`20000`

. So we have to do something like this:4) When working with the

`tmp`

variable, it is necessary not to forget that it cannot be more than`m`

, so it is necessary to set conditions that`tmp <= m`

everywhere you are increasing`tmp`

.The answer is

`n * tmp`

or`n * m`

if there isn't solutions with zeros at the end.181495692

can someone tell me where i went wrong in problem B . my approach was 1<=i<=m add b[i] and s = total sum.

find if there a n such that n(n+1)/2 == total sum . then yes or else no .

why wouldn't this work

Take a look at Ticket 16433 from

CF Stressfor a counter example.my solution for 'A' was very stupid can it pass hacking and system test or will it get tle (due to using substr) can someone try to hack it?

My code return negative value in problem D but i am doing only multiplication in my code. can someone please help me with that

here is my submission 181501309

When you are multiplying it's crossing the limit for integers. Try with long integers

What you are facing is integer overflow, try using 64-bit integers.

That is something you will often have to care about.

Thanks, i will take care about it from now on

how to solve F?

Observation 1. The maximum answer is $$$p-1$$$, One case where this happens is where the number is a single digit.

Observation 2. A full carry across one or several digits happens

only once at most.Now think of it like this. How many steps do we need when our number is 0? That's $$$p-1$$$. Now, how many steps do we need when we have a few more digits existing in the number, smaller than the last digit? Might be smaller than that, but we may still have to run a full cycle. For this reason, we manage two variables, $$$\mathbf{pp}$$$ and $$$\mathbf{pk}$$$. $$$\mathbf{pp}$$$ is the smallest number connected as a continuous group with $$$p$$$, $$$\mathbf{pk}$$$ is the same but it is for the last digit. These two variables serve as boundaries. To be precise, $$$\mathbf{pp}$$$ serves as the boundary before the carry, $$$\mathbf{pk}$$$ serves as a boundary after the carry. So, you should be very well able to case-work with $$$\mathbf{pk}$$$ on whether we need a carry or not. Now the rest is just implementing the carry, and fiddling with these two variables (and the last digit).

phew, i was scared that you abandoned CF after 4 days of inactivity

Solution for F:

Note $$$k=a[n]$$$,consider $$$k->k+1->...$$$

Case1:

$$$k->k+1->...->p-t$$$

In this case, there is no carry bit.$$$0,1,..,k-1$$$ must appear in $$$a[]$$$.

$$$p-t$$$ is the largest number that does not appear in $$$a[]$$$.

Case2:

$$$k->k+1->...->p-1->0->...->k-t$$$

In this case, you should recalculate $$$a[]$$$(consider carry bit),note it as $$$b[]$$$.

$$$k-t$$$ is the largest number that does not appear in $$$a[]$$$ and $$$b[]$$$.

Great Contest!

Accidentally hacked jiangly submission with Ticket 16429 from

CF Stress:)coooooool!

jiangly's algorithm: the greatest element of an array is always the last element.

How to solve D ?? Made me drop the contest..

Just find out how many 2 and 5 you get from m and (1<=x<=n) if(x==1) answer will be m*n; otherwise m*x;

Sorry can you elaborate ??

You'll get $$$n$$$ 0's in the suffix of the answer if the factorization of $$$n * k$$$ ($$$1<=k<=m$$$) contains atleast n fives and n twos. Greedily check if you can get $$$x$$$ zeroes in the suffix of the answer for every x uptill about 18 should suffice (since n * m is atmost 1e18).

For example, if you want $$$x=5$$$ and $$$n$$$ contains $$$4$$$ 5s and $$$3$$$ 2s in it's factorization. Then you require such a k that has atleast $$$x-4$$$ 5s and $$$x-2$$$ 2s

~~~~~

~~~~

first i'll find out no of 5 and 2 in m; while(m%5==0) { m5++; m/=5; } while(m%2==0) { m2++; m/=2; }

then i have to find a number x. let x=1; x<n; i'll try to equalize no of 2 and 5. ex. if(m2>m5) multiply 5 with x;(x<=n) same for m5.multiply 2 with x;(x<=n) if both of them are same multiply 10 with x;(x<=n)

Here's my Submission of problem B. Getting WA in 69th test case which can't be seen. Can anyone help me providing me with a test case that will help me find out the loop-hole of my code?

My idea:

Smash meFinding the missing number up to the maximum value of the given array. Then trying to add max++ and checking whether it is equal to the s. If it is equal then YES, or if it gets greater it will be NO.

Take a look at Ticket 16430 from

CF Stressfor a counter example.wait the sum of the elements he found + the sum of the lost elements must be equal to some n where n = n(n+1)/2 this should do right ?

I think more people will solve E if E and D swap position

Could someone please explain why my code for E fails on test case 3? https://codeforces.com/contest/1759/submission/181515762

I have used the following approach, there may be at most 3 orders of choosing potions whenever we encounter an astronaut with health greater than the humanoid's health, and they will be GBG BGG GGB

The final answer should be the maximum of the number of astronauts we can defeat using one of the three orders. Please point out the logical inconsistency in my approach or provide an example where my code fails.

(the recursive way to solve this did strike my mind, but I decided to do it this way as it seemed simpler)

As the humanoid can gain more health as it consumes an astronaut, you should iterate until you cannot consume an astronaut anymore instead of binary search the position at the beginning. After that you will drink a potion then try again that strategy. There are 3 ways to drink potion (2,2,3), (3,2,2) and (2,3,2) so just try all of them.

brb gonna go slap myself a couple hundred times.

As I mentioned in the comment, I have tried all three ways BGG GBG GGB.

NeverCompromise I think you can look at the submission of mine https://codeforces.com/contest/1759/submission/181620640 . Its similar to your approach. I just have done the implementation in the other way which is shorter

Yeah that's a neat way to implement i, I had made a really dumb mistake at the start of my code. Fixed that and got it ac yesterday. Thanks for trying to help anyways.

Can somebody tell me what's wrong with my B code? (Pretest 4 WA)

your intuition ?

I figured it out, I thought the original permutation's size can't be over 50, but the problem statement never actually said that, only the found elements can't be more than 50

omg I just noticed, thats cheeky.

Personally to me, it felt more like a div4 round. The maximum difficulty rating for the problems of this contest should be around 1700 or 1800 I guess.

Agree 100%, nitin sir orz

Someone please tell me how the humanoid can absorb

the astronaut with power 15in the case 7 of Problem E. It says that the humanoid can absorb an astronaut with powerstrictly lesshumanoid power. It confuses me a lot.Use blue first.

Oh!!! I think the solution is greedy, which is not valid. I think too simply. Thank you very much !

Take a look at submissions of the rk8 NightSky_Yozora.

Obviously, there are 4 coding styles in his submissions:

Problem

A, B, Cis solved by the first person;Problem

Dby second;Problem

Eby third;Problem

F, Gby fourth.it's a violation of the rules, isn't it?

Sir, what about Sneh_Patel_0701 he has cheated in previous rounds too. But didn't get caught. Now he is going to become expert. Is there any punishment for them?

Can someone point out what I may be doing wrong? What edge case I may be missing?

https://codeforces.com/contest/1759/submission/181520511

Take a look at Ticket 16432 from

CF Stressfor a counter example.you missed

elseafter your lastelifThanks I was actually printing -1 in the else case but should've printed 2.

1759D - Сделай круглым

Greedy approach : We will try to append as many 0's at the end within given constraint How to find it ?

Let's take input examples : 6 11

Now we will build our answer by checking if I append zeros can I get some k such that it falls under constraint, once we can't append 0's we will break loop

How to do it? How to find possible values of k? Well it's easy, we can use some mathematics

Here's my code

Thanks :), You can ask doubt

can u explain little more how exactly u getting logic to finding k?

k = x/gcd(x, n) x goes from 10, 100, 1000, ...

now i get it thanks

I tried doing this but it was always giving wrong answer in C++ (probably some overflow issues or something like that) but it worked fine in Python

Hi, @djay24 What's your intuition behind (m / k) .?? like, you used k = (11/5)*k, why are you doing so, can you explain.?

Thanks in advance.

1 <= k <= m Initially k is minimum possible value it can take within given constraints So k can take values from k, 2*k, 3*k, 4*k, ... z*k, I have to find such z such that z*k is maximum possible but less than equal to m, so I am multiplying by (m/k) factor... That's why initially k = 5 but after we can see k = 10 is best answer so k = 5*(11/5)

Is there a problem with hacks in problem F? Any hack in this problem still in the waiting process.

Also Some of them give unexpected verdict. MikeMirzayanov

Upd: Hacks still in waiting state after 6 hours problem F hacksWhat is wrong with my B task submission ? Every time I generate the sum of an arithmetic progression using 2*a1 + d(n-1)/2 * n and check if they can be equal

It fails for this TC

Thank you, I have just fixed

wait the sum of the elements he found + the sum of the lost elements must be equal to some n where n = n(n+1)/2 this should do right ?

why wouldn't this work?

Binary Search in Problem B.

wait the sum of the elements he found + the sum of the lost elements must be equal to some n where n = n(n+1)/2 this should do right ?

E and G are far far easier than D (did'nt read F yet)

Has anyone solved E with a memorized dp? the recursive approach is very simple but i cant seem to get my head around implementing a dp solution.

What is your state? I had an iterative dp where $$$dp[i][j][k]$$$ denoted the max cost achievable till index $$$i$$$ if you take $$$j$$$ green potions and $$$k$$$ blue potions in total. While transitioning, I tried to take $$$l$$$ green potions and $$$m$$$ green potions in the $$$ith$$$ step. If the cost achievable without taking these $$$l$$$ potions and $$$m$$$ blue potions was greater than $$$a[i]$$$ then I greedily add $$$a[i]/2$$$ to my answer and then multiply by 2 and 3 on the basis of how many $$$l$$$ potions and how many $$$m$$$ potions I took.

Submission

Thanks, makes a lot of sense! i had the same 3 states but had trouble differentiating the cases where the potions were taken before or during the round.

177670435``

````````Please, Who can help me to find out the queetion of problem D. I guess my code's part of maximal price of several possible variants is wrong.because the Checker Log told that expected: '874999993000000000', found: '749999994000000000'.

hello！can i get some help please

guy you... have a mistake in function qpow. Remove

`else`

, because you should increase`a`

in every iteration.oh！thank sincerely！！！，I finally solved the fuking problem， -----love from china

My feedback to the round authors (just for problems A-D (problems I managed to solve in the contest)):

A: Easy implementation problem, the problem statement could've been written more simply because I needed to look at the examples to understand the problem without spending like 5 minutes. It shouldn't be like that since this is Div.3 A problem.B: OK problem, nothing else to say.C: Good problem, even though I don't prefer problems with $$$l,r$$$, and many if statements like this one. I like the problem.D: Best problem (rating $$$\leq{1400}$$$) along with 1748B - Разнообразные подстроки in the past few weeks in my opinion. It's always interesting to solve number theory problems.yeah D was interesting

The feeling when Div. 3 round was more problematic for you than Div. 2 round XD

i have a problem.In problem e ,java,i use the same algorithm to solve this problem,but three different sorting methords(priorityqueue, Collections.sort,Arrays.sort),the first two is accepted,but the third one has TLE.Can anyone figure it out?

Arrays.sort uses a dual-pivot quicksort algorithm. Unfortunately, its worst case time complexity is $$$O(n^2)$$$.

got it

What could be the rating of Problem D?

Solutions were already there on YouTube I check the timings. People just copied from there. So unfair

Dont worry,Mike will ban them

Goof luck everyone

luck goofs

can anyone please explain me the approach for problem C? I can't figure it out!

you only have the following options:

1. if

`a`

is already at`b`

-> answer is 02. if difference between

`a`

and`b`

is >=`x`

-> answer is 13. from

`a`

go to`l`

(if possible), then to`b`

(if possible) -> answer is 24. from

`a`

go to`r`

(if possible), then to`b`

(if possible) -> answer is 25. from

`a`

go to`l`

(if possible), then go to`r`

(if possible), then to`b`

(if possible) -> answer is 36. from

`a`

go to`r`

(if possible), then go to`l`

(if possible), then to`b`

(if possible) -> answer is 37. if none of the above is possible, then answer is -1, otherwise it is the minimum of all the cases

for steps 3-6, we chose either

`l`

or`r`

and not any other temperature because its always efficient to use an operation go to the farthest possible index.got it , thanks a lot

I like this competition very much. The F and G questions are very simple, which is simpler than the level questions in div2. They do not involve advanced algorithms and data structures, and their meanings are very clear.

i wasn't able to solve F without implicit segment tree with lazy propagation and binary search on it(got MLE 2 times xd)

I remember it was a line segment tree binary, someone did it, maybe you wrote it wrong.

yeah, memory consumption was huge because i didn't use shared ptr or vector, just raw pointers..

Can someone please clarify if it was rated? I participated in the contest and solved A but I was not rated. What are the rules for being rated?

The contest doesn't even show in my contest tab. What is the reason for that? Can someone please clear up the confusion?

The full system test hasn't been completed yet, how can it be rated?

Oh ok thanks

As a pku fresher,I bravely tried. Then made my ability clear yesterday(doggy).

NightSky_Yozora was obviously cheating, plz do something.

Is this contest was rated for participant whose rating < 1000 ??

< 1600

is This contest was rated for participant with rating < 1000 ??[contest:834 div3]

There are still about 40 hacks stuck at

"In queue"or"Unexpected verdict"state, despite the fact that the final testing has already been conducted. This is not the first time such a situation arises.I think that such occasions should be automatically reported to contest managers and, in general, should block the final testing unless fixed.

It's showing unrated, does anyone got rated ?

Why my rating is still not updated until now?

Oh, it's already updated but I didn't noticed it because it shows "unrated". Why?

ALL MY RATINGS SHOWS "UNRATED"!

because you are unrated

Impossible, I solved 2 problems in R833, and I solved 4 in this one. My solutions are not hacked.

everything is possible

Maybe you're right. Evidence

Is this system error?

And why a point is in the year of 1970? Evidence2

i think because you participated in 1 contest that time

Codeforces was founded in 2011(maybe) and I was born in 2010.

maybe you were born in 1970 and you just forgot about it

How possible? And there was nothing called Codeforces then! You can ask MikeMirzayanov for it.

I asked him and he said codeforses was founded in 1920

I think it is, because the rating is updated in both competitions now.

how long till the tutorials are posted

Now it`s time for the Ultimate Question of life, universe and everything:

Is it rated?At least not now.

All Questions were mathematical as always. No concept used of DSA. Disappointing contest

MathforcesApart from D, I don't see how the others are mathematical (don't know about F, since I didn't solve)

Well, D may be a simple math problem, but all other questions are focus on basic algorithm ability. Though I'm a newbie, I think this round is of good quality.

Bruh only D was math

Nice problems! I love the round

I hope that I can be specialist xd

Why it is not showing any updated rating ? I am not seeing any rating change and even my previous contest rating is showing unrated . Even though I got rating before

Waiting for rating change

Has this contest turned unrated??? Why the ratings are not updated yet

Wait. But y will get + 47 raiting for this contest

Please update the rating

chill they will be updated ~8pm

No

Hello, 2 hours ago I got a message that tells my solution for problem C matches with some other participants. Also, this round will be unrated for me and it shows I am out of competition. My solutions are skipped also.

I checked their submissions and found some similarities in logic and I think this is normal as thousands of people are participating here. But their code writing style and solution for this problem are totally different.

Also, this happened to me for the first time. I don't know what to do! Help me if anyone can! Thanks in advance.

Sorry to hear that. You may send emails to the codeforces official account and provide exact evidence to show you didn't share code with other participants. You may get the round into a rated one finally if you are resonable.

Hope that will help you.

Thanks. Can you give me the email address?``

You will see it in the message from codeforces noticing you are skipped. I just cannot find it, sorry for that. Wish a good result!

Okay. No Problem! Same to you!

Codeforces@codeforces.com maybe.

One contest doesn't mean everything.... Try in next contest.

First of all Thanks. I know that but the message says my account can be blocked. That's why I am so tense.

I have been skiped , i have not cheated , what i should do to prove that

You have been skipped several times. Just admit it and don't cheat again.

noooooo he didn't cheat of course he just keeps getting accidentally skipped /s

Yes i have skiped before but not this time

When is editorial going to be posted?

I liked this round, would enjoy to read the editorial too.

When will the editorial be posted?

I have recently received a plagiarism warning in Codeforces Round #834 (Div. 3) due to which you have dropped my rating. please review my code. I have not copied it from anywhere. I solely came up with that solution. I spent full 2 hours trying to solve these problems. please revoke this penalty. what is my fault is someone else had the same solution popping up in their mind. It's really demotivating. Please do not do this. Vladosiya

https://codeforces.com/contest/1759/submission/182444103 I am Getting Runtime Error in Problem F but not able to find it , It would be great if anyone can help .