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Hello! Codeforces Round 656 (Div. 3) will start at Jul/17/2020 17:35 (Moscow time). You will be offered 6 or 7 problems (or 8) 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 rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

You will be given 6 or 7 (or 8) problems and 2 hours to solve them.

Note that **the penalty** for the wrong submission in this round (and the following Div. 3 rounds) 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 participants of the third division*, you must:

- take part in at least two 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 my work. Thanks to my good friends Daria nooinenoojno Stepanova, Mikhail awoo Piklyaev, Maksim Neon Mescheryakov and Ivan BledDest Androsov for help in round preparation and testing the round. Also thanks to Artem Rox Plotkin and Dmitrii _overrated_ Umnov for the discussion of ideas and testing the round!

Good luck!

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**UPD**: Also thanks to infinitepro for help with statements and testing the round!

**UPD2**: Editorial is published!

will it be rated??

I hope so.

Answer will be available after the round.

`<too-old-joke-about-rated-contest>`

Third unrated round for me in a row :(

codeforces nowadays .

xD

Hi, i'm new, how do i know if a round is rated or not?

It is written in the contest announcement, and if the round becomes unrated there will also be an announcement saying that. But most rounds held are rated.

Monogon contest again got rescheduled :(

It's good that it got rescheduled. Would have been sad if it became unrated like the previous two rounds.

I think they are running this round because they think there will not be a crash and an infinite queue. looking forward to it

Codeforces is my new School ... teaching me and hosting Contest :XD Thanks a lot CodeForces!!

Edit: Sorry for the spelling mistake

грамматика

Can you please translate it to english. I don't understand russian

Grammar*

It should be ""Russian" ,not "russian".Grammar Nazi

I don't think this is a grammatical error.

Your profile picture correctly depicts you :)

In English : Mike says "do not spoil my idea with tell everyone"

"the round will be rated for you." Hope this time,,, this line will be true for me..Btw Thanks for your hard work..

Deleted

Then participate virtually.

Hopefully this wont be unrated !!

I was eagerly waiting for the announcement :)

Let's all pray for smooth conduction of this round.

It was better make a test round before real contest :D

There is one testing round arranged, it will start in ~45 min. check here

This will be my first ever time competing.. any pointers?

Nope, use Java.

Why Java? Use any language you are comfortable with!

You didn't get the joke ;)

Damn, that was a nice one.Tubelight moment for me :)

Hahaha, Java doesn't use pointers.

Situation or system in last 2 rounds.

hope all will be fine this time :)

Why there is unusual time for round 657 ?..

May be authors comfortable with that time.

Codeforces contest time are generally xx:05 or xx:35 according to GMT. But that rounds scheduled at 09:00 GMT(not 09:05). So that i assume that round is tied with some other contest(may be some onsite contest).

on july 19 there is the cookoff on codechef. maybe to avoid a clash

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not too old for me :DTo qualify as a trusted participants of the third division, you must:

>>> 1900 <<<or higher in the rating.I believe this is incorrect :)

Edit: Nevermind, just noticed it's the same in previous rounds. There's a different limit for being a trusted participant and being ranked. Sorry for the noise.

Hoping that this contest goes on smoothly

_____Who could tell me how to remove or report that comment?so dirty..

Ok,that comment has been removed

i'm coming back to this site after a break of 3+ years. so i'm new to this "12-hour open hacks" system.

firstly, is it something new? or has it been a regular thing in contests recently?

also, does it mean there will be a kind of "hacking phase" which lasts for 12hrs after the 2-hour "coding phase" has ended?

It's a regular thing for

`div.3`

and`Educational Rounds`

.Yes.

oh this is new for me ... i guess we would need to solve the problem (or atleast, pass the Pretests) in order to hack other participants' solutions?

Nope. You can hack any problem, even if you didn't solve it.

What's more, sources can be copied for local test.

dolphins returning to Italy and footballers returning to CodeForces, Cool!

Hoping positive delta change for all :p

That literally can't happen afaik.

What If everyone has exactly the same rank(same score with same penalty).

even then the higher rated participants will have a negative delta while lower rated ones will have positive

Everyone will get 0 rating change.

I believe there should be a half-hour unrated short contest some time

beforethe actual contest, to test the system's current load handling capacity.As a tester, I won't say anything, since you already know how good vovuh contests are!

Hope atleast first problem do not take time.

really excited for this contest. I just hope it doesn't get ruined unlike past few contests. Looking forward to learn a lot from vovuh's tasks.

Try not to tag the person unless it's needed

why is the time extended 15 min?

Good luck and high rating to everyone! :)

I hope i will be able to change my profile lol

Hey, I don't think this is right:

Although you can find this sentence after it:

if your rating is less than >>1600<<, then the round will be rated for you.I think this may puzzle those people who are new and without any experiences before in Codeforces.

Seems more like a Div2 than Div3

Not really. Problems A-D were very doable for anyone <1500

question D also??I am not getting any clue on how to implement question D

Contest is still running

he tried to pull off a cheeky one yeah lol.

Simple bruteforce . set l=0 , r=n-1 , mid=(l+r)/2 , at each moment you can either change range [l,mid] to current character or change [mid,r] ; Time complexity : O(2^log2(n)+26n) which is O(n) Ps : O(26n) to compute prefix sums of occurences of each character .

You don't even need prefix sum. Naive bruteforce works https://codeforces.com/contest/1385/submission/87119176

Great contest, thanks a lot for this, now I see what vovuh hype was about.

Deleted

Just realized 2^26 is more than 200000 lol the meme is pointless

living in a country with a different number of alphabets surely make you confuse

how to solve question D???????????????

Just brute force with recursion.

what will be the time complexiy in that case

O(2^logn) i guess

Simple bruteforce . set l=0 , r=n-1 , mid=(l+r)/2 , at each moment you can either change range [l,mid] to current character or change [mid,r] ; Time complexity : O(2^log2(n)+26n) which is O(n) Ps : O(26n) to compute prefix sums of occurences of each character .

O(2^logn)=O(n)

yeah exactly... O(2^logn)=O(n)

With every time using "count", it should be O(n*2^logn) which is equal to O(n^2) right?

Actually length of string is also halved each time. It could have been precalculated,it thinked this way also, but realized that count within solve() is better in term of time.

Divide and Conquer, complexity should be O(n), because each character would be counted at most twice. (I am not sure about the complexity part tho)

yes, it´s $$$O(n)$$$. Look: let $$$T(n)$$$ be the amount of operations for an array of length $$$n$$$.

$$$T(n) = 2\cdot T(\frac n 2) + O(1)$$$, for $$$n > 1$$$.

$$$T(1) = O(1)$$$.

if you solve this: you´ll get that $$$T(n) = O(n)$$$

My solutions (untested, but overall ideas should be right):

AReorder x, y, and z such that x <= y <= z. It is clear that doing so does not affect the answer. Consider the value max(a, b, c). Clearly both y and z has to be equal to this value (because two of the max expressions include it). Therefore, if y != z, the answer is NO. Otherwise (if y = z), output a = x, b = x, and c = z.

BClearly each value from 1 to n occurs exactly twice in the sequence a. Delete the second occurrences of each value from the sequence a to get the answer.

CSuppose that in an array, there exists three elements a_i-1, a_i, and a_i+1 such that a_i-1 > a_i < a_i+1 (in other words, there is a local minimum). If such 3 elements exist, the array is not good, because either a_i-1 or a_i+1 must be taken before a_i, violating the requirement that c be non-decreasing. If such 3 elements do not exist, the array is always good (in fact, a good array is always composed of a nondecreasing sequence followed by a nonincreasing sequence; simply take the lowest element at either end each time to obtain a nondecreasing sequence). Insert elements from right to left until the array is no longer good.

DPrecalculate prefix sums in order to allow counting the number of letters of a certain type in a particular interval in O(1). Do dynamic programming on all intervals that could be involved in the definition of a c-good string (the set of such intervals has the same structure as a segment tree; there are only O(n) of these intervals). dp[i][j] represents the minimum cost to change interval i into a j-good string. To make transitions, decide which half is filled with the same character and which half is a j+1-good string.

EIf the directed edges already form a cycle, the answer is NO. Otherwise, do a topological sort considering only the directed edges. Then direct the undirected edges from the node that comes before in the topological sort, to the node that comes after.

FIf the degree of a node is not equal to 1 (mod k), it can never be deleted (because an operation preserves the degree (mod k) of all non-leaf nodes). Call any node which is impossible to delete a blocked node. Any node which lies on a simple path between any two blocked nodes (in other words, any node located in the minimum connected subgraph containing blocked nodes) is also a blocked node, because they can never be turned into leaves. Except in the special cases where there are no blocked nodes at all, or if k = 1 (in these cases, we can always keep deleting until there are no more than k nodes left), it can be shown that all nodes not directly connected to a blocked node can be deleted (by defining a recursive "turn-into-leaf" function); nodes directly connected to a blocked node might also be deletable in groups of k if there are k or more such nodes attached to the same blocked node.

GClearly each value from 1 to n must occur exactly twice in the table in order for the answer to exist. Construct a graph where each node represents a column of the table. Imagine coloring the nodes black and white, where black indicates that the column should be swapped, and white indicates that the column should remain the same. For each value, we want the two occurrences of it to be located in different rows (this condition is both necessary and sufficient). Each value thus imposes a constraint on the two columns where the value can be found; either they must have the same color (black and white, as described above), or they must have different colors. Consider each constraint of this form as an edge. For each "same color" constraint, merge the two nodes at the ends. Then test whether the graph formed by the "different color" constraints is a bipartite graph. If it is not bipartite, the answer is NO; otherwise, for each connected component, choose the smaller part of each bipartite component to be black.

in D do you need the prefix sum? since it will iterate in ln(n) anyway?

I did D without prefix sum

no, it's not needed (although I used it), look:

This has complexity $$$O(n\cdot \log n)$$$, so it is fine. If you want to do it linearly, use a prefix sum to compute the cost.

is DP was necessary in question D or normal recursion will work

Bruteforce recursion works in just 30ms

Actually, simple recursion does work. Oops.

I am not understanding why my solution to D is giving TLE when I don't use s.substr() and instead pass the limits in the function(Link to submission) and gets accepted when I pass the substring to the function(Link to submission). Can anyone tell me the reason for it?

Passing a string to a function passes it as a copy if you don't take the reference as the attribute and hence it leads to a O(n^2*log(n)) solution which is TLE. Instead, when you pass the substring, it is of the same order as the number of operations in your for loop and the complexity is not affected. You can instead do:

`int func(string &s,int st,int ed,char c)`

Thanks!

Thanks sayam! I had the same doubt!I even used prefix sum still I got TLE, was unable to understand the issue.

Can you please elaborate?

I got TLE on my first submission which i thought was having O(2*logn) as time complexity, after reading your comment i just changed (string s) to (string &s) and boooom the sol got accepted :(

Doesn't that change only affect the space complexity? How time complexity is getting affected?

creating a copy of a string of length n takes O(n) time and O(n) extra space. (this happens if you call by value) which when done over n times gives time complexity O(n^2).

& operator in parameter name helps it to pass by reference, (as if using the same string but with a different name) so no time or space required.

Normal recursion will work

For D you can also try every possibility of the good array and find the minimum answer from all possibility. The total possible good array will be n

Solution

How did you come up with the number of good arrays for a given n? I thought there is only n good arrays for a given n...

Yeah! Thank you for pointing out. I just recalculated, It should be n. My math sucks.

I used the local minimum approach on C but getting WA on test 876 pls help

Try this test case.Your code will give wrong output. 1 5 3 4 4 3 3

It will 2 only right?

Array is — [3 4 4 3 3]. Correct answer must be 0.

https://www.geeksforgeeks.org/assign-directions-to-edges-so-that-the-directed-graph-remains-acyclic/ Question E

Was G component wise 2-SAT or there is a easier approach to it?

The easier approach is just bipartite coloring. You just need to build a bit special graph to apply it.

Thanks for such an amazing contest, enjoyed it a lot. :)

That was an amazing contest! Thanks Vovuh

Problem E explanation can be found here : https://www.geeksforgeeks.org/assign-directions-to-edges-so-that-the-directed-graph-remains-acyclic/

Its Problem E I think. Also it shattered me when I cam to know that a good problem can be googled easily.

Ohhkk.. So whosoever read that article just had to implement it.

Is it me or the contest was at a bit higher level than usual Div3?

This can help you with first 4 problems. https://docs.google.com/document/d/1Fq4UAJBh9Tn48hiF6bLzwvl7ce5XikqWeFmFfw1Q0OE/edit?usp=sharing

Help me with the last 3 problems.

How to solve E ?

https://www.geeksforgeeks.org/assign-directions-to-edges-so-that-the-directed-graph-remains-acyclic/

First remove all undirected edges and check if graph is cyclic. If it is cyclic, answer is NO otherwise answer always exists. Now, topologically sort the graph and assign undirected edges from vertex appearing first to vertex appearing after it in topologically sorted order.

G was implementation heavy i believe.

No!

Okay , my rating got fucked up again anc I need a quick editorial now.

My Solution to C , i try to provide a proof of the solution too . Hope You like It.

in D I forgot to call the string by reference and got one TLE omg

Just realized what I was doing wrong after reading your comment. Big facepalm! I usually always pass things by reference.... F

And here I am trying to use DP, thinking my solution is too slow instead of passing string by reference :(

Did anyone solve F with dp on trees + rerooting technique ?

I solved it by using the same technique . You can check out my code.

I think there was typo while naming the round It should be Codeforces Round#656(Div 2)

https://codeforces.com/contest/1385/submission/87160668

Can anyone tell me what's wrong here? I guess the complexity O(n),still giving TLE.

What is your expected time complexity for memset?

First thing is you should pass string by reference otherwise even if it is right you will get TLE, but even after that your code gives TLE, so there must be some other problem.

Here's your AC, don't use memset in this case as there are many test cases and you are resetting the whole prefix array. Also pass string by reference. https://codeforces.com/contest/1385/submission/87164492

Is there anyone who solved E without topological sort? I tried with dfs but getting WA on test 2.

I did. First check for cycle with directed edges. If not found you can always direct the undirected edges and still have an acyclic graph. I had two adjacency lists, one for directed edges and the other for undirected edges. Have 3 states each for visited, completed and not visited separately. In a dfs run from a vertex, if any of the undirected edge is directed to a completed vertex, then this can never be a part of the cycle. So remove this vertex from the adjacency list of the completed vertex. Code : 87141053

D is annoying complecated statement. Once understood the defintions it is "implement what statement says".

Not sure why such problem is in the set. From my point of view the most idiotic kind. It is not more than decypher obfuscated problem statement.

Tasks were interesting, more div 2 than div 3 in my opinion.

Only thing I do not like, that you are allowing more than two occurrences of same element in $$$G$$$. That is not funny joke.

Yep, I thought it could be stupid because it is the only case when $$$-1$$$ appears, but now it's too late to change anything.

Problem G test case:

ans:

why it is wrong? help please :)

If you swap columns $$$1$$$ and $$$2$$$, you get $$$a_1 = [1, 2, 5, 1, 4]$$$ and $$$a_2 = [5, 3, 3, 2, 4]$$$. Both rows are not permutations.

see the other way, I am swapping 5 from 1st with 2 from 2nd i.e [2,3,5,1,4] n [1,5,3,2,4]

You are not allowed to perform such operations. The element can't change its initial column.

Wasn't that included in the samples?

Yeah, you are right. I have noticed it after contest. I had another case when I expected ans to be $$$-1$$$ (and that situation is impossible, except case where more than two numbers are equal). I need to check samples more careful.

Anyway, still it is not funny :)

What is the proof of E sol.?

Consider an undirected edge $$$(u,v)$$$. Suppose making this edge directed, gives rise to a cyclic graph, then both the directed edges $$$(u,v)$$$ and $$$(v,u)$$$ must belong to some cycle with the directed edges. But in this case, even after removing the undirected edge, the graph is still cyclic. Otherwise, either $$$(u,v)$$$ or $$$(v,u)$$$ (or both) could have maintained the acyclicity. Thus when the graph is acyclic with the directed edges, we can always direct the undirected edges and still get a $$$DAG$$$.

can someone till me the difference between this submission 87132940 and that 87161171 I thought that they are the same idea, but I don't know why the first submission failed

My solutions :

A. Iterate on all possible triplets from set of numbers 1, x, y, z, 1e9 for each a,b,c, and check if a given triplet exists that satisfies the given condition

B. Take only first occurrences of each number while iterating make the permutation

C. One obvious way to generate array c is to always take minimum end element of the array and generate the final array. So as we move right, we get a smaller array and better odds to generate a valid array c. So binary search on the answer from range(0,n-1) and check if array C formed by above method is correct.

D. To build the answer for (1,n,c), we obviously need the answers for the range (1,n/2,c+1) as well as for (n/2+1,n,c+1), so either use memoization dp or just build the answer for each character for a range recursively, ie, for (1,n) for 26 characters, build the answer for (1,n/2) and (n/2+1,n) for 26 characters each. Then generate answer for each character for (1,n) using conditions described in the question.

E. An observation was that if there is no cycle in directed graph using only directed edges, then all other edges can be directed to avoid a cycle. One such orientation is to direct edges on the basis of topological sort. So check if graph built using only directed edges contains cycle, and if it doesn't ,generate a topological sort from that, and order edge nodes on the basis of topological order.

F. An observation was that if at some time we have k leaves for a node, or k leaves for many nodes, it won't affect the answer's maximum if we decide to do operation on any such valid node because if leaves of that node stay, then that node is as good as unusable node at any point in the future. So a possible algorithm is to take the node with highest leaf count, remove (cnt — cnt mod k) leaves and hence do floor(cnt/k) operations, where cnt is the count of leaves attached to that node. We can do this by maintaining the adjacency list, adjacent leaf nodes list, and leaf frequency list and a set to store the node with highest leaf count and updating these as we pop elements from the set.

G. I did this using bipartite graphs. First we check if each value from 1 to n has exact frequency of 2 for making 2 valid permutations. If so, build a bipartite graph where nodes are pair of (row,column) for each value. 2 nodes are in disjoint parts(of bipartite graph) or have edge in between if they are either in the same column or have the same value. Build this graph, and check if it is bipartite. If it is, then an answer exists. To find the optimal answer , let s just find the cost for a connected component. It is the minimum of the cost of moving one disjoint part of the component to row 1 and the adjacent part to row 2, and vice versa. That is where implementation might become a bit heavy.The total cost is the sum of cost for each component.

master_noobie How do we prove this observation for

`E`

?Consider the graph with only the directed edges. If there is a cycle, then obviously we can't remove it, so the answer is NO.

If there isn't a cycle, then there exists a topological ordering of the nodes, that is an array $$$pos$$$ such that if there is a path from node $$$u$$$ to node $$$v$$$, $$$pos[u] < pos[v]$$$.

Now consider an undirected edge $$$(x, y)$$$ and assume WLOG that $$$pos[x] < pos[y]$$$. Then if we direct the edge from $$$x$$$ to $$$y$$$ and add it to the graph, the topological order of the new graph remains the same as the topological order of the old graph (there are a few cases here I guess, but nothing difficult). Thus, since the order still exists, there cannot be a cycle in the new graph. Thus after directing and adding all the undirected edges in this way, the resulting graph will have no cycles.

First result of google search adding directed edges without creating a cycle

Dammn! so I don't know how to google properly xD

Thanks for such a nice problemset vovuh :)

Here are my solution ideas for all problems if you are waiting for the editorial. (video published after the contest ended of course). It is possible that my idea for F isn't actually right; I don't have a proof for it yet.

Why is this submission wrong? 87125487 It passes the tests if I clear the dp array, but it should not affect the answer, if I understand resize() correctly.

Resize does not reset the existing values in a vector. It only expands/shrinks the tail to match the new size.

Thank you, I found the bug. I kinda knew that, but for some reason thought that it only concerns values (not data structures, and new values are initialized, but old values stay the same), but vector is also a value. This bug has ruined several contests for me, and I am very glad I finally tracked it down.

I will be adding detailed editorials for CF problems with appropriate proofs and implementations on my YT channel

Editorial for problem C: https://youtu.be/vjRHaZb16bU

Editorial for problem D: https://youtu.be/9gVpnosPKec

Editorial for problem E: https://youtu.be/yn6ZPaqwlso

Enjoy watching!!

Thanks for the contest, in Problem D — I did learn that passing strings to recursive function can lead to TLE, so pass reference :(

pass the substring works fine

Wait is it unrated??? when did it change to unrated???

There's no reason for this round to be unrated.

why is it unrated??

Very interesting tasks!! Almost every task required something new. Now after seeing solutions of E, i am awestruck with the use of topological sort in it.

can anyone tell me whats wrong with my code? Problem C https://codeforces.com/contest/1385/submission/87169051

3 2 1 1 1 2 3 -> answer : 2

Please post the link for your submission, not your code, that makes the comment section messy.

can someone explain why i am getting TLE in problem D here is link to my code https://codeforces.com/contest/1385/submission/87163991)

Make string global.

thanks bro this just ruined up my whole contest i just forget that it will be taking time to copy the string in recursive calls

:)

same

vovuh nooinenoojno awoo

d_rushabhis hacking the solutions that he submitted using his fake accounts. Remove him from the contest.+adding similar contestant: mihirarora

I think he doesen't know that in div. 3 rounds you don't get points by hacking. If he hack all his alts, I guess there is no problem then

i_say_what_i_want Where is that mentioned?

In div. 3 rounds you don't get points if you hack someone. Also, if all his alts gets 0 problems solved, then they won't change the rank of contestants, then, it wont change other contestants performance or rating change.

I have tried to explain my approach for E , basically explaining the code for topological sort .

Can anyone tell what is this code wrong for solution of Problem B? I take all integers in permutation1 until next integer of permutation2 appears. But it throws wrong ans.

There is a possibility that when you increase the value of j, it might become more than the last index of B and the value at that memory location might be equal to the next A[i], in this situation your code will skip that value and might give you the wrong permutation.

8 bbaaddcc explain this pls

aaaadcbb / aaaabbcd .. Need 5 change ..So answer is 5.

Isn't it strange today's contest E question is my doubt post on codeforces only that has been answered ? https://codeforces.com/blog/entry/68765

It is imposible to check your problems against all other problems in the world, especially if they weren't actually published in an OJ. Of course this shouldn't happen, but in general it is common for easy problems to collide with each other.

Can somebody help me in question E .My code is giving error on test case 2 and the error says jury has an answer but your code doesn't .Pls help link to my code https://codeforces.com/contest/1385/my

Please use the correct link to your submission! I'm not sure what the issue is yet, but judging from my experience with this question, have you cleared the adjacent matrix for the vertices after each test case? That might have caused your code to give a wrong answer on testcase 2.

My code is here 87179787 (I hope it helps)!

Can anyone please help me on this in C problem? Getting RTE on Test 2.

https://codeforces.com/contest/1385/submission/87169468

I looked at your submission. For starters, I think you have overcomplicated your code. However, here is why you're getting your run time error.

Try running this as a test case.

If you dry run your code, in the second iteration of your while loop, you'll see that, you have popped all elements from your deque, but you're still running the loop and comparing the elements, which throws an exception.

I have not corrected your code, but a simple approach that will help you is this. The solution idea is to look for a suffix of the array such that it's one of the three types. 1. All numbers are non-decreasing 2. All numbers are non-increasing 3. At first the numbers are non-decreasing, then they are non-increasing.

Among all such suffixes, you needed to find the one that had largest length (ensuring smallest prefix to be removed). I did it in a simpler way. See it here 87122310

Again div3 with the same difficulty as div2 bruh.

Really? I can't think of making the problems easier than this without compromising the quality of contest. If anything, its the div 2 rounds that seem more like div 3 nowadays.

Div2.5

Can someone help me debug this https://codeforces.com/contest/1385/submission/87198122

Rated.

Hello i'm newbie, can someone explain why is my contest doesn't get rated?

This is not my first contest, I've participate more than 2 contest. I solved atleast 1 problems for each contest.

The past contest you attended became unrated, due to server problems. The rating for this contest will get updated in some time.

This was my first RATED contest. I'm a new user. I participated in last two contest Codeforces Round #655 (Div. 2) and Educational Codeforces Round 91 (Rated for Div. 2), but somehow both the contest got unrated. So in this #656 Contest I solved 3 question. Few moments ago I got my rating, which surprised me.....only 460. How can this be possible? Please help me understanding this.

It will take 4-5 contests to stabilize....

A and B should have been swapped :P

Edit : Wrong Post

edit: commented on wrong post

Hello, I got quetion, in the problem D the test case include el case asdfghjk and the answer is 7, but the answer shouldn't be 6?. The string of the answer would be ddddgfee