Hello Codeforces!

On March 06, 18:05 MSK Educational Codeforces Round 39 will start.

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

This is a special Educational Round — **Top three contestants among those who are eligible for participating in ACM ICPC** will be invited to attend the Hello India x Russia Programming boot camp under the Codeforces + SDV banners, on a full team sponsorship.

Social Discovery Ventures (SDV) creates, develops and funds Internet projects specialising in communication platforms that enable people across the world to get closer and expand their social networks.

The winners will be fully sponsored with their boot camp participation fees, accomodation, and meals for the India location!

This round also has a special regional prize — the company Phaze Ventures, a platform founded to transform the region by unlocking the untapped potential of our youth, startups and corporates, will be sponsoring the **top 3 participants of this round from Oman (also among those who are eligible for ACM ICPC)** — further sharing the importance of supporting talented programmers and sending them to these types of events throughout the world.

The round will be **rated for Div. 2**. It will be held on extented ACM ICPC rules. After the end of the contest you will have one day to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given **7 problems** and **2 hours** to solve them. Problem authors are PikMike, Vovuh and me.

Good luck to the participants! Hope you will enjoy the contest.

**UPD:** The tutorial has been published.

It's not in the upcoming contests list?

Will it be on Codeforces? Because it's not on the contest list of Codeforces.

Are you

blind?Nope, but I think you need some memory refreshment as this contest was added in the list 1 day ago. My comment was before that time. Thank you.

Wow, what a pleasant surprise.

Acceptable start time for Chinese users:)

Vote for the contest and it will appear on the HOME Page?

Unfortunately, we cannot add the contest to the registration form right now, and when we will be able to do it, it might be too late for some participants.

We've decided to postpone the round. It will start on March 06, 18:05 MSK (exactly 24 hours after initially announced time).

is it cause you forgot to thank mike

Mandatory comment- "Hope the problem statements are as short as the announcement!".

take my mandatory downvote

Black.Monster XD

It's actually long. So are you hoping long problem statements?

Please rated for Div.1 and Div.2

no, please, no

Maybe Div 1.5 (purples included)

http://codeforces.com/blog/entry/49826 (14 months ago)

http://codeforces.com/blog/entry/58153#comment-418381

I hope it will be rated personally for you and not to your fake accounts (http://codeforces.com/team/37327) .

Stop crying dont be jeloussojfsijfops

gotem

Do you want to come back tip of Div.2?

Удачи

"Educational Round — Top three contestants among those who are eligible for participating in ACM ICPC will be invited to attend the Hello India x Russia Programming boot camp under the Codeforces + SDV banners, on a full team sponsorship. Social Discovery Ventures (SDV) creates, develops and funds Internet projects specialising in communication platforms that enable people across the world to get closer and expand their social networks."

I have a question top winner are always top why you don't give a chance a pupil? they haven't anyy quality? Do a Randomly Winner.It will be help for him and her -_-

Losers are losers

But you are a pupil.. do you believe that you are a loser?

This is the bug of competetive programming — only the world best participants can win prizes, it is dissapointing -_-. A lot of newbies quickly lose interest -_-. In any other sports participants are divided according to their age/gender/rating etc. and it is possible to win the prizes in any competition, if you are good for it.

Isn't that with many things? -_- gotem

From where do we register for the contest??

Clarification: everyone is eligible to win invitations or that is for div2 participants only?

there is no contest on contests page..

Believe me, you are really blind!

I know man. That's why I'm sometimes confusing your mom with my girlfriend.

I feel like educational rounds are kinda easier than regular CF rounds. Is that true?

"Educational"Of course they are supposed to be easier

Educational doesn't mean easy. It is better to upsolve.

Indeed, it's always better to upsolve.

The contest starts in 23 minutes, but I can not sign up...

Contest will start on Tuesday

when can I register?

Before system testing!

Hope the problem statements are not as long as the announcement

I wish everyone hacks himself!

please hit the like button of my comment

thanks all !

This is not Facebook

I wanted to change my Contribution. 0 to anything.

School 23:00 lights out,but the content begin in 23:05.so I urgently hope to 2 hours ahead of the game

“Top three contestants among those who are eligible for participating in ACM ICPC will be invited to attend the Hello India x Russia Programming boot camp under the Codeforces + SDV banners, on a full team sponsorship.”

Are the finalists meant? Or any contestant who can participate in next ACM ICPC is eligible for the invitation?

Any of seasons 2017-2018 or 2018-2019.

Is it

guaranteed?Russian translation states that it is possible to hack your own solution, while english original doesn't. Is this true and how would doing it affect your score?

It is

"Educational"you don't take any score for hacks!What if someone has already paid for this? Can we get a reimbursement?

Hi evil666man!

Let's discuss this after the results of this round.

rating predictor is not working!!

Now it does. Too late

Extremely bad experience on problem C.

13wa on problem C. Maybe I should go to sleep...same here :(. wasted more than 40 minutes!

How to solve D?

we can use dp. dp states are dp[i][j] --> number of minimum hours he has to stay in the college if he skips 'j' lectures till the ith day!

Calculate pre[i][j] = Minimum time spent in university on i'th day after using j skips on this day.

Then calculate dp[i][j] = Minimum time spent in first i days after using j skips.

Answer is dp[n][k].

Transition is dp[i][j]=dp[i-1][z] + pre[i][j-z]. 0<= z<= j

how to get pre[i][j] ?

spend time in university = r — l + 1; we can calculate cost[i'th day][spend time in university] = skip time then pre[i][skip time] = spend time in university

Observe that its useless to skip any lecture which is not at the beginning or the end.

So, we should always skip some prefix and suffix of lectures.

Let x be number of skipped prefix lectures and y be number of skipped suffix lectures.

Try all x,y such that x+y<= k .

pre[i][x+y]=min(pre[i][x+y], val)

(pre[i][x+y] is initialised to inf)

So calculating pre[i][j] is a n^3 operation?

Both Pre-calculation and main DP are O(nk^2)

What's val in the above expression?

The time spent after skipping first 'x' classes and last 'y' classes.

It can be calculated in O(1) if you store all lecture times in a vector.

Let size of vector be vs.

val= 0 if x+y>=vs

val= v[vs-y-1]-v[x]+1

You may have an initial greedy idea to skip classes on endpoints of a day if skipping that class saves you the most number of hours. Unfortunately, this does not work because of the case

2 8 2 10111101 11000011

Greedy solution would skip first and last class on first day, but optimal solution is to skip first and second class on second day.

We can make sets

s_{ i}for each dayithat contain the hours that have classes. Then, for each day, we can calculate the minimum number of hours to attend class that day if we skipjclasses on that day, and calculate this number for alljsuch that 0 ≤j≤min(k, |s_{ i}|).Once we have that, we do a simple dp, where our dp array has

nrows andk+ 1 columns.dp_{ ij}will contain the minimum number of hours to attend class after dayiwhile still havingjdays left to skip. Then, for the first row, we fill it up with the values we calculated in the last paragraph (i.e. dp[0][k] will be total number of hours of day 0, dp[0][k-1] will be minimum number of hours on day 0 if we skip 1 class, dp[0][k-2] will be minimum number of hours on day 0 if we skip 2 classes, etc.). For rowsi> 0, we just test the possibility of skipping 0 ≤j≤min(k, |s_{ i}|) days from each point in the previous row.Then the answer will be the minimum number in our last row.

any idea what is test 16 for D?

Totally random test with n = m = k = 500

Shouldn't it be prefix/sufix calculation for each row then dp? I couldn't figure why my solutions fails for over an hour :D

It should not, it can be arbitrary substring in a day

i think the test 16 is something like when we skip on ith day some starting lectures and some lectures from last for an optimal solution. e.g 2 6 2 101101 000000 ans=2

Very probable, as I made the same mistake of not including that case :c

Yes exactly !! I forgot to consider that case when some prefix as well as suffix is skipped

I also got WA on test 16, mostly it would be the case of having some prefix and some suffix skipped on a particular day, as I was considering either of them on one day not both.

D is harder than usual

Well, I solved E in about 15 minutes, and still wondering how could so many people solved D successfully.

Can anyone give me a hint? ;)

http://codeforces.com/blog/entry/58155?#comment-418680

Btw can you please explain your approach to E?

First of all, here is my submission (36014589), in case my explanations are unclear. ;)

With each test, provided that the number in the input is guaranteed to have an answer, we are sure that there is an answer with (|

S| - 2) '9' digits (perhaps not the most optimal one, just get moving on).Now to find an answer with |

S| length.First, we can use an array to store the number of appearances of each digit. Along with that, we will know the number of digits that has an odd appearance count.

For example, with

`28923845`

, we havecount(2) =count(8) = 2 andcount(3) =count(4) =count(5) =count(9) = 1.Since the answer is strictly lower than the given number, we can iterate the string from right to left. With each digit position, we can do the following:

If an answer has been found before, terminate all iterations.

If not, iterate all digits lower than the current one in a decreasing order. This time, all digits to the right of the digit we're iterating can be arbitrarily large without violating the "strictly-lower" criterion, therefore, an answer will be found if:

There are enough digits to the right to compensate with the current number of digits with odd appearance count.

If 1 is satisfied, the remaining digits to the right (after excluding the one used for compensation in step 1) must be an even number (to not creating another odd appearance count again).

If both are satisfied, then an answer is found: which is all digits to the left of the iterating one (order remained the same), the value we're working on with that digit, and the digits to the right, which are modified according to the above criteria to keep the number "beautiful" and make it as large as possible.

Taken the example of input

`28923845`

again.Starting at digit

`5`

. As we're not working with this digit any more, decreasecount(5) by 1, i.e.count(5) = 0 now.Since we still have 3 distinct digits with odd appearance count (3, 4 and 9), it is certain that we can't find an answer here. So we move on to digit

`4`

.Again, we have to decrease

count(4) by 1, i.e.count(4) = 0.Let's replace that digit with '3'. So,

count(3) will be increased by 1, i.e.count(3) = 2.Now, only digit 9 has an odd appearance count, and we have exactly one digit to the right to spare, so we will set the value of that digit to '9', so all digits will have even appearance count. The answer is

`28923839`

.P/s: What if the input is

`44400000`

?Since no digit is lower than

`0`

, we will surely start at the rightmost`4`

.Let's replace it by

`3`

. Of course, we can replace a digit on the right by`3`

to compensate, and there are 4 digits left un-handled. Since it is an even number of digits, we can safely replace all 4 of them with`9`

.The answer is constructed by the following: first two

`4`

digits on the leftmost (their states don't change at all), the digit`3`

(replaced from the initial rightmost`4`

), and on the right, we have four`9`

s and one`3`

. We will write them from left to right in a descending order to make sure the number's value is largest possible.The answer for this input should be

`44399993`

.Hope that this help you! ;)

The answer for 2892384500 should be 2892384493, not so?

Ouch, my bad :-P trying to clarify the case with exceeded digits but looked like I used a wrong case.

Will fix it soon enough ;)

UPD:I have fixed it! :Dthanks a lot mate :)

Many thanx to you!

After reading your explanation — ez-AC :)

Also, thanks for the hint for D ;)

I got stuck with greedy for about 30 minutes, and after solving E, for some reasons my mind just froze and couldn't visualize a clear DP approach for it :-P

Didn't solved it but seen the same type before usually solved by DP+Bit

C is very easier than usual and D is harder than usual...

Was D meant to be a solved in a Knapsack Problem way?

Damn Codeforces, it became too slow towards the end

I am not a grammer nazi, but this —

`If you can gen the string that will be obtained from the given string and will be contain english alphabet as a subsequence, print it.`

UPD: It's fixed.

Hoyl, why did nobody tell us this during the contest!? Sorry, it should be fixed now, statement will get updated in a couple of minutes.

How to solve C if we are asked to find the minimum number of operations (for any string s and t)

DP[i][j] indicates the minimum way to solve the suffix ending at index i, with j as the letter we are currently trying to assign (starts at 0, goes to 25), S is the input string

DP[i][j] = Min(S[i]-j + DP[i+1][j+1], DP[i+1][j]) — if S[i] <= j

DP[i][j] = DP[i+1][j] ------------------------------------ if S[i] > j

O(N*Alphabet)

I think cases should be swapped, but the idea is same. Thank you

Oh, right, thanks I'll fix it.

is there anyone else who cannot see others solution to hack!!!??

Same issue

same issue, I am also not able to see solutions to hack!!

Same here...

~~I'm investigating it. Thanks.~~Fixed!How come some are hacking and some of us are not able to see the solutions..

Up until now, people are only hacking their own solutions (the only solutions they can see right now).

yup got it, in fact the problem has been fixed now

How to solve problem E?

How can i hack others solution ?

It's sad that my F couldn't get accepted because I didn't initialize 2

^{ fib 0}and 2^{ fib 1}:(So I ended up writing a dp solution for problem C which finds the minimal number of moves. And after the contest I noticed in the problem statement "Your target is to make some number of moves (not necessary minimal)..." :|

Implementation issues in D :( Good problem, though, finally a DP problem I was close to solving.

Is it possible to view other's solution ?

seems most of them are hacking them self!lol XDXDXD

only a few who hacked others!! haha

Still, i am not able to see any other solution to problems. Ha Ha Ha

Hack to other solution how possible without a see code.

What is the test hack for problem C? I understand now.You are allowed to do multiple replacements for the same position of the string.

For the last question. maintaining prefix LIS and suffix LIS in same segtree and function as sum of both.(we subtract i from each a[i] initially and coordinate compress the array)

Can we do something like doing rollbacks on a segment tree in one direction and normal lazy updates in another direction. Does this work?

I'm not sure whether we can do it using only one segment tree, but model solution uses two segment trees with almost the same approach.

I have used one segment tree. There was no need of rollbacks, lazy updates, dynamic segment tree whatsoever. Only range query and point updates.

Solution

How decreasing memory to solve D.

http://codeforces.com/contest/946/submission/36016554 can anyone provide a test for which this fails?

Not allowed to view request page :/ Pastebin/Ideone?

thank you

I had like a minute left and read problem G and BS'd an answer that was basically copy-paste from LIS template.

Threw it together in <30 seconds and submitted without compiling, WA on test 20 lol 36021949

EDIT: I realize the error and I'm surprised it made it as far as test 20.

In Berland students study in the university 500 days a week

How did http://codeforces.com/profile/assbb solve D and E but not the first 3? Seems a bit curious to me.

How this output is ok for problem C?

UPD:Fixed and this solution got hacked.Because the output is not "-1" maybe.

"You are allowed to do multiple replacements for the same position of the string."

We are investigating the issue.

Why do so many people have weird barely readable aliases in their solutions, even for simple problems? Like this, for example.

So that hackers stay away?

Extract from CF Contest Rules

"It is forbidden to obfuscate the solution code as well as create obstacles for its reading and understanding. That is, it is forbidden to use any special techniques aimed at making the code difficult to read and understand the principle of its work."

& I think nobody follows this

I guess these short alias namings just a bit quicker to write

Did someone solve E using Binary Search by fixing the length of the prefix then generate the appropriate suffix?

Yes, I did.

36015253

I've just been hacked on problem A, Actually I discovered that I understood it wrong ... but my solution passed during the contest. I don't think that you should put so weak tests like those so solution for another problem pass!!!

Hacking is for some tricky case or clever case that makes the solution wrong but not for making absolutely wrong solutions pass during the contest and then hack them after the contest. I really think you should rethink about that.

The tests are not final. All the solutions will be re-tested on successful hacks after the hacking finishes. One of the points of competitive programming is that you should test your solutions better yourself.

I said I understood the problem wrong ... is it logical for a solution of a problem pass another one!! That's my point, I'm not saying that they should put all the tests, but totally wrong solutions shouldn't pass!!

Yes, actually I too wanted to raise the same issue, the testcases should not be so weak that it passes absolutely wrong solution, like at least it should cover all the basic cases.

Test: #1, time: 15 ms., memory: 1908 KB, exit code: 0, checker exit code: 0, verdict: OK

Input

aacceeggiikkmmooqqssuuwwyy

Output

aacceeggiikkmmooqqssuuwwyy

Answer

abcdefghijklmnopqrstuvwxyz

Checker Log

ok Ok

why checker log is ok ok? for problem A this submission

really weird.

this comment also says the same. It seems like a lot of them would be hacked

If the judge status were correct then the contestant could have a second chance.

Same thing happened to me in C.

SUBMISSION LINK:http://codeforces.com/contest/946/submission/36018053

It got accepted during the contest but now the same solution is giving "Wrong answer on test 1."

how to approach B. My solution gives wrong answer for large test cases.

brute force. take mod while a>=2*b and b>=2*a

There is an "else if" missing in Line 15 in your code, without this "else if" the while loop breaks if the condition on Line 15 is not satisfied, however, your code won't pass even if you fixed this line because of the input constraints (10^18) and will give you TLE.

The right approach is to subtract the quotient of the division of a by 2*b from a and subtract the quotient of b by 2*a from b while you can instead of subtraction.

My AC Submission : Link

What is the solution for F?

I did not solve it incontest and nor have I submitted it later, but pretty sure this is the solution (Confirmed by comparing it with accepted codes).

Basically notice that when you make the ith fibonacci string you already have a concatentation of

F(1)...F(i-1)present so itsF(i)...F(i-1)F(i-2)F(i-1)An important intuition is that the number of occurences of the string s in F(1)...F(i) can be broken as number of occurences of some prefix of s in F(1)...F(i-1) and number of occurences of the remaining part of s in F(i). [We multiply these].

To do this, we maintain 2 types of DP.

till[i][l][r]stores the number of ways to get s[l...r] as a subsequence till the ith fibonacci string.curr[i][l][r]stores the number of ways to get s[l...r] as a subsequence in the ith fibonacci string. Notice that for each way intill[i-1][l][m]we can match it with a way incurr[i][m+1][r]. Thus we multiply these. We apply a similar logic for the transition of curr. In simple terms these events are independent so they're multiplied.Transition:1.

curr[i][l][r]+=curr[i-2][l][m]*curr[i-1][m+1][r] for all m in [l, r)2.

till[i][l][r]+=till[i-1][l][n]*curr[i][n+1][r] for all n in [l, r)The final answer is

till[x-1][0][n-1]if fibonacci index = x and length = n (0 based indexing)Firstly, for every subsequence that is exactly the pattern, it will appear in 2^(positions free to the left) * 2^(positions free to the right).

left[i][l][r] = sum of 2^(positions free to the left) for every subsequence that contains exactly characters [l, r[ in the pattern

right[i][l][r] = sum of 2^(positions free to the right) for every subsequence that contains exactly characters [l, r[ in the pattern

sub[i][l][r] = number of subsequences that contains exactly characters [l, r[ in the pattern

sub[i][l][r] = (sum l <= m <= r of sub[i — 1][l][m] * sub[i — 2][m][r]) + sub[i — 1][l][r] + sub[i — 2][l][r]

left[i][l][r] = (sum l <= m <= r of left[i — 1][l][m] * sub[i — 2][m][r]) + 2^(size of f(i — 1)) * left[i — 2][l][r] + left[i — 1][l][r]

right[i][l][r] = (sum l <= m <= r of sub[i — 1][l][m] * right[i — 2][m][r]) + 2^(size of f(i — 2)) * right[i — 1][l][r] + right[i — 2][l][r]

The subsequence either has parts in f(i — 1) and f(i — 2) (the summation part) or is completely inside f(i — 1) or f(i — 2) (the rest). But this isn't the complete answer because it has 2^(one side). So the answer comes either from f(i — 1) and f(i — 2) or from one of them so:

ans[i] = ans[i — 1] * 2^(size of f(i — 2)) + ans[i — 2] * 2^(size of f(i — 1)) + (sum 0 < m < n of left[i — 1][0][m] * right[i — 2][m][n])

This was 2012 ACM ICPC World Finals problem D. A thorough explanation can be found at:

https://www.youtube.com/watch?v=dPFq4OLc-_Q

Did the image change just now?

In D, how could O(n*m²) solutions run in less than 100ms for some people?

n = 500 with o(n ^ 3) just so fast. Btw, for O(n ^ 2), if the operations are not complicated, it works for n = 10000.

On problem A ,I think the statement was very confusing. What was meant by "You may partition this sequence into two sequences"?? It seems the sequence should be partitioned into two contiguous part. Here 'partition' and 'sequence' have created confusion. The author should chose words carefully. I misunderstood it. Unfortunately the test cases was so weak that they passed for my wrong solution. Who always read explanation of test cases on A !!

I wasted 10 minutes on this problem and got a penalty of 20 minutes because in my first submission i thought that I must maintain the order of the input and build a prefix sum array to know the best position of splitting :\

Problem A was written in a very bad way that I didn't know whether should I treat the input as a contiguous sequence or 2 subsequences after splitting and that cost me 30 penalty points.

Problem C test data was soooo weak and that resulted in hacking a lot of submissions including mine

I hope you consider these 2 issues in the next round, the weak test data and the bad problem formulation

It says "in two sequences". I think this is pretty clear. They would say subsegment if they meant contiguous numbers

PikMike

Vovuh BledDest

/*

Input

aacceeggiikkmmooqqssuuwwyy

Participant's output

abcdefghijklmnopqrstuvwxyy

Jury's answer

abcdefghijklmnopqrstuvwxyz

Checker comment

ok Ok

*/

How did my solution for C got accepted verdict? It's not giving correct answer even for the 1st testcase.

Here's the link to my accepted solution(during the contest, it got hacked now): http://codeforces.com/contest/946/submission/36018053

Another thing, now if you try to submit the same solution, it gives "wrong answer on test 1".

As you can see it clearly gave wrong answer for the test case but the verdict is "ok,ok" :(.` Can you please fix this? this is a real serious bug. Btw, it really screwed my contest. Now,I am literally inventing ways to screw a contest :(.

/*

Input

aacceeggiikkmmooqqssuuwwyy

Participant's output

abcdefghijklmnopqrstuvwxyy

Jury's answer

abcdefghijklmnopqrstuvwxyz

Checker comment

ok Ok

*/

Thanks!

There was an issue in the checker, for some tests it returned OK instead of WA. It is really unfortunate but it is not a dramatic issue because you can always assume that there weren't such tests during a contest. Obviously, this argument can't be applied to wa1, though our estimated number of people affected is not that great. The participant can literally check his output for the sample case himself. We are still really sorry about it, the checker is fixed now and we will do our best in the following rounds to eliminate the possibility of similar mistakes.

Hi Vovuh

Shouldn't this round be made unrated for all the users who experienced the same problem?

Thanks!

`Subsequence of the string is the string that is obtained by`

deleting`characters at some positions`

Did this note mean any thing to the solution in problem C ?

because it mean we can just print

`abcdefghijklmnopqrstuvwxyz`

or im wrong ?For example, see this test case (note the z at the 2nd position)

`azbcdefghijklmnopqrstuvwxyz`

There is a subsequence valid if we delete the first z, but we didn't use any change operation so the correct output is:

`azbcdefghijklmnopqrstuvwxyz`

The output must contain the changes you made for the string to be ok, but deleting a character is not a "change".

Hope this helps you!

But he didnot say that i must made a change he just need the final subsequence

Actually it says "You need to print the string that will be obtained from the given string and will be contain english alphabet as a subsequence" and "If you can get a string that can be obtained from the given string and will contain english alphabet as a subsequence, print it."

It says to print the changed string

that containsthe alphabet as subsequence, it does not say to print the subsequence.I agree that it could not be 100% clear at first sight :p

Please put stronger pretests in futur contests, We really don't want this awesome website to turn to Hackforces instead of Codeforces, this is really bad I mean I understand that this whole pretests thing is necessary due to huge number of participants but really try to make sure you put some strong pretests (sometimes a completely wrong solution can pass pretests... ) . In addition to that the hacks idea is really good but I think the hacks should not be very easy like they are now ( some people don't even read the code they just try limit input data and see if they get lucky hacking the solution )

After doing 108 successful hacks and then later know that it doesn't give you points in educational rounds..

its doesnot give you point but if your rank is 300 and you hacked user with rank 299 you will be 299 so its good to hack

I was thinking to send a message to inform you but I said maybe he do it for a purpose.

I enjoyed it :D, hacked many div1 users.

When should rankings for div 2 update?

How is penalty calculated in educational rounds?

Sum of 1)Time taken for all questions from start of question to accepted solution 2)20*no. of wrong submission

Oh thanks for that. 20 seems quite high tho, considering around 1000 people solve 3 in 30 mins.

Yes, in my opinion in should be lowered, as it is blown out of proportion against time penalty in educational rounds.

Why during system test of ACM-style contest pending runs are shown with

`-x`

instead of`?x`

? And API also says that result is final, but some submits are pendingWhy the rating has not changed so far ?

BledDest

This is with regard to question C 946C - String Transformation. I have solved the question C but after system testing i got TLE.

But to my surprise when i resubmitted the same code after the system testing it got accepted . I am posting the links of 2 submissions here one with TLE and other with Accepted.

one with TLE — 36008794

one with Accepted — 36055044

and both solutions are same , you can check yourself by comparing diff(they have no diff)

Can you please see this issue?

These are my hacking testcases.. I did 108 hacks with them for B:

3 1000000000000000000

1000000000000000000 49999999999999999

49999999999999999 1

1 1000000000000000000

When rating update will happen for div 2 :(

So u were waiting :D You became specialist. Congrats (Y). I became green :D

thank you <3 good luck next contest

thank you <3 good luck next contest

My cf predictor shows my rating will increase by 125(it has not been wrong yet) . This result was shown to me just after finishing the system testings. As cf rating changes has not been updated yet, may i assume that cf will use a new formula this time to calculate new ratings? Some of the results by cf predictors looked unfair to me. So new formula won't be bad if authority approves.

When rating update will happen for div 2 ^_^

do you copy past codes like comments? so .. you are just green :/

my_comment

i just saved my comments_copy_rights c.

no i write my code ,but i too lazy to write comments when it written already

I think that the reason that the rating is not updated yet is because of the error in the previous test of problem C.

rating has just updated now wow ;)

nvm

Who won the special prizes?