Hello, Codeforces!

I am very glad to invite you to the Codeforces Round #843 (Div. 2), which will take place in Jan/10/2023 14:15 (Moscow time). **This round will be rated for the participants with rating lower than 2100.**

My sincere thanks to:

Wind_Eagle, BaluconisTima, kartel и Aleks5d for the tasks ideas, and also Wind_Eagle, BaluconisTima, kartel, VEGAnn и Vladik for preparing the tasks.

andrew, gepardo, 244mhq, vilcheuski the rest of the author team, without whom this round would not have happened.

BaluconisTima for the splendacious pictures, which complement each problem.

MikeMirzayanov for Codeforces and Polygon platforms.

You for participating in this round.

You will have 2 hours and 30 minutes for solving 6 tasks, **one of which will be divided into easy and hard verions.** The round is based on the problems from the first day of Belarusian Regional olympiad. **We kindly ask all Belarusian pupils who participated in this olympiad, to refrain from taking part in this round and discussing the problems publicly before the round ends.**

I hope you will enjoy the round!

**Round testers (not complete list):** MathBoy, FairyWinx, nnv-nick, K1ppy, olya.masaeva, Septimelon, PavelKorchagin, 4qqqq, kova1.

.**Score distribution: (500+1000)-1000-1500-2000-2000-2500**

**UPD.** Score distribution was changed: **(500+500)-1000-1500-2000-2250-3000**.

**UPD2: Congratulations to the winners!**

**Победители**

Vote for your favourite task!

Task А: Gardener and the Capybaras
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Task B: Gardener and the Array
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**UPD3: Finally, here is the editorial:** press here

omg subtasks for A round

NOTE THE UNUSUAL TIMINGS!! Wish you positive delta!!

omg no green circle

I hope this round will be better and more interesting than before.

Which is better: Div. 2 or Div. 3 for those whose rank is higher than rookie?

Div.2

omg subtasks for A round

omg subtasks for A round

omg subtasks for A round

omg subtasks for A round

omg subtasks for A round

Hey! silly i broke your chain!

omg subtasks for A round

omg subtasks for A round

omg subtasks for A round

omg subtasks for A round

I am breaking your chain down vote me if you can.

omg subtasks for A round

omg subtasks for round A

omg subtasks for round A

Just curious what's the meaning to reply this message recursively

omg subtasks for round A

omg subtasks for round A

omg subtasks for A round

Please Downvote me. And not feeling sorry to break the chain.

omg subtasks for A round

I think (500+750) is better than (500+500).

Thank you, AHMADUL.

what is your age bro ?

My birthday was some days ago and now I am 13

I'm amazed though seeing u became purple at such age, I didn't even knew computer when I was 13 to be honest. Nice work, keep working hard.

I wish Belarusian students good luck on Belarusian Regional Olympiad.

I hope tasks will be interesting for you :)

## Thanks for inviting

Please no more problems like edu 141B

Delete this they will surely give it then

That's was a good constructive problem. I like those problems because even newbie can solve it and i have seen even master are sometimes not able to solve them. Try till you get the right approach.(It's just how i see them haha).

Well I literally hate problems like edu 141B and div2 842C. I feel like these problems are all about guessing. The faster you guess the better rank you get

could u post a blog about this server i recently got to know about this where mass copiying is happening https://discord.gg/6RskCNmG

bruh imagine advertising cheating

is there anything we could do?

they will create new accounts after bans anyway, so best we can do is not advertising them

i think u can't do anything, but sadly they didn't improve them self

As a tester, I'd like to say that problems are interesting imo and I recommend you to participate!

hope this time problem statement is clear :/

Bro there is now way you are now reaching pupil after 1 year. You are surely doing something wrong. Please analyze it quickly or else it will be too late. 3 months are more than enough for pupil if you are serious that is.

thanks bro.

https://codeforces.com/blog/entry/110965.

Hope This blog helps you Sir.

First time see a problem A with hard version

Good luck everyone!

for the first time i think there would be subtasks for a this is very exciting!

Omg! Yellow round.

Omg! Yellow round.

Amazing.

Yes,its really amazing contest.

As a contestant I think that most of the peoples have positive results.

Good Luck to everyone.

So What?!

So it seems like from the score distribution, the whole problem A has the same difficulty as B right?

Yes, with a 50% sub-task scored as well. AgreeDisagree

OMG,there are subtask for problem A.

Dont include Mathematical Problems please.

Will this contest be as beauty-full as the last one.XD.

Finally a contest with friendly time to Chinese coders! Gl, hf!

Actually, 22:35(UTC+8) is not TOO unfriendly for who are used to staying up late. It's better called friendly time to Australian coders.

I remember that once there were two contests held at about 15:35(UTC+8) and 18:35(UTC+8) on the same day.

Exactly, very friendly time! Hopefully, this is also a fantastic round!

Of course it's a very friendly time.I can't stay up late because my father don't allow me to do that:)

nice

i give up :(

Div2 A with subtasks? Sounds interesting,but I don't know if it will be a good problem.

OMG unusual starting time and subtasks for A!! Wish you positive delta!!

harder version? omg with the same rating

Finally, the opportunity presents itself. I will become a newbie again.

Unusual time :(

I hope that this round will be a starting point for me to succeed in the new year 2023 ! And I wish good luck and success to everyone this year.

Hello can you upvote me to help me with my contribution.

No! I prefer Downvoting you.

Ok as you like but i don't think i did something that i should be minused

You act like a girl while being a boy

I down voted you, to show world doesn't work that way. :)

Plan to take part in this contest. Hope $$$\Delta>0$$$ for me, and you too!

OMG, subtasks in problem A!

OMG, subtasks in problem A!

OMG, subtasks in problem A!

OMG, subtasks in problem A!

OMG, subtasks in problem A!

NOTE THE UNUSUAL TIMEGood time for Chinese.

wow

wow

wow

My GPA is crying out

prove your loyalty to codeforces

8 pm is best time . We have classes at college .

nice u got a reason to skip classes :)

Hope it will be a wonderful contest with fantastic problems!

As a Chinese,I usually have to stay up late for the cf round, but today I can go to bed earlier, so enjoy it and have fun!

Ok so in this round I will not able to do problem A as well (interesting).

I suggest to do more A , B subtasks in the futureTrue Mate.

After this contest, I have to disagree.

Problem A2 is scaring me. I wouldn't have participated if I cared about my rating.

orz 244mhq

Why Delay 10 min Sir ??

Why Delay 10 min Sir ??

I have the same question.

Same question

Maybe rating update

Contest start time changed?

10 minutes postponed!

Yes, postponed by 10 mins. Good luck!

Thanks! I thought my country decreased its standard time by 10 minutes

bruh, save your logics for the contest lol

10 min. delay!

DelayForces

Now it's delay-master-forces.

Will there be second delay if the rating update of last contest still uncompleted?

I did not understand at all what to do in the task B

I love this world.

Love this contest! As I expected from CF Round #741's author.

Mee too :D

Thank you for participating! Round 842 was great too!

F is 100491E - Expedition to Mars on higher constraints but it's possible to adjust the intended solution to 4e5.

Actually, I agree with you it's just a matter of time to solve this version if you have a solution for that gym problem and 144 participants solved that problem already I guess so it's easy for them.

Tbf $$$n$$$ up to $$$500$$$ allows a lot more different approaches. When my team and I were solving that gym contest, we went for a solution that isn't really viable for $$$4 \cdot 10^5$$$. Well, at least I failed to fix it :(

The most balanced contest

I kinda feel bad for spiders with one leg(

You are not worried about $$$3\cdot10^5$$$-legged spiders, right?

legend says that spider is still tying his shoelaces.

For C when i run the code in my system it was giving correct output for testcase 1 , but its giving wrong output when run on codeforces for testcase 1 . Any Idea why so ? https://codeforces.com/contest/1775/submission/188754539

Nice problemset..

How to solve C?

If $$$x = n$$$, the answer is trivially $$$n$$$ itself.

If $$$x = 0$$$, this means that the most significant bit of $$$n$$$ must become 0 at some point. If this is the $$$a$$$-th bit from the right, then the first number to flip this bit is $$$2^a$$$ (single 1 followed by $$$a$$$ 0s). So the AND must include $$$2^a$$$ and we can see that $$$x$$$ AND $$$2^a$$$ is 0, so the answer is $$$2^a$$$.

Now $$$x$$$ must have at least one $$$1$$$. Find the last $$$1$$$ in $$$x$$$, and let's say it's at position $$$\ell$$$. Because bit $$$\ell$$$ is a 1, this means it was never flipped while incrementing the numbers, suggesting that all bits to the left of bit $$$\ell$$$ remain unchanged. So the prefix of $$$x$$$ up to bit $$$\ell$$$ must be equal to the prefix of $$$n$$$ up to bit $$$\ell$$$. If there is a mismatch, the answer is $$$-1$$$.

We can now write $$$x$$$ as some $$$prefix$$$ (which ends with a 1) followed by all 0s, and $$$n$$$ as the

same$$$prefix$$$ followed by some $$$suffix$$$ (whose length matches the block of 0s at the end of $$$x$$$). There are two further cases to consider:If the $$$suffix$$$ begins with a 1, then we have a problem. Since this bit is 0 in $$$x$$$, it must be flipped at some point. But flipping this 1 will also flip the 1 to its left, i.e., the last bit of the $$$prefix$$$. But we know that the $$$prefix$$$ cannot change, so this scenario cannot happen, i.e., answer is $$$-1$$$.

Otherwise, find the first 1 in the $$$suffix$$$. Let's say it's in position $$$c$$$ (from the right). This bit is 0 in $$$x$$$, so it must be flipped at some point. When this bit is first flipped, the bit at its immediate left (position $$$c + 1$$$) becomes 1, and everything after it becomes 0. This number is $$$x + 2^{c + 1}$$$, i.e., the common $$$prefix$$$ followed by all 0s except a 1 in position $$$c + 1$$$. Conveniently, applying AND between this number and $$$n$$$ will already yield $$$x$$$ (the only $$$1$$$ after the prefix is at position $$$c + 1$$$, which is a 0 in $$$n$$$), so this number is the answer.

My submission: 188732906. I converted the numbers to binary first, and worked from there, converting back when printing the answer (except when the answer was $$$n$$$ or $$$x$$$ itself).

"I converted the numbers to binary first, and worked from there, converting back when printing the answer."

Conveniently for me, my computer stores numbers as binary internally so I never need to convert back and forth.

lol, yeah, I know I can use bit shifts to process them faster, but binary strings are more readable imo. I don't have enough experience with bit shifting to be confident that I can code the correct solution with lower thinking + typing time.

Answer will always lie between n and INT64MAX and bitwise and from n to k >= bitwise and from n to k+1 therefore you can apply binary search . To find bitwise and from n to k you can search on net and get method easily .

"bitwise and from n to k >= bitwise and from n to k+1"

How did u arrive at this property? Can you please explain/elaborate as i am unable to understand.

just try to analyze the truth table of AND.after applying and operator you will realize that, it keeps value as it is or make it decrease.

1&1=1 0&0=0 value remains as it is in above two cases.

0&1=0 1&0=0 value decreases in these cases.

"Time limit exceeded on pretest 12"

no need to look at one prime number more than once during bfs (this helped me with test 12)

~~I hate A so much that I mistook subarray for subsequence in B.~~OMG I didn't read this in problem A

SpoilerThe name of each capybara is a non-empty line consisting of letters "a" and "b".

Problem A-D is interesting.

How to solve F?

I suspect giant casework.

Actually, no. The short idea is that all the possible answers are just rectangles with their angles cut somehow. You need to make a DP on how you cut the angles and iterate over the possible rectangles.

For example, if $$$n=7$$$, then the optimal perimeter is $$$12$$$, and you need to try out the rectangles $$$3\times3$$$, $$$4\times2$$$ and $$$2\times4$$$ and cut their four angles somehow. A cut of an angle is just removing a stair-like figure, so it's calculated with DP.

Ah got it, problem seems interesting now!

The idea is actually obvious, can you tell how to calculate the DP?

You may precalculate $$$dp_{n,k}$$$, that is the number of stairs from $$$n$$$ blocks which are $$$k$$$-block tall. This DP can be computed in $$$O(n^2)$$$, but, since the number of blocks to be cut from angles is no more than $$$O(\sqrt n)$$$ in total, it's OK.

Then, you just need to write another DP which decides on how much blocks to cut from each angle. It's $$$d_{i,j}$$$ where we have decided about the first $$$i$$$ angles and have already cut $$$j$$$ blocks.

Note that all these DPs can be calculated in the beginning and used to answer all of the testcases.

Got it, thanks!

But how do you check that the stairs in different angles do not intersect while calculating $$$d_{i, j}$$$?

You don't have to check it.

If they are going to intersect, then the perimeter is just not optimal.

Oh yeah, fair enough. It was the only part I was struggling with and this observation is so easy (( Thanks a lot!!

They will not intersect. Because if it is possible, you can decrease the perimeter of your figure.

Any idea of B? I've tried many times but always got TLE

Also please update the rating of last educational round quickly

What I have done, is to count the no of occurrence of a bit throughout the whole array. Then for each element in the array if any bit which is set for this element is also set for any other element, Then the ans is always yes otherwise no

Let me explain the reason with an example ,consider a no whose bit 2 and 4 is set. Another no whose bit 2,1 and 3 is set and another no whose bit 4, 5 and 7 is set. Since For First No, every bit that is set for this, also set for either b and c, so we can always take Or of every no in the array and Or of every No except that No(First one in this case). In Other words f(a) = Second No | Third No, f(b) = First No | Second No | Third No. Hence Ans is always possible.

Here is my submission 188728251

If bits set in one number is a subset of another number , then the answer is always possible.

what about 4 number {1,2} , {4,3} , {1,3} , {2,4}

Yes in that case also the answer is possible.

We need to check for every number if the occurrence of all the bits in this number is > 1. If it is satisfied, then the answer is Yes.

Looks like you cheated in this round.

Nah dude , I just failed at explaining my idea xD.

Did you by any chance used a fixed-size frequency array? freq[200001] * n <= 10e5 = TLE... I should know from having fallen for it 5 times this contest.

So we need to use map for frequency?

Yes, I changed to map for frequency and got AC with 300 ms

OHHHHHH I just skipped B......

Yes, map for frequency.

I've thought Double linked-list and priority-queue could solve E, but got WA.

when I used vector to store the frequencies I got the TLE, but then changed to unordered_map it passed. not sure why vector[i] is slower than map[i] :(

I used int[200001] and got TLE 5 times because I forgot n <= 10e5 so I was doing 200000 operations per test case just setting up a frequency array.

now I get it, thanks

Two cases:

1) if second character is ‘a’, then place s[0] s[1] rest

2) if second character is ‘b’, then place s[0] allCharactersExceptLast s[n-1]

Hint: the first subsequence is the whole array, the second is the whole array except only one number.

I've noticed that but my approach got TLE

It seems that we need to store frequencies by map instead of array

You can use array, but it should be global. Furthermore, you should only set the elements you touch to 0, not the full array.

How to solve E?

Let $$$s_1$$$ be the maximum sum over all subarrays and $$$s_2$$$ be the minimum sum over all subarrays, then the answer is $$$\max(s_1,-s_2)$$$.

Lower boundClearly you can change the sum of some subarray by at most 1 per operation, so the answer can't be lower than $$$\max(s_1,-s_2)$$$.

Upper boundFor simplicity we suppose $$$s_1>-s_2$$$.

Consider the following rules:

Adjacent positive elements are automatically merged into their sum. (The same for negative elements)

Zeros are automatically removed from the array.

It can be shown the answer won't change if we apply the rules.

After merging, we can repeatly try to change every element closer to 0 and apply the rules after each operation. It can be shown it's optimal. The sum of the greatest subarray will always reduce by 1 after each operation. (If we can't at some moment, then it's not the greatest subarray.)

If the answer is greater than $$$s_1$$$, then there exists a subarray with sum greater than $$$s_1$$$, which is a contradiction.

currently, your comment helps pretty much noone except those who just who want to see a AC and be done with the problem.

What was the idea? how did you come up with it?

Errorgorn's solution makes much mose sense in that regards

The intuition is that after each operation, the total sum of the elements changes by $$$-1, 0$$$, or $$$+1$$$. So if the current sum is $$$s$$$, then the answer is at least $$$abs(s)$$$.

The second thing to note here is that the answer for the whole array is at least the answer for some consecutive subarrays. So the answer is at least the maximum of the absolute value over all subarray sums.

Seems like this is achievable. But I currently don't have a proof for this. I have solved it using the same approach that errogorn used in the next comment.

Sorry, it's my bad. I've added some details.

We will process elements left to right. When processing some element, we either create new operation or extend some old operation.

Intuitively, it makes sense that we would not have to using operations to both add and subtract from an element. So a simple $$$O(n)$$$ greedy works.

codeThank you very much!

Is there any formal proof? (or an outline of a formal one)

YesThe red and blue are different operations. If we delete the middle operations in the green box, we can merge the left red operation with the right blue operation and also merge the left blue operation with the right red operation.

This way, we will never both add and subtract from the same element in an optimal solution.

Think about a sequence of prefix sums and observer how such operation affects this sequence of prefix sums.

If you analyze carefully enough, then you'll see that one of the operations just adds $$$1$$$ to any subsequence of prefix sums, and the other operations just subtracts $$$1$$$ from any subsequence of prefix sums.

So, AFAIK you just need $$$\max(0, \max(p_i)) + \max(0, \max(-p_i))$$$ where $$$p_i$$$ is the array of prefix sums.

Thank you for the second solution!

I take back my words, it was a nice round.

F: Why did you set two types of questions as one problem? The difficulty levels of the two questions are different, so I think they should be set as separate problems.

A problem of calculating the number of ways was very interesting!!

I think subtask 1 is as easy as problem A...

You are right probably, it should be split.

By the way, the original contest is OI-style, so there two problems were combined as one, with different groups for each of the subproblems. For Codeforces, the problem was just retained as-is.

can we use binary search for Problem C? since & is decreasing function, bitwise and of all number from n to m will be higher than x initially. We need to find first position where bitwise and of all number from n to m is x

Problem E almost = This Problem

Yes, you are right, sorry for that :(

It's interesting that the round with the similar problem is also of Belarusian origin, but the authors came up with the idea independently AFAIK, so it's an unfortunate coincidence.

Very nice contest! Congrats to the authors!

lol, img when i got wrong ans on test 15 problem D and it was only 5 minutes left

bruh I feel like a total loser, but at least I'm trying and participating lol

In problem B, I read "exclusive or" instead of "bitwise or" and was thinking how to find out if the set of sparse binary vectors is linearly independent and why it is Div2.B

I did this too xD...

String consists of letters 'a' and 'b' only :) could have written it in black :)))

WhiningWow, now I do remember that I did read that line, but after reading that whole description, I actually forgot about it :)

Moreover, the divided strings were ofcourse named

`a`

,`b`

and`c`

,which even cut the possibilities of me thinking about the string only consisting of`a`

and`b`

for once :)Well, I just knew about it know.....

I think B problem cannot be solved by people who use JAVA because my O (n * k) solution is getting TLE even after using the fastest I/O operations. Link to my Submission :- https://codeforces.com/contest/1775/submission/188733443

change your "int[] freq" to "map freq" and try, it worked for me in c++.

Fixed-size frequency array is unsafe, n <= 10e5 so setting up a frequency array for each test case will instantly TLE you. USe map/hashmap because k constraint is guaranteed to be <= 10e5

You are saying that as t <= 10^5, so the freq array of size 2 * 10^5 (max p) will be iniatialized for each t which will give TLE and everytime initializing a fixed length array is not good beacuse there may be a case when max p is very small but still I am iterating till 2 * 10^5 every time which is unnecessary and the reason for TLE. But HashMap will not give TLE beacuse when we initialize a HashMap its initial size is very small due to which it will not give TLE. Right?

Yes

if i solved a problem correctly in contest and later during contest , by mistake i submit wrong code for that problem then is question points are added or not to my points

how to solve problem b??

hint1denote bitwise-OR of the whole array as S. if there exists an element x such that removing x from the original array won't affact S, the answer must be an "yes"

hint2if there isn't such an x, the answer is "no". figure out why.

Can u share submission plz

I can't understand the

Inputof B :(in problem b n is size of array and n line is given in each line first no is the no of set bit in that number and after that positions of set bit is given

I still don't get it, but thanks anyway.

The very first number is the number of test cases.

The first number of each test case (say k) is the number of elements in the array of that case.

Now k lines of space-separated numbers follow — each line is an element of the array represented in this form: the first number (p) is the count of set bits of that particular element of the array.

The numbers that appear after p (there are p of them) are the positions of each of the set bits.

Edit: Essentially, each individual line represents an element in the array.

If a line is 3 1 2 5,the binary number is 10011，to decimalism is 19. In the contest I understand it this way，Is it so?

Yes, 3 1 2 5 implies 10011

Wow!!

The "Note":In the second test case, one of the possible answers are following subsequences: the subsequence a formed by the element at position 1, and the subsequence b formed by the elements at positions 1 and 2. 2 2 1 2 1 2 This is a array{3,2}，but 3 != 3 ^ 2.this is why？ Can you tell me? Thanks.

I don't completely understand your question, but I can explain the second test case:

like you said, this represents the array [3, 2].

The given solution for this case is that the two subsequences would be [elem_at_pos_1] and [elem_at_pos_1, elem_at_pos_2].

So the text claims the subsequences would be

`s1 = [3]`

and`s2 = [3, 2]`

.And this is true because the bitwise OR of

`s1`

= 3 (as there's only one number) and the bitwise OR of`s2`

= 3 | 2, which is also 3. Clearly, they're both equal.I think you may have misunderstood the bitwise OR operation to be something else.

Oh, I see. I mistake bitwise OR as XOR. QwQ My English is not good.I didn't expect to make such a mistake. Thank you very much.

No worries, cheers!

In task A, either the string b is lexicographically not smaller than the strings a and c at the same time, or the string b is lexicographically not greater than the strings a and c at the same time. @authors kindly look into the language

Screencast with commentary

What IDE/Editor/Plugins do you use for contests?

how to solve B? also in C is there any relation between x and m?

didn't read "The string consists of English letters 'a' and 'b' only." this part in A, So I solved it for every character and I had to use Z function for prefix comparison XD

Task is easily solvable for the greater alphabet. Soon we will post the editorial.

Now that I have though about it more, you're right

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

realized a b only after reading this comment

I didn't read that either. Here is my solution: 188713747.

Upd: it is wrong

Accepted in last 4th sec.

Can't understand why for B:

answer is YES?

Next follow ki

distinctintegers pi,1,pi,2,…,pi,ki (1≤pi≤2⋅10e5)Lol, didn't see that.

Ki integers are guaranteed to be distinct

This input is not valid, since the set bits must be distinct. For example, in the line

`3 8 2 2`

, the given set bits are 8, 2, and 2, which is not a valid input. In the line`3 8 3 3`

, the given set bits are 8, 3, and 3, which is also not a valid input.If you actually meant the following:

then the answer is NO.

Nice problems. Hope to see more contests like this in the future

Will this round be rated for who has reached 2100+ from 2100- in the last educational round?

Overlooking "the characters are A or B" from A2 took me 1hr...

I can't proof my solution for A2 (but it's allowed to use A-Z) so please hack it: 188722831

This is actually a correct solution, if I'm not mistaken. For an arbitrary alphabet, if answer exists, then there exists such answer that |b| = 1. You can now solve task by bruteforce in O(n).

I saw the A or B part, but didn't have a very good proof, so I solved it the most stupid way imaginable.

submission: 188710406

I also overlook the

aorbterm which waste my 1.5 hours. And if am not wrong, the sentence which told that specificaorbterm inA2is added later on the contest. I have solved this through an observation that excluding the first and last character, if there exists anathen if I take it as the second word then it remains the smaller or equal the other two word which are the left and right part of the taken second word. If excluding the first and last character, there not exists anathen it should be the occurrencesband if I take those consecutivebas the second word and the first and last character respectively as the first and last word then the second word remains bigger or equal than the first and last word. In such these way, under the constraints of the problem there always exists a valid answer.And if am not wrong, the sentence which told that specific a or b term in A2 is added later on the contestYou are wrong and just try to justify yourself.

Thanks for the confirmation. It is honor for me to your reply. I don't try to justify myself. Everyone do mistakes. From next I will be aware of this.

Intrestingly, this is the original problem that was given in the olympiad. It took me 1h40m to solve it, and when I saw codeforces standings, I was absolutely shocked by amount of people who solved it that quickly.

How to solve B and C ?

didn't notice the unusual timing :facepalm:

For C when i run the code in my system it was giving correct output for testcase 1 , but its giving wrong output when run on codeforces for testcase 1 . Any Idea why so ? https://codeforces.com/contest/1775/submission/188754539

In my system, your submission is giving wrong answer only for the last test case.

what about the output here . https://www.onlinegdb.com/online_c++_compiler

That's weird ;)

You're doing

`ll rem=(1ll<<i)-(i?pre[i-1]:0);`

inside a for loop where`i`

can equal`0`

, meaning you're trying to do`pre[-1]`

which points you to other memory. This is why the code doesn't work or works differently on different machines.But i am checking for the case i==0

Oh, how did I not see that. Sorry.

Maybe an overflow is happening. Change LONG_MAX to 1e18+1

And don't forget, that in the case of n and x being both zero, your pre will be 0 sized, which gives an error

Thanks Bro .I changed LONG_MAX to 1e18 and it worked . But any idea why so ?

It seems, that LONG_MAX doesn't work on cf. It returns 2147483647(int_max) on cf, and 9223372036854775807 on onlinegdb. This wasn't an overflow then

Best ever standing and solved problem C in Div. 2 the first time! But I didn't have any clue of problem B :(

HintSee the count of the i-th bit appears in the input.

SolutionCheck if for every number there exist a bit that appears in the input count is 1 than the answer is "NO" else "YES".

but how are we sure that if we cant remove any element such that or of remaining element is same then answer will not exist?

If for every number there is a unique bit that only appears into that number than to make the difference in the sequence you need to not take this one. If all the numbers contains a bit that is unique in them you have only two options either take all of them or nothing. To make them different we take both as these are the only options but there OR is not same.

It seems that we should check every number in array c, if there exists one i, every bit in c[i] appears 2 times in the whole input, then the answer is "YES", otherwise "NO".

poor English :( Obviously the maximum set is a1 or a2 or ... or an .So let p is this subsequence.

To find q, the number of elements in q must be at least 2.In this case, without q the p has no elements left.

The set is in binary system.（二进制）

Some discussion for E:

The main purpose of operation is to decrease the value of sum(|ai|) (the sum of the absolute values of ai), to achieve this purpose, we need to subtract from positive numbers, and add to negative numbers.

So every time when we choose subsequence, we can assume we always choose it with alternative sign (which like {+a1, -a2, +a3, -a4...} or {-a1, +a2, -a3, +a4...}), because if not, we can remove 2 adjacent element with same sign, and the total change of sum(|ai|) remain the same.

Therefore, we can always choose the longest subsequence with alternative sign. We can regard some consecutive elements with same sime as one element, and regard zeros as not exist. So we can rewrite the initial array as some non-zero numbers with alternative sign, each element represents the sum of a maximal block with same sign.

So we can get the answer: In each step, we do min(|ai|) times of operation, and all |ai| will be decreased by that value, then we remove zeros from the array, and merge adjacent elements with the same sign. However, this naive approach will be O(n^2) and get TLE. How to optimize it? Notice that if we do min(|ai|) times of operation, some elements will become 0, and it's adjacent elements will be in the same "block with same sign" now. So we could store the initial array into a double linked-list, whose node is like (long:value, Node:prev, Node:next, boolean:valid) and put each node in a priority queue. Each time we poll out a valid node from the priority queue, we merge it with node.prev and node.next to get a new node, link it with node.prev.prev and node.next.next, add it to the priority queue, then mark node.prev and node.next as invalid. Then we could get an O(n*log(n)) solution. (Also do not forget check nullity)

Update: Now my submission:188766704 has got AC. Unfortunately, I've not implemented it properly during the contest. Sad!

Mmmmm

Right after the contest :D

Wonderful round with great problems, it seems they all have clean solutions, but require "cute" observations. A2 is harder than E for me though (failed system testing on A2).

To problem B: Why is this case no？ 1 3 2 1 2 2 3 4 2 5 6

Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

But the rating of last educational round has not updated yet! Please update it first! I should have been div1 in this contest......

Oh, sorry for that, our fault. Will fix soon.

Thanks very much (for saving my rating)!

fixed, you got your rating back :)

OUR fault :>

I might be wrong, but I think whether the contest is rated for you or not depends on which rating you had

at the time of registration. Even if the rating for the edu round was updated before this contest but after registration, this contest would still have counted for you.In fact, this is why some users deliberately register for as many contests as they're able to when they're close to the border, so that they're all counted even if they cross the border sometime in the middle, e.g., earning Div2 rating boosts even after reaching Div1.

So regardless of when the ratings are updated, it should not change the fact that this round will be rated for you. In the future, please consider your rating

at the time of registrationbefore you decide to register for a contest, since rating changes that were not yet applied would not be considered for the contest you're registering for.Huh, thank God I took this div2 seriously :) Otherwise my M would have lasted less than 24 hours.

Problem D was nice.

I agree. It was a very nice combination of Sieve and BFS.

Wait, can we use DSU for problem D ?

I don't think so as DSU can tell about the connected component but we need shortest path.

I liked the problems themselves, but I have two minor issues with the presentation:

For Problem B, the input format was rather unusual, so I think it would have helped immensely if there was an example of a line in the given input format with an indication of what number it represents (preferably in both binary and decimal). As it is, I think the problem looks rather intimidating, despite actually being very simple, especially since participants do not even need to be aware of how to code bit operations to solve this.

For Problem C, specifying that the answer is guaranteed to be bounded below $$$5 \cdot 10^{18}$$$ is rather misleading, and notably suggests that a standard 64-bit signed integer type (like

`long long`

in C++) is insufficient. However, the actual upper bound is much lower (I think it's $$$2^{60}$$$), so 64-bit signed integer type would actually have been fine. But having the constraints specify $$$5 \cdot 10^{18}$$$ can scare away participants who memorized the upper bound for 64-bit signed integers before they could figure out that the upper bound was highly overestimated or that an unsigned 64-bit type would have worked regardless.These are minor issues, but I think they serve as obstacles for less-experienced programmers who might otherwise have been able to solve these problems more smoothly, and I don't think early problems like B and C are good places to punish them for these kinds of little details.

The max value of signed int64 is 2^63-1=9.223*10^18, which is definitely larger than 5*10^18.

Oops, sorry, you're right. I was mistaken in thinking that was the unsigned upper bound. That being said, I don't think I'm alone in making such a mistake, so I do still think the overestimated upper bound can be intimidating for many, regardless.

Most problems I see that involve

`long long`

calculations use an upper bound of $$$10^{18}$$$, or (more rarely) $$$2 \cdot 10^{18}$$$ if intermediate values are not expected to overflow. An upper bound of $$$5 \cdot 10^{18}$$$ is quite unusual, especially for a C problem.How to solve C?

Hint 1Find a way to compute the value of x & (x+1) & .... &(x+m) in O(lg n).

Hint 2Binary search

what is wrong with this submission 188779455 for problem C. Time limit exceeded.

TLE is due to

`1 << msb_p1`

, it is fine for`32`

bit numbers but not for`64`

bit numbers.You may see my submission

Thanks coderdhanraj

you can either use binary search or can use maths!

check this

How to solve D? I tried to construct and graph using prime factors and then apply DFS but got WA.

DFS ? I think you mean BFS

Use BFS to get the shortest path. DFS is not for shortest paths.

I believe dfs (or DP) can be only used for Directed Acyclic graphs to find shortest path between nodes. anyone correct me if I'm wrong.

Actually both are used to find the shortest path in DAG. First using DFS for finding the topological sort and then according to topological sort apply the dp.

Is the jiangly's answer corrent for this case of problem A?

1 abcbbc

The string only contains letters 'a' and 'b'.

thanks bro.

The string only consists of "a" and "b".

String consists only of letters 'a' and 'b'.

The string consists of English letters 'a' and 'b' only.it is mentioned in problem statement.

I wasted 10-15 minutes because of not reading this carefully.

I think this solution 188779455 works in O((log(n))^2) for problem C

Time Limit Exceeded on problem C