Hello, Codeforces!

We invite you to Codeforces Round #765 (Div. 2), which will take place in Jan/12/2022 15:05 (Moscow time). Note the unusual contest start time. **This round will be rated for all the participants whose ratings are lower than 2100.**

You will have to solve 5 problems in 2 hours. One of the problems will be divided into easy and hard version. The round is based on the problems from 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.**

Thanks to everyone who helped to prepare this contest:

- Our coordinators, KAN and isaf27 for good coordination.
- Wind_Eagle, gepardo, VEGAnn, andrew for help in preparing the round.
- programmer228, Vladik for testing the round.
- MikeMirzayanov for making nice systems Codeforces and Polygon.

**The scoring distribution: 500-1000-1500-2000-(2000+1250)**

Good luck to everyone!

**UPD**: Finally the editorial is available.

Wish you all good luck & high rating!

thanks, you too

I laughed at this more than I should've, XD. Newbies are so cute at times.

Hey can you help me, I accidentally submitted code for A problem while thinking about pretests and now the time shows after 2h, but I had solved it before 10 min. It's a huge decrease in rank. Please a help will be great, I just commented after the contest so it can be considered. Please if that submission could be removed and my score revaluated.

Don't be too much greedy with ratings.

Yeah sorry, I didn't mean it but just wanted to know if there was such a correction or not, it didn't feel good to go down this way. Sorry, but isn't it too harsh to get so many downvotes asking for help :(. Anyways won't do this again.

Good luck! I wish every grey become green!

Can you please also wish for green to become cyan?

OK! I wish every green become cyan!

Can you please also wish for every cyan to become blue?)

Ok, i wish every cyan to become blue!!

and how about for every blue to become purple?

I wish Every blue becomes grey. Every cyan becomes grey with lower rating and every green becomes grey with 386 rating .

I wish every P_key25dec become red.

I don't want to go back to cyan!

Thanks, I join this wish.

shrek tomorrowDon't know about greys but i was green and now became grey.

Does it mean that the round has many OI-style problems?

If someone can answer me, thanks in advance.

I think it will have standard codeforces problems, but the ideas of the problems are based from the Belarusian Regional olympiad

Yes, the contest will have a usual Codeforces format. For example, on the Regional Olympiad we have partial scoring for the problems, and there will be no such thing on the round itself.

Finally a good old div2 with 5 problems. Can't wait for it!

Wish everyone good luck

THank you CODEFORCES for round with 5 problems.

Wish everyone good luck!

It's a pity that I can't participate, good luck everyone who can!

Oh, i will solve this problems tomorrow at 9:00 AM(i from belarus):D

Good luck then :)

Two contests on the same day damnn!!

what is the second contest?

One of Codeforces and other of Codechef

Looking forward to participating and performing well.

Every Monogon follower after seeing a recent CF Round Announcement: 'Not gonna lie, I spent the last one hour thinking of a smart comment that would get many upvotes.'

Together we can stop this. Donate by upvoting my comment.

(I may or may not be a Monogon Follower)

Yey,First div2 contest in the new year,Wish everyone good luck!

Memeall Belarusian specialists can do whatever they want

Also div1 pupils.

Why has the Magic tab not disappeared yet ?

it's gone, hope you're happy now :D

Are the problems completely the same as the Belarusian Regional Olympiad or they are just based on them?

Great! The score distribution announced very early!Please, make problem B easy, so newbies can solve it

You will see on the contest whether it's easy or not :) We cannot change the problem, as it is already prepared.

Really hope it's not about bitmasks or dp

B problem rare bitmask or dp

LOL! A was bitmask.

Wish you all good luck & high rating!

Nyaharoo~~

essayforces!Am i the only one who read it as easy forces.

who all like non-theme based contests wherein the problem statements are easy to understand and one dont have to spent extra 10-15 just to grasp the statement fully?

It really took me around 10 mins just to understand the statement of A....due to which this contest became the first div 2 contest in which I solved B faster than A xD

I want to bid for the modern art painting of Problem C.

What was that with problem A man? Almost gave me a heart attack xD

Essay Forces!!

I participated in olympiad, this problem there today, so it's really easy

Also, I hate Jupiter & its moons now XD

Why so strict memory limit for problem C? :(

not able to solve single problem :/

ruined by kids downvoting and crying over it

Participating in contests is a form of practice.

ruined by kids downvoting and crying over it

what do you mean? div2 is rated for newbies to

ruined by kids downvoting and crying over it

You're not enough eligible to advice others, go practice yourself and gain some ratings. And about demotivation, it's your theory that you are explaining to others, maybe for others things doesn't work that way. When i lose rating i feel little happy about it coz in some other contest i'm gonna get a huga +ve delta and also for current contest which i couldn't solve, will give me something new to learn.

Losing ratings is just because you did not perform well enough during the contest and also because of some hard problems in it. That might be a motivation for you to keep practicing and getting better to solve these hard problems to gain more ratings.

And about quitting CP just because you can't solve problems and depressed because you have lost your ratings, I think it didn't really come from ratings, but you didn't keen on CP enough.

in this contest really bad problem statement , thats the main reason :/ hopefully in future contest not like that long essay.

You didn't solve the problem not because of statements. Don't lie to yourself.

We had only 45 questions in a contest, which is far less than the average. Statements were crystally clear.

I hate long statements

thanks to problem A i could now probably go to sleep with 0 submissions

I love pretest 5 on problem C

I love pretest 5 and MLE on problem C.

Wish the memory limit were 512Mb.

Tried really weird things to make my 3D dp ~476Mb solution pass (array of maps and maps of vectors, lol).

This was made on purpose to make you write 2D DP.

Sounds quite complex. The intended solution uses just a 2D array, and the memory limits were adjusted for this solution.

Really enjoyed that, being new to DP, it was fun to optimize from 3D to 2D. Great contest you guys :) Thanks!

Long statements really consumes lot of time.

Yes, but they are more clear and contain more explanations, so you most likely will understand the problem from the first read.

actually they are not more clear the story is not useful, i mean shorter statement is better

Would it be better if we rewrote the statements as

I tried my best to make the rewrite above as short as possible, without omitting essential details.

Exactly, but this applies only if statements actually contain explanations, rather than random planet's moon.

I know that I don't hold the popular opinion here. Still, I want to show that the problem A mostly contained detailed explanations, rather than "random planet's moon". See yourself:

ImageAs you can see, the legend part is no more than a third of the statements, while the other part is actually explanations.

To prove my point about clearer statements further, I want to say that this contest had much smaller amount of questions than other recent rounds I helped setting. The number of questions was even ~1.5 times smaller than my old Codeforces Round from five years ago, where the number of participants was ~3 times smaller than now.

Anyone else ?

problem AThat's why i solved it in 7 minutes xD

Why is the memory limit for C so strict?

ReadAlotForces

Is it just me or contest these days are really hard?

Is there any greedy solution for problem C?

With a simple greedy solution (eliminate a sign which is causing the largest delay at each step), I came till pretest 8.

Can anyone tell what is wrong in this approach? Link to my solution by this approach: https://codeforces.com/contest/1625/submission/142540240

I don't know such solutions. Use dynamic programming.

Greedy is not feasible for problem C.

try this test case:

10 18 3

0 1 3 4 6 7 9 10 11 12

2 3 2 3 2 3 2 3 4 3

the answer is 42, meanwhile greedy approach is likely to give 45.

Take a look at the illustration:

So use dynamic programming, I hope my submission 142583078 may help you.

How to solve D?

Special case: if $$$k = 0$$$, the answer is all indices from $$$1$$$ to $$$n$$$ inclusive. Hereafter we assume $$$k \neq 0$$$.

Partition the values $$$a_i$$$ into groups according to the prefix of bits higher than $$$k$$$'s most significant bit. Observe that you can consider each such group individually, as any two values from two different groups will xor into a prefix that's greater than $$$k$$$'s most significant bit.

Now in each group, we can always choose at least one member, and at most two members, because if we choose more than two, there will be two values that will xor to zero in the position of $$$k$$$'s most significant bit, thereby producing a value less than $$$k$$$.

To find out if we can choose two values from a group, we find the pair with the maximum xor and compare it with $$$k$$$. Finding the maximum xor of two numbers in an array is a well-known problem (but you'd need to modify the algorithm on this page so that you know the indices of the pair) and can be done in $$$O(m \log k)$$$, where $$$m$$$ is the number of values in the group. Therefore the total time taken over all groups will be $$$O(n \log k)$$$.

can u give me an example how to partition such groubs. I think in each group the xor between its elements must make the bits that greater than most significant bit in k equal to zero, right???

Say $$$k = 5 = 101_2$$$, and $$$a = [27, 31, 17, 13] = [\color{red}{11}011_2, \color{red}{11}111_2, \color{red}{10}001_2, \color{red}{1}101_2]$$$. The first and second element would be in one group, the third element would be in another, and the fourth in yet another, according to the bits highlighted in red, which is anything above $$$k$$$'s most significant one-bit.

I get it thanks a lot

I created strings for those bits like "11", "11", "10","01" and stored them in a map so that only unique ones are stored and then printed all the unique strings' corresponding index, but it didn't work, any idea why? 142643010

From what I can tell you are indexing by bits $$$\geq msb$$$ rather than $$$> msb$$$, and I don't see anything recognizable as a max-xor-find there.

142559786 Here's my solution for reference, it's in Rust though.

Damn, I was stuck at that "well known problem". Thanks for the clarification!

Should we also add one number whose MSB is less than k's MSB? And also one number whose MSB is equal to k's MSB?

EDIT: nevermind. I guess these numbers will be in the group with prefix 00..0

Observation is that a minimum value of $$$a_i \oplus a_j$$$ is achieved at some two adjacent numbers after sorting.

After sorting, you can write simple dynamic programming, $$$dp_i$$$ — the length of the longest subsequence that ends with $$$i$$$. Transitions are very simple $$$dp_i = max$$$ {$$$dp[j]+1$$$}, where $$$a_i \oplus a_j \geq k$$$.

It can be optimized using trie.

But why only $$$a_i\oplus a_j\geq k$$$?

It is possible that whatever subsequence $$$dp_j$$$ is storing in has a xor value $$$x$$$ and $$$x\oplus a_i\lt k$$$.

Does this have something related to the optimization using trie?

It is possible that whatever subsequence.No, because of the first observation.That 128MB memory limit on Problem C was tricky.

What's the expected time complexity for problem c ? my O(n*k^2) dp solution was giving tle.

my solution passed with this complexity.

I have $$$O(n^2 \cdot k)$$$ and I passed pretests. $$$500^3$$$ should fit if you don't do expensive operations.

A was easy, It was based on bits and count of which is more 0 or 1, My submission My sub

Man , I am saying Statement is way too complicated for A . BTW I have solved it

I read problem-A for 20 mins..... So so complicated and there are lots of new words for me.

Me reading all problems after 2 hours

Can someone please give me hint for C?

Hint : Try to think of 3 state Dp transition and optimize space complexity

dp[i][j] = minimum time to reach L from ith speedpost (this speedpost is not removed) provided you are allowed to remove j speedposts in the middle.

So.. is dp[i][j] = min (dp[i+1][j], dp[i+2][j-1], dp[i+3][j-2],... dp[i+1+j][0])? That's O(n*k*k) right?

Dynamic Programming. Lets iterate over the stops only. M(i,s) = minimum time to reach ith stop. ans = min {M(n,s')} where n-k <= s' <= n.

why aren't you allowing $$$log(n)^2$$$ solutions to pass in D?

it depends on how efficient your implementation is. I'm pretty sure that my 142503668 is $$$\log^2$$$, and it got AC in 1.6s

Has anyone solved B via Binary search ?

142482987 Here

Why did I pass the code sample locally, but not in codeforces ?

Spoilerproblem B

what ？

change ~~~~~ int n, t[N]; ~~~~~ into ~~~~~ int n, t[N + 1]; ~~~~~

Some hints for D please.

bit trie

can anyone of you please tell in

`problem B`

why it is sufficient to check only two consecutive indexes?because you can use all the elements before the previous one to be your prefix, so at any position it's always better to maximize that number.

Let's say two numbers that are equal have index $$$l1$$$ and $$$l2$$$.

Consider $$$l1$$$<$$$l2$$$ Now we'll see what is the length of the segment, for number at position $$$l2$$$ to be at the $$$l1$$$ position in its segment, segment must start at index $$$l2-l1$$$.

As we have to maximize the length we can take the endpoint of the segment to be $$$n-1$$$.

Now length of the segment becomes $$$(n-1) - (l2-l1) + 1$$$. You can see that if $$$l2$$$ and $$$l1$$$ will not be nearest index of some number then answer will become worse.

I think the below approach is correct.

SpoilerLet $$$t$$$ be the minimum $$$i$$$ such that $$$k$$$ < $$$2^i$$$.

Now, let's divide a number $$$a_{i}$$$ into two parts: prefix and suffix

prefix: $$$a_{i}$$$>>$$$t$$$

suffix: $$$a_{i}$$$%$$$t$$$

Now, it's easy to see that there can only be almost 2 numbers in our result with the same prefix. If possible, find those two numbers for every prefix.

Thank you!

Yes you are right! After seeing your approach, I modify my code and it AC, thank you!

I like problem D, but why $$$k=0$$$...

damn, I don't know how I missed that :(

I hope this is not my mistake

A hint for Problem E2:

SpoilerFor operations of type 1, there're only '.' between the left and right brackets.

(Many of my peers didn't see this and so didn't me. T_T)

Maybe an interesting extension for type 1 queries is to guarantee that $$$s[l+1 \dots r-1]$$$ is an RBS.

SpoilerThen instead of removing a leaf, we may have to delete a node and reattach its children to the parent. Thus we need to update who is the new parent of these children efficiently. If we keep a node's children in ordered sets, we can merge smaller sets into larger ones and have the merging take $$$O(n \log^2 (n))$$$ overall.

I think this should work but I could be wrong.

fun fact: actually expending queries of type 1 to even not guaranteeing the deleted brackets are paired can still be reduced to the problem you said above

Can you solve E1 with MO's algorithm?

Yes, here is my code: 142512011

Could you please explain the approach? How does the add/remove function work for this problem?

Cheater in my room. She his/her code for C: here

The most miserable thing for python users during the contest: Figure

Exactly, I also faced the same issues. in the end, I had to write the code in C++ ( which has the same time complexity and space complexity as of python code) and it got AC.

A very nice contest ruined with long problem statements

Can anyone help me in problem B,wrong answer at pretest 4Basically selecting minimum difference indices of same character than adding count of elemnts before first appearence and after elements of second appearenceans = x[0] + n — x[1];

CodeI didn't get why you want to maintain the size of map entry to 2.

I just collected all the indices and did the max among every two consecutive indices https://codeforces.com/contest/1625/submission/142480649

yeah i did the same after my frnd told me to simply run for all, basically looping for all indices am already selecting two indices of minimum diffrence

This is the issue I guess

Let say indices for a number are 1 3 10 11

U will initially store 1 3 then u will ignore 10 as 3-1 < 10 -3

But u need 10 to take 10 -11

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh, bruh marry me

The idea of Problem D is good. Not only observation but also a specific technique is required to solve it. However, I think it is too hard for Div.2D. Due to the difficulty of this technique, it should be somewhere around Div.2E. Also, the corner case $$$k = 0$$$ is tricky.

The technique required to solve D0/1 Trie

Trie is not needed if you make use of the observation that choosing elements with different msb (most significant bit) are independent and to consider which elements of the same msb should be chosen, we just need to switch off the msb and consider the subproblem recursively.

At the last step, when the msb becomes smaller than or equal the msb of $$$k$$$, we can make use of this to check whether there exist a pair of numbers whose xor exceeds $$$k$$$, and if otherwise we just pick any random number.

To handle groups of elements with MSB <= k, I also used the technique in the GeeksforGeeks article you linked during the contest, but I got TLE. I even tried replacing the

`set<int>`

that they use with an efficient hashmap, but still got TLE. In order to get AC when upsolving, I realized it was faster to use a 0/1 trie to figure out what the maximum XOR of any pair of numbers is. See the`maxXor`

function in this solution.My submission 142523115 was able to pass in 1560ms using

`set<int>`

though.That is very close to the time limit, so my guess is that you have lower constant factor than my solution did, since you are using

`printf`

/`scanf`

instead of`cout`

/`cin`

and also you are using`__builtin_clz`

which I didn't know about before. Thanks for sharing your solution.What is wrong with this logic for D: let i = leftmost set bit of k, right shift each element by i-1. Output the no. of unique elements after the right shifts.

Hint — you can take at most 2 elements from a group of unique elements, not 1. Think about it.

Can anyone, who was able to solve Problem C walk me through the recursive->Memoization->Top-Down/Bottom-Up solution? I am learning DP right now and could not understand the code of other participants. Would really help me a lot...

Hope this helps https://codeforces.com/contest/1625/submission/142530775

Let $$$dp(i,j)$$$ = min time to reach $$$d_i$$$, by removing $$$j$$$ signs. Assuming $$$d_n=l$$$, the answer is minimum of $$$dp(n,j)$$$ over all $$$j<=k$$$.

Recursion:Consider computing $$$dp(i,j)$$$. One candidate for $$$dp(i,j)$$$ is $$$dp(i-1,j) + a_{i-1}*(d_i-d_{i-1})$$$. But it's possible that we removed the $$$(i-1)$$$th sign, so we drove at speed $$$a_{i-2}$$$ from $$$d_{i-2}$$$ all along, so $$$dp(i-2,j-1) + a_{i-2}*(d_i-d_{i-2})$$$ is another candidate. Continuing with this reasoning, and also noting that we could not have removed more than $$$j$$$ signs, we get the recurrence:$$$dp(i,j) = min_{p<=j} dp(i-1-p,j-p)+(d_i-d_{i-1-p})*a_{i-1-p})$$$.

Base case is of course $$$dp(0,*)=0$$$.

pretests were deadly :(

Why is my solution for C giving WA verdict? 142521905

Among all possible

`(ans, prev)`

in`dp[i-1][j-1]`

, there might be such pair that the`ans`

is bigger while the`prev`

is smaller and it is better to transition from that pair.Will you please elaborate more? Will this case arise only when temp1==temp2 has been happened previously?

Your dp[3][1] should be 8(5+3, removing the 2nd), because if we remove the 3rd, the ans will be 1+2*5, which is worse.

And prev[3][1] will be 3 because the 3rd is not removed.

But the best transition is when dp[3][1]=11 and prev[3][1]=2.

Thank you Ji_Kuai, I was also struggling with the same thinking what's wrong in it. Thanks to nyet for replying me with the link of this comment.

But I am curious, like can't we do anything to this approach to get it better. Like at any (i,j) if we store all the possible (ans,prev) pair and using the one to calculate for further states. Tho it's more complex and inefficient. But can we do that? Will that give the right answer?

And making the conclusion that "for calculating some new dp state, the stuff we are using for it must be unique in the sense that that stuff should be only valid stuff for that previous state" is right or wrong?

For the first question. It is obvious that for any pair (ans, prev), (a, b) is always better than (c, b) if a < c (the ans is smaller and prev is equal).

If we use dp[i][j][prv] instead of dp[i][j] and store the smallest ans, it will be correct(But this will get MLE in this problem).

Okay, one more, Just the last one Ji_Kuai, please! Why can't we make

`dp[currBoard][prevBoard]`

storing the minimum time currently on currBoard and the last board taken be prevBoard &&`maxCan[currBoard][prevBoard]`

storing the k value when we were there. Now my claim is that`dp[currBoard][prevBoard]`

will update only when currMaxCan is greater the`maxCan[currBoard][prevBoard]`

or when it is not updated yet!My code link -> 142507920

I think your claim is correct. But it doesn't mean that you can return

`dp[curBoard][prevBoard]`

when`curMaxCan`

is less than`maxCan[curBoard][prevBoard]`

.When the

`curMaxCan`

is less than`maxCan[][]`

, the return value for that should be greater since you can not close as much station as the stored dp value.Ratings updated preliminarily. We will remove cheaters and update the ratings again soon!

If you were stuck at a problem <=C, Here are the video solutions,

D: Stop makingORIGINAL PROBLEMS.I think testers should realize that, because this problem is very classical and is obviously an original problem.

for red guys only :)

I don't think problem setters should avoid classical problems in div2 and div3. Div 2 and 3 are "educational" contest. Although problem D uses classical idea, not everyone (especially experts and below) have seen it before, and they can learn in this contest, which is good for them to study.

The problem is good to use if only you are not able google the problem by the statement online.

I thought Educational Codeforces rounds were created as "educational" contests, not the regular Div. 2 rounds.

Yes, there is educational contest, although the frequency is not much. But since problem D is a master level problem (difficulty rating > 2100), I think it's OK to use it in the div 2 since red and orange users are unrated.

Hello, no, I don't realize that. When I solved this problem, I did not remember anything like that.

Moreover, I liked this problem more than all the others I have solved in the original contest.

In case anyone else is struggling with TLE in problem D with Java, I needed to do both of the following to make it fast enough:

LONG STATEMENTS made bad reading experience.

A good contest anyway.

These are original statements from BelOI.

To be more precise, Russian version of the statements is the original version, and English statements are just translation near to the original. Also, there was partial scoring on the olympiad, but it was removed for Codeforces to allow only full solutions.

Video Editorial for Problem C: Road Optimization

I explain my 2 dimensional dynamic programming solution along with the intuition for why DP is used (and not greedy) and how one can come up with the recurrence.

https://codeforces.com/contest/1625/submission/142524915

Want to know what is wrong with my approachMy approach for C was as follows:

I find the index of all the signs, which when removed

decreasethe total time.Now I just need to choose at max

kof thesecandidatesfor removal.this is where

DPcomes in, I apply it on the preselectedcandidates.IFwhile choosing value of currentk>0than I have 2 option1.)either choose to remove(insert index in a set named removed)

2.)Ignore and move to next candidate

ELSEIgnore and move to next candidatefor base case after completely going through all the candidates , I have a set of chosen sign's indices(i.e signs which should be removed). So considering that these set of removed signs as not being present, I calculate the total time taken and return it.

I memozized this using a dp[501][501] array.

If $$$a=[1,99,3]$$$ and $$$k=2$$$, do you include $$$3$$$ in your

candidates? (Clearly, removing both 99 and 3 is optimal.)YES

Here is slightly harder version of problem D. Source: this comment.

printf("Codeforces Round #765 (Div. 2)");

Are u okay?

bruh imagine waking up at 4am on US west coast to do div2 contest, just to give up after 15' of trying to read problem A.

That's similar to my experience: I woke up at 6:30 AM (normally I wake up at 8 AM or later) on US east and after reading problem A's long statement I started feeling dizzy and then gave up.

You didn't see statements with really long legends :) The problem here is in Russian.

How do you quickly find out whether we should use greedy algorithms or DP for optimization problems like C? The only clue that I found for C is the input size: 500. This might imply that we can use O(n^3) DP. If I don't know this size, I might spend time on thinking about possible greedy solutions.

(By the way, I didn't participate in this contest but I read and solved A, B, and C. I think they are interesting problems.)

Solve dp problems on timus for 3 days straight prior to the contest lol, totally not me

The simple answer is that if you can't prove your greedy solution, you can't be sure it works. In this case it's reasonably easy to construct a counter-example; in other cases it's more difficult. If you can't prove greedy and you can see a DP solution, it's likely to be safer.

Problem E2 is very interesting! I love this problem.

where tutorial? in tourist's stream?

Can problem C use Convex Hull Optimisation to solve it?

Yes, I solved it with convex hull after the contest.

Code : https://codeforces.com/contest/1625/submission/142559648

dp[i][j] is min time required to reach ith city using j road signs. So the transition is like dp[i][j] = min (m<i) dp[m][j-1] + (d[i]-d[m])*a[m]

The transition part can be written as a line like y = dp[m][j-1]-d[m]*a[m] + d[i]*a[m]

The slope is a[m] and the intercept is dp[m][j-1]-d[m]*a[m]. We can pass in d[i] and get the minimum using convex hull trick.

thrilling round :)

Spoilerexcellent round with strong samples and pretests.

I could attempt only 2 problems and now my rating went down ;-;. How2 time management on easy questions. Aaaaaaa I wanna get to green so bad.

Try this

can anyone find problem in my code for problem D i am getting WA on test case 14. I am using bit trie, sorting and segment tree. after sorting if i am at indx i than first i will find the maximum element such that it's xor with current element is greater than or equal to k. Now i used dp since we can select any element which is less than maximum element(mx) so i made a segemnt tree on values of dpi's. dp[i]=max(dp[j]+1) for all j such that a[j]<=mx Then to update value of current element we just query of the prefix from 0 to mxid(index of maximum element) dp[i]=query(0,mxid)+1;

Code

where can i get editorials to this contest

Can anyone guide / help / suggest me on how to reach

expert(1600+) rating at codeforces...... I am currently specalist.Just try to solve 3 problems in div 2 contest.

After the round read the editorial and solve D and E problems.

Keep doing this, then you can nearly always solve at least 3 problems in div2 contests.

You can be blue even purple then.

in my experience, you need to consistently solve ABC problems div.2 in 1 hour

Could someone please explain why greedy wouldn't work for C?

About why greedy is not working on problem C, take a look at my reply: https://codeforces.com/blog/entry/98913?#comment-877347

I hope this will help.

Some hints for people stuck at debugging problems

CandD.anonymouspegion

InputExpected OutputYour OutputCommentThe optimal move is to remove the signs at the index $$$1$$$ and $$$2$$$, resulting in the signs/cost of

The cost would then be $$$(4 - 0)*1 + (5 - 4)*2 = 6$$$.

Dev_Manus

InputExpected OutputYour OutputCommentThe optimal move is to remove the sign at index $$$1$$$, resulting in the signs/cost of

The cost of travel would then be $$$(3 - 0)*1 + (4 - 3)*1 + (5 - 4)*2 = 6$$$.

_Enginor_

InputOne Valid OutputYour OutputCommentSince the pairwise $$$xor$$$ should be at least $$$3$$$, and $$$a[1] \oplus a[3] = 2$$$, you violate the constraints.

@variety-jones Can you please work your magic to show what goes wrong with this submission (failing on test 14). Thanks!

So I'm a personal assistant now? XD.

InputA Valid OutputYour OutputComment$$$10 \oplus 9 = 3$$$, hence an answer definitely exists.

Apologies if it sounded like I took you for granted. You are the savior!

Any hint for Problem D

You can see my solution:-> https://codeforces.com/contest/1625/submission/142628388 I have provided necessary comments to make it easy to understand.

when is the editorial? i think beluorusian regional olympiad has it, because for example in russia, we have editorials right after the contest

Yes, but we have only Russian editorial now.

so u are going to wait for the official English editorial instead of publishing your own?

There is no such thing as "official English editorial" in our Regional Olympiad, so we need to translate it.

The editorial is going to be published today (considering UTC+3). Sorry for long delay.

sorry but it's two days after the contest and there is still no editorial.

coue any one explain how to solve E2, please?

We hope to publish the editorial today (i.e. before 00:00 UTC+3). Sorry for long delay.

Thank you for your effort anyway.

Can't wait to read the editorial

It's available now. Thanks for your patience.

I don't know if the authorities can see it. It is the first time FOR me to play CodeForces, because I handed in the same code with this number and my trumpet and was targeted by the authorities. I hope you can listen to my sincere explanation.

I don't know if the authorities can see it. It is the first time FOR me to play CodeForces, because I handed in the same code with this number and my Tuba and was targeted by the authorities. I hope you can listen to my sincere explanation.

Is it possible to do C in O(n^2)?

Where editorial ? Where banana?

Apparently all their limited time got used up in storytelling about Martians. Priorities, you see.

Why is the editorial so slow?