Hi Codeforces!

Dio, Keshi, Tet, alireza_kaviani, Davoth, AliShahali1382 and I are delighted to invite you to participate in Codeforces Round 722 (Div. 1) and Codeforces Round 722 (Div. 2), which will be held at May/24/2021 17:35 (Moscow time). Each division will have **6 problems** and **2 hours and 15 minutes** to solve them.

The curse has finally been lifted! We are proud to announce that antontrygubO_o didn't reject even a single task from the Div. 1 part!

Huge thanks to the following people:

antontrygubO_o for being an outstanding coordinator; This round wouldn't have happened if it weren't for him.

Our ever-growing army of testers gamegame, Amoo_Safar, Atreus, dorijanlendvaj, aarr, Monogon, EnEm, N.N_2004, AaParsa, _Seyed37_, Roham, Clix, Narut, SinaSahabi, sinamhdv, arvindr9, shokouf, munghatekartik, Ali_Tavakoli, vodacbaoan, Shahraaz, -Suto-, coderz189, kassutta, the_borono, saarang, iANikzad, Etherite and -this-is-obd- for testing the round and providing invaluable feedback.

MikeMirzayanov for the amazing platforms Codeforces and Polygon!

Last but not least, to all the participants of this round!

For the sake of having short statements, this round does not have a theme. The characters featured in the statements are: Soroush (Tet), Sifid (Davoth), Parsa (Dio), Nima (N.N_2004), AaParsa (AaParsa), Haj Davood (davooddkareshki), Kavi (alireza_kaviani), Keshi (Keshi), Mashtali (AliShahali1382) and AmShZ (AmShZ).

Please read all of the problems and their notes, enjoy your time and solve as many as you can! Good luck have fun to everyone!

~~The scoring distributions will be announced later.~~

**UPD:** Here are the scoring distributions:

Div. 1:

**750 1000 1750 2000 2750 3000**Div. 2:

**500 1250 1750 2000 2750 3000**

**UPD:** Editorial is out!

Being a setter, I fully understand the pain of other setters while writing statements.

Thank you for your contribution.

"Tet" is Vietnamese traditional new year holiday. :D See you, I think about Tet :D

davooddkareshki orz

Mehrdad_Sohrabi orz

omg saarang tester orz

what is orz ?

imagine O as the head of the man bowing , r as his hands , z as the rest of his body . Its used for giving ovation

woahhhh. thanks man

as a teacher and as a friend, I am sure we will see a great contest :)

<3

<3

<3

<3

<3

<3

<3

Finally wait for Anton round is over, would be a great round for sure!!!

its a trap

Not in this contest :(

But its_Atrap

As a tester, the problems are very nice and interesting! I recommend everybody to participate!

As a tester, I agree with vodacbaoan!

orz <3

As a participant, I agree with you guys:)

How to be a tester?

You should either be friends with setters or have a history of being a setter.

It's too hard for me

Sir, How can I be a tester? :P

vodacbaoan is the official wife of Monogon. Upvote if u want them to marry.

so you have downvote hell. Enjoy

Do u have a crush on vodac ?? xD

Yes

Woow !! Finally again an iranian round :))

I'm so interested to participate in my compatriots' round !

Hope to reach pupil in this round, Thanks AmShZ and other dear authors.

Many times I see people commenting it's an Indian round, it's a Chinese round.. etc and now you are saying it's an Iranian round. What difference does it make? I don't think there is some pattern in the problem set associated with the region or something like that.

We respect to our compatriots ! :)

It's just that participants get excited because they have the same nationality as the authors.

Not sure if it's true for Iran, but I guess that some countries have a few problem-setting-eligible users and those users are well-known among the people of that country. Wouldn't it excite you to see an acquaintance of yours set a round?

Agree with you

I definitely would, but expressing it in that way looks inappropriate. Mentioning country looks like drawing borders to me.

Maybe I'm over reacting. Btw thanks for the contest.

Nah I feel the exact same way as you. But then again, different people have different opinions and beliefs, so I dont mind it.

Chinese round: Maths, implementaion,data structure

Russian round: constructive, observational

Iranian rounds: balanced perfectly

Indeed it was balanced perfectly.

i will add something about your comment after the contest

Hello, what is it?

ok as i said there are some few things you can find out when the contest is made by irainian:

1) there are always some graph problem and most of the graph problems are either shortest path or tree problmes

2) there are lots of dp problems

3) there is no complicated greedy problems because most of irainians hates greedy

4) they also hate geometry so you 99% can be sure there is no geometry

5) there are usually hard combinatorics & DS problems

these are the things that ive learned because im either student / classmate / friend with them : )

"There is no complicated greedy problems because most of Irainians hate greedy" =(

I dont know who likes greedy, atleast no one I know.

i like greedy

I dont know you. :/

i still like greedy

I love greedy.

as an author, give me contribution!

as a no one , sik tir :)))

:(

as a Participant, I have given you contribution

orz

antontrygubO_o orz.

Its a really nice round with great and beautiful problems. I hope everybody can enjoy this round.great effort from all of the setters to finaly make this round.good luck everybody and have fun. (Dont read kostia's comments :))

As a tester, the problems are interesting and the statements are short. Hope you enjoy the round :)

Such boys !

round after another 2 days:( can't wait

me seeing the contest announcement

Frankly, you shouldn't count it as a ghaazzz round. And this way we can still expect a real ghaazzz round.

agree.we can't count it as a ghaaazzz round,but as a ghaazzz round.

As a tester, I am way too late to post an as a tester comment.

(insert all the normal tester comments about how good the round was here)

as a tester, orz

The best adhoc cooontest")

Whenever We open Codeforces and see a contest announcement on home page, Inside us

So that you can cheat ??

I hope Question C has no nonsense

Good Luck Everyone !!

Thank you boii.

Length of the round is set to be 2 hours on the contests page, please fix it. AmShZ

May I ask what was the reasoning of having exactly 2 hours 15 minutes, and not 2 hours, and not even 2.5 hours?

Whenever you give a 2hr contest and submit one of those messy problems in the last minute and get WA on pretest 2 and then wish you had only 10-15 minutes more. These are those 15 minutes.

Benq is going to participate in this game, I think he will play a very good result in this game.

Finally another Div.1!!!

But what a pity that the contest is held on Monday! I think many people like me can't participate in. sad to see that

dblark

Google Kickstart(which on Sunday) may be the reason.Auto comment: topic has been updated by Tet (previous revision, new revision, compare).The blog mentions the length of the contest as

2 hours and 15 minutes, while the contest page is showing2:00.We don't have editing rights yet, it'll be fixed soon.

Ever since I became candidate master(more than one month ago). It's first Div. 1 contest Hope to have good contest Thank you very much Haj Davood :)

Which one won the game Div 1?1) Benq 2) Tourist 3) RadewooshGuess?None of the them :P

Which one won the game Div 1?1) Benq 2) Tourist 3) RadewooshGuess?

He used a very high level technique called "guessing"

Why codeforces is not taking contest frequently ? The wait after every contest is lasting soooo long :(

Though codeforces is kinda better, but you can also participate in other online contests. Such as atcoder, codechef and even yukicoder. These websites combined you'll have at least one contest every other day(maybe everyday).

Ya , but the happiness of increase in rating on codeforces is of different level. :)

I agree. The same is true for atcoder, but still not as much as codeforces. Besides, competing is great itself.

Don't know why, but I feel good about this contest. Feel like I'm going to get to almost reach pupil.

antontrygubO_o orz.

antontrygubO_o orz

Indeed.

good luck, have fun :)

Dio If I surpass you, will you give that username?

IT WAS ME DIOi dont think so...

The problems are always interesting <3

Iran is an actually a good country :D

I agree with you

and if i say orz i get negative contribution "(

Hi !

When will the scoring distributions be announced ?

We are still discussing it :(

I'll post them the instant we agree on something ...

B part has a score of 1250 instead of 1000 or 750, I hope I become a specialist.

Best of luck! :)

You have solved 393 problems in 3 months!! Great work bro. All the best!

Thanks a lot :)

how does the scoring distribution of the problem matter? What does it actually mean?

It gives you an idea about the level of the problem. Though it is not always accurate but most of the times it is. For example B problem might be a bit hard today as compared to other div-2's B problem but again it might be easy too. You can never tell you can only guess.

Div 2 3rd problem is of same difficulty of Div1 3rd problem wow :(

what algorithm did u use to find that out

Seeing the score distribution, contest will be hard

time to skip after seeing the score distribution.

you should never skip

good luck to all!

Ghaaz = Goose(in Persian)

rainboy orz

Only way i can solve Div 2 C is, if it is given as captcha to view anime online.

how to solve C?:)

dp on trees, and only $$$l_{i}$$$ and $$$r_{i}$$$ are important in the range.

I did exactly the same, still I'm getting TLE!! :(

can solve by dfs. just handle parent and child relation and add every child value with parent then answer is root value.

SpoilerFor those who solved Div2C / Div1A, could you guide why DFS is needed (as opposed to BFS)?

Any traversal technique works ig. I solved using dfs though xD

Actually I had used BFS, which was incorrect. I think I now understand why. DFS is a must here since the max value of the root node (you can pick any node as root) depends on the max values of its children, which in turn depend on their children, etc.

So once it's clear that you have to pick either L or R for each vertex (and no value in between), then it's a matter of solving subproblems first — recursion with DP does that effectively.

Yeah, it doesn't I overlooked it. Basically it's like we should find the answer for all nodes in the last level and then the previous level and so on. So we can somehow store answers for higher levels first and then for lower levels using BFS. But we have to store the nodes which are at a particular level and it is implementation heavy.

So I think BFS is something like top-down traversal so it's better to use DFS as it's a bottom-up traversal in such cases.

Is O(N^3) meant to pass div1D?

Why is a simple DFS getting TLE on Div2 C??

No its not

https://codeforces.com/contest/1529/submission/117252283

Use fast IO in C++

I think the large input gives you TLE. Try using ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); this at the begining.

You mean to say the reason it's TLE has got nothing to do with the logic?? It was due to slow I/O?? I'll cry the whole night today :(

Yeah but all of us got TLE on some problem before this and learned from that to use this "trick". At the end of the day you learned something new so is a good day!

Because it has O($$$n^2$$$) time complexity

It happened to me too. Simple O(n) recursive dp. TLE on test 4. I have absolutely no idea why.

What happens if you AC a problem twice in pretests? I resubmitted one of the problems due to runtime and got -50 penalty but if the first solution passes systests, will I still get penalty?

I don't think the first submission even gets run on system tests, so you will still get the penalty.

I like meethi lassi and chole bhature.

Oh ok, sad

Only the last submission will be judged against systests. (The first one is given a verdict of 'Skipped' after systests.) You will still get penalty.

Is D on OEIS? And how to D it?

You need this one https://oeis.org/A051950

I used dp. categorized the length of pair (L) into two parts:

for each n, define f(L) as:

if L <= n, only those n % (L-1) == 0 will contribute. I count the frequency of each prime factor of n (p_i) and do a product of (p_i + 1).

if L > n, it will be a fully covering case, and the inner part will be dp[L-n-1]. Those can be calculated use prefix sum.

so each dp[n] = sum_{L=2}^{2n} f(L).

I'm not sure if this is good for system test, and sorry for the brevity of the description

Very hard to demonstrate without paper.

[...] [...]

Suppose those two arches have the same length. Then we know nothing can belong inside of both of them, as each subarch isn't inside of the other (place inside of the left, not inside of the right), etc. And because the subarch must be smaller, then neither condition can be satisfied.

Now, we know that must be a be a thick arch. Like for example, an arch of "thickness" 2 cross weaves 2 archs, if you know what I mean. And now you can place subarchs inside, this gives a relation dp[i] = sum of all j <i dp[j]. Since there is only one unique way to place this thick arch (otherwise it because multiple archs of varying sizes, but you just want that 1 thick arch). This thick arch can be anywhere from 1 thickness to the entire rest of the length so that's why we go for all j<i.

The tricky thing is to handle the no-archs case, where all lengths are equal and you can place them wherever. The answer turns out to be the number of divisors of the number, though I have no idea why (I saw the potential pattern but no proof as to why it works), but proof by AC I guess. So add on the number of divisors at the end of every precalculation DP. And because it's prefix sums just keep a running prefix to add on every time.

Holy fk I passed D pretests in the last seconds! Shoot ...

Hints on problem C, please.

I used DP actually. Try to solve the problem (maximum sum) for each subtree. And for each node u, you can calculate what the result will be if you use the left or the right.

The transition for one node using its left value will be

dp[u][L] = max(|l[u] — l[v]| + dp[v][L], |l[u] — r[v]| + dp[v][R])

Let me know if you need more spoilers

I'll try. Thanks!

Solved it! Also learned a new way to implement DFS from your submission. Thanks a ton. Edit: Will this implementation work on any graph or is it exclusive for trees?

Hi there

I learned this from https://cp-algorithms.com/graph/depth-first-search.html :)

I think the one I submitted only works for trees since it only checks whether the parent has been visited or not. You may an extra

`visited`

flag to work with general graph.I am not familiar with the rules. I have submitted a correct solution at 0:05. But at around 0:35, I submitted another correct solution. I wonder why my score for that question is based on the finishing time of the latter one? (0:35 instead if 0:05)

The second one is scored.

I understand the rule right now but I wanna know the rationale behind this rule. Why don't it count the first correct submission that able to pass the system testing.

If you want the first to count, then there is no need to submit the second. So, you decide.

It's just because I didn't know this rule before this contest. I would definitely not do it again in the future contest. Btw I resubmitted it just because the second one has a higher chance to pass the system testing.

Then isn't that a valid a reason to not consider your first submission ? As you were not confident that the first one would pass the final test cases and modified it to pass the system cases.

I think its kinda synonymous to getting a WA and then debugging and resubmitting the correct solution .

In case of multiple correct submissions (for same question), only latest one is considered. All previous submissions get skipped and don't appear in system testing.

Does D need some special DS or math technique?

DP and number theory is enough.

My approach on C Code

Can anyone tell me what is wrong with it.

I have taken 4 critical numbers for each vertex

1) li

2) ri

3) (li+ri)/2

4) (li+ri+1)/2

The best result for each of them in dfs is returned in a array in the same order .

In the end when recursive calls finish we output max of all 4

****

EDIT NOW WORKINGWE need only left mini and Right maxiDfs Based Solution of C Parsa's Humongous Tree without DP based on the fact that during path traversal on a tree from root , A particular node is reached only once.

We return a pair {a,b} from every node.

A has the best solution from that node to all nodes below it when left_mini is assigned to that node.

B has the best solution from that node to all nodes below it when right_maxi is assigned to that node.

We store best possible values of the current node from all children node i.e. one with maxi on all possible calls and other with mini and then return it.

In the end when recursive calls end print the maximum of the pair of values returned from the root.

Solution

D < C :((

No,D>C.

It took me more than 1 hour to solve D, whereas less than 10 mins to solve C...

B is not nice problem, also misplaced.

B>D>C

I guess you also thought that subsequence must be continuous.

No, I think I understood the promblem correctly. But there seems to be some more or less obvious observation needed that is invisible to me.

From other submission it seems that we can use all negative numbers (ok, understood), and then all positives up to the first bigger than the smallest diff so far.

But why, why cannot be there ohter solution? Or did I get it wrong?

Have you considered the case when any element <=0 is repeating twice?

This has caused me 5 wa and 40 minutes.

Note that the difference between two positive numbers is strictly smaller than the larger of the two numbers, so any solution can have at most one positive number. On the other hand, we can take as many non-positive numbers as we like. So the question really only is about whether we can take a positive number in addition to the non-positive ones, which is possible iff a positive number at most as large as any difference between the non-positive numbers exists.

"...so any solution can have at most one positive number."

Ok, thanks, that is the missing observation. What a stupid problem :/

Indeed, it was stupid.

I found B to be hard to understand, not from the language, but from the definition of the promblem itself.

I had to read the definition of a strange seq at least a dozen times, and actually now I am hardly able to tell from head (just did read it again some seconds ago).

So, basically that problem is "somehow get that wiered definition into your head, then you will quickly see the solution." I am ok with a complecated, or strange definition if it is a complecated problem.

But a very simple problem with a complecated definition is a bad problem.

You can see that there can be at most one positive integer and all others are 0/negative and their minimum abs difference must be greater than the positive number so we choose least positive number. If that number is still greater than minimum abs difference we remove that number. Since we are taking all nonpositive integers there can't be another solution. If we remove some negative integer to increase difference, ans will be less than number of nonpositive integers.

Problem Div2B was constructive at best, it felt like you cater to the observation required and it should work.

I am also curious about other approaches as well :)

How did you solve B? Tried for two hours to Frankenstein it with all of the possible cases I could think of, using different sizes of negative, zero and one positive number.

You can prove that there can be at most 1 positive number. Then it's easy.

Take all negative + zero numbers or take the smallest positive number also. For the second choice check whether it is possible.

i tried with 5 case and take the maximum.

1. all zero and all negative.

2. only one zero and only one positive

3. only one positive

4. all negative and one positive (for taking positive and negative at a time check minimum of positive >= max difference of any 2 negative numbers)

5. all negative and one zero and one positive (And taking all pos, neg and zero checked minimum of abs(negative value) >= min positive value and must checked 4 condition also)

And taking all pos, neg and zero checked minimum of highest negative value >= min pos value and

Very easy solution:

SPOILERSLet's notice that the order of elements doesn't matter. Sort the array. Then in a loop calculate the least difference and check — is it possible to add a new element?

Used deque for interval minimum with correct complexity in div1D but TLE, unfortunate.

Was your code $$$\mathcal{O}(nm)$$$? Because even $$$\mathcal{O}(nm \log n)$$$ passes in less than a second: 117224037

117252128

mango_lassi I think it is O(N^3) or O(NM) if you like, I will take a look again.

Looks like $$$\mathcal{O}(N^3)$$$, so I guess the constants are just too high.

Kinda sad if they were trying to cut $$$\mathcal{O}(nm \log n)$$$ and this happens :(

It doesn't, consider this code : Link

I guess I am just unfortunate.

I think you can avoid re-computation using deque by doing it before only and storing it?

Yep you are right. But you wrote in the smart way (avoid recomputing) and still took almost two seconds, which suggests that the deque solution probably has a large constant factor. Considering that most people's submissions ran in 1~2s, I think there is a good reason that I avoid unnecessary deque next time.

Still, thank you very much for your suggestions.

mango_lassi

117254437 Didn't use deque and simply passed, but still used more than one second though.

Wrong submission? That's a solution to div2A

Oh sorry, this one 117255437

I will take a look at your solution... less than one second is interesting.

such nice problems! good job :)

I found that my $$$D$$$ is on the edge of the TL much latter in the contest but decided against improving it(I hope it passes), how to calculate number of divisors for each [1,2n] fast, I did $$$O(sieve*logn)$$$?

You can do it by iterating for each number from 1...n by its multiples.

it's very similar to sieve. This has complexity NlogN

Yeah, very stupid of me being unable to find such simple algorithm given that I had used this before more than once

Can you explain the basic idea of how to solve D.

@dhruv7888 It can be solve by DP. Ex, N = 3: dp[3] = dp[2] + dp[1] + dp[0] + number of diviors of 3

Because when you know all the good pairs of N = 2, to form good pair with N = 3, you can add a pair bound all pairs with N = 2, similarly add 2 outside pair to N = 1 ...

Does anyone solve problem D with DP formula, dp[i] = prefix[i] + number of Divisors of i? I got TLE, could you share the way to count the number of divisors?

If you have factorisation in form $$$n = \prod p_i^{a_i}$$$ then number of divisors is equal to $$$\prod (a_i + 1)$$$.

Factorisation can be done pretty quickly by precomputing all primes, or just using linear sieve.

Thank you, the factorisation is really good idea, I have leant new thing. What the time complexity of the factorisation? Does Linear sieve mean that iterating for each number from 1...n by its multiples? And it will take O(NlogN)?

https://cp-algorithms.com/algebra/prime-sieve-linear.html

Either $$$O(primes < \sqrt{n})$$$ or O(prime factors of n) depending on what you use. Both should pass pretty comfortably.

Yes. I used dp[i] = pref[i — 1] + number of Divisors of i.

I never thought 6 months ago that I'll be able to even maintain a lower specialist. But alas my dream come true and If all my 4 solution passes, I'll be an expert.

My B failed (-_-)

but congo on solving C and D :)

Solve C 10minutes, B 2hours and failed :(

How to solve C?

Observe that foreach vertex it is allways optimal to use either l[i] or r[i], never some value in between.

Root the tree somewhere, do a dfs. Result of the dfs is foreach vertex the max value we can get if we used l[i] or r[i] for that vertex.

To find the two values for the current vertex, find the values for all adj vertex, then try both values and record the max value possible.

can u explain more?

Can you give feedback in a way that I have sufficient information to write an answer fitting your needs?

Just another day using javaVery nice problems! Thank you for the contest!

At first, I tried some greedy solution of C, and it looks scary...Then I glanced at D, it looks scary too...So I waste about 1.5 hours trying to solve E...

This story tells us that don't solve problems out of order if you're not sure you can solve all of them :(

I decided to abandon python after this contest. Using python in codeforces is totally a disaster and will ruin your rating. Many problems, especially graph problems, like problem C in this contest, using python will cause you TLE, TLE, TLE...... while others languages can pass easily with same algorithm. It takes me 5 minutes to think of problem C, 15 minutes writing and compile, and an hour to submit, TLE, optimize my code, submit, TLE, optimize my code .....It's very very very frustrating to make tens of attempts but didn't pass, and see my current ranking drop from 300 to over 2000. Goodbye my rating, and goodbye python.

I am getting TLE in C.Time complexity of my code is O(2*(v+e)) but I facing TLE, please help me. 117248045Edited- This issued was solved, instead of cin, I used scanf. It's working properly now.

Use map<> instead of unordered_map

still getting TLE

Actually it's caused by read large amount of data by "cin", in other words, you get TLE because you spend too much time on I/O

It means I should use scanf instead of cin, right?

Yes.

Thanks MonkeyKing, got AC

removed

Not knowing C++ grammar very well, but:

is O(1), while:

gives CE on my computer.

You are right, arrays work like pointers, nothing is copied.

So it might be undefined behaviour caused by negative value of parameter p which is used as array index.

Can you help me to findout why I am getting TLE ?

deleted

I hate it how cf will timeout even with the correct time complexity. I gave up the problem after 2 submissions knowing the time complexity is correct for my solution and I just need to define my variables globally

I gave a google interview and the interviewer asked me not to declare variables globally but pass it by reference

Defining your variables globally is not the issue; you TLE because your IO is too slow. Adding cin.tie(0)->sync_with_stdio(0); to your code gets AC (scanf/printf would probably also pass).

:( Can anyone show me some tips for solving A and B super fast?

observe observe observe.

I'm such an idiot. I solved div#2 C but was using int instead of long long. Thankfully it struck me just 10 minutes before the contest ended.

And it wasn't the first time this happened to me. :(

define int long long gang

sometimes it will cost a TLE :)))

Lol man, exactly what happened to me. First time I solved DP problem in a contest, first time I solved C in division 2. Lesson learned man, I changed my template to replace ints with long longs by default now. I do not want this to happen again, feels really bad

Lol same, first time dp, first time problem C!

Solved div 2 B with ternary search .Can anyone explain a simpler solution for it..

the simplest solution I found is take all -ve and all 0, then try to take as many +ve as possible

You can take atmost one positive

Simple solution is to take all the non-negative numbers. After that, it is pretty easy to observe that we can take almost one positive number. So start by finding the minimum difference in the non-negative elements already selected and try to pick a positive number greater than or equal to the minimum difference. I observed this after solving a few cases on paper.

How did people solve C div1? The solution boils down to running a DFS on the first graph, adding/removing vertices from the second graph and counting number of nodes from the second graph which currently have no descendants. What is the good data structure for this? I used some preorder segment tree with rollbacks, but there might be an easier solution...

I used a binary indexed tree to check if a given vertex has any marked descendants, and a link cut tree to find the deepest marked ancestor of a given vertex. 117214561

I used just a set. If you run a euler tour on the second tree and map each vertex to a segment, it's equivalent to finding the longest subsequence of vertices on the first tree such that none of their segments intersect, and the subsequence is part of a path from the root to some vertex in the first tree.

Basically you store a set of all the segments that you take from the root to a parent of a vertex; when scanning the vertex in, check if a segment intersects it in the set. If it does, remove that (because it's never optimal to take a larger over smaller segment if both intersect, and if two segments intersect, one must completely overlap the other since it's from a euler tree). Then, add the segment belonging to the current vertex in the set. The answer is the maximum size of the set at any time.

The only data structure I used was a set. If you record the in and out times using DFS in the second tree, we have each node in the first tree is labeled with an interval $$$[in, out]$$$. We want to consider all intervals on a path from the root to a leaf, and try to find the largest set of nonintersecting intervals. But since the $$$[in, out]$$$ intervals either don't intersect or contain each other, we can simply use a set to do this.

Submission

When is rating normally distributed?

Is there a proof of correctness for Div2 problem C ?

Considering the right or left value on each node will yield the maximum value.

Say you choose some value

`a`

such that`l < a < r`

, and you've already chosen some value`b`

for another node. If`a ≥ b`

, then adding 1 to`a`

will increase`|a - b|`

. Otherwise, subtracting 1 will increase it. You can keep adding/subtracting until you get to`l`

or`r`

, and get a max value.Consider $$$v[i]=l[i]+1$$$ to be a better solution than $$$v[i]=l[i]$$$ Then this is because there are more adj vertex with $$$v[adj]<v[i]$$$ than there are with $$$v[adj]>v[i]$$$

If this is so, then $$$v[i]=max(l[i],r[i])=r[i]$$$ is true, by induction.

I did not pass problem C because I used int instead of long long. Lesson learned. That would be the first time I solved problem C :(

In problem D, when I used vectors for adj lists it TLEd and when I used 2D array for weights, it ACed. This was not fair T-T T-T. It costed me "Grandmaster" today.

I think vector is not as fast as array for n > 1e5 :)

[Deleted]

A non recommended approach for Div 2 problem $$$D$$$

Write bruteforce solution and generate first few terms of the sequence, here they are $$$1, 3, 6, 13, 25, 52, 102, 206, 411, 823,...$$$

As this can't be found on OEIS, make an observation that $$$dp(i) = 2 \times dp(i - 1) + c_i$$$

Search for sequence $$$c$$$ : $$$1, 0, 1, -1, 2, -2, 2, -1, 1,...$$$

which is A051950, so $$$c_i = \tau(i) - \tau(i - 1)$$$, where $$$\tau(i)$$$ is the number of divisors of $$$i$$$.

AAAAaAAaaaAAaaaaa I didn't submit D1D because I fucked up one line of my dijkstra and an "i" was supposed to be a "j", and last round I FST'd because a "<=" was supposed to be a "<"

why am I so cursed

Feedback on the contest:

A: Clean nice easy (and a bit standard) problem.

B: I am not a fan of counting problems where there is fundamentally zero computer science involved.

C: Cool problem. The two crucial observations (finding a polynomial solution, and then finding a pseudo-linear solution) where equally hard for me. The difficulty of the problem was really on the spot.

D: Cool problem. The fact that, in the end, this is just Dijkstra is very nice.

E: Once again, I am not a fan of counting problems where there is fundamentally zero computer science involved (I could not get AC during the contest, so maybe my solution is flawed and thus my comment does not make sense).

Overall I really enjoyed taking part in the contest since the two problems C, D were cool! Thanks to the authors, nice contest!

C is a cool problem, but for me, D is just a very straight-forward problem as it is just a very simple shortest-path problem.

Nah, you are correct, E is just a counting problem where zero computer science is involved.

Not sure if that is the case for F, but really, 3 counting problems in a single round?

thats fine, did you notice there were 4 graph problems and 3 of which were trees!

Are you counting E as a graph problem? I wouldn't.

C was a nice problem though :)

anton round orz

To not keep you waiting, the ratings updated preliminarily. In a few hours/days, I will remove cheaters and update the ratings again!

Thanks !

thanks mike

thanks mike

Weak test cases on problem div2 B, solutions are getting Wrong answer on:

1

3

-2 0 3

Come on! Let me become purple it's just a single point! :"( My curve does have limit in 1899!!!. Wish to become candidate for the first time after removing cheaters :)

wish to see you in red babe ツ

Can anyone please explain how the Problem D(DIV-2) boils down to just finding prefix sum of number of divisors?

Foreach group of i pairs we can build "packs" of size k if k divides i. So, that number contributes to the solution. Also, we can pair the 2,4,6,... outermost points in combination with the previously calculated result for the i-2, i-4, i-6,... inner points (see third picture of examples). So we need to add up these prefix sums.

117249412

Got it thanks:)

Motivation

Is there a way to solve Div2C using BFS instead of DFS? I could find the relation to find the maximum sum to reach any node from some initial node say 1, but got stuck after this since I was using BFS and could not think of a way to find the total maximum sum.

the approach for this problem is bottom-up, which means that you need to solve DP for the leaf nodes, then accumulate to the root node

I think it's more convenient and makes more sense, and DFS is often used for DP-on-tree problems

You should do a bottom-up dp which can be done with either dfs or bfs, but dfs is easier to implement.

During the contest, I made this observation for D — the range of distances for each city is $$$[minC, minC + n - 1]$$$ where $$$minC$$$ is the minimum traveling time of a cannon in that city; to remove the $$$log_n$$$ from the Dijkstra complexity.

I thought that was pretty neat until I read other people's codes and realized that we could just run a simple $$$O(n^2)$$$ Dijkstra...

For the D problem, what should be the starting point to think in order to solve the problem?

Read Problem And Go To Editorial :)

Why is this (https://codeforces.com/contest/1528/submission/117194884) solution so much slower than this solution (https://codeforces.com/contest/1528/submission/117196442). The only difference is that in one case I used vector with assign() (it should have linear complexity I think) and in the other case I used a plain array.

EDIT: nvm it's o(n²), I'm blind

You are assigning lr vector inside a loop so overall its N^2 complexity

UMM no since its the sum of values of n, which is fixed

117259248 Ur code works with moving the assign out of the loop

Oh yeah I'm blind... Thanks

good contest !

To the setters, or to setters in general:

Just out of curiosity, do you test standard solutions on languages other than C++? I didn't compete today because I was busy, so there's no bitterness over rating here, but I'm curious as to why a 1 second time limit was necessary for Div 1A/2C (for example). A DFS solution should be perfectly adequate, but it will not pass in pypy without some trickery (e.g. coercing to floats, flattening arrays, etc). Clearly an N^2 solution would not pass in 2 seconds even in C++, so why the need to make it so difficult for a standard language like pypy? To my mind if you implement a clean DFS solution, it should have the chance to pass without all the extra little hacks. I know using C++ is quicker and always will be, it just seems like 1 second versus 2 is unnecessary here, and in many other cases.

For reference, here is an example of a DFS solution that fails at the 1 second mark

Honestly even an badly optimized C++ O(n) solution doesn't go through, once I used vectors with assign and once plain arrays and one passed but the other did not: https://codeforces.com/contest/1528/submission/117194884 https://codeforces.com/contest/1528/submission/117196442

EDIT: NVM stupid mistake

Yikes. That's unlucky.

nvm, I just made a stupid mistake (actually squared complexity lol)

Hi, Can someone explain me why we add the number of divisors of i to dp[i] in Problem C of Div 2 ? Thanks in advance .

Number of divisors will be the count of ways to make i segments of equal length. Where length of each segment would be divisor of i.

Thanks ayush2hack.

One of the best contest I ever wrote, almost got D at the end. Thanks Iran. Thanks to the setters.I hope to see more rounds from you all!

In problem B what is the answer for this test case 3 1 0 1 I think the answer is 3

You can have at most one positive element in a strange sequence.

let a and b be positive integers, then max(a , b) > |a — b|, thus [a , b] is not strange.

in this input |1-0| = 1, |0-1| = 1, |1-1| = 0 and the max element is 1..

No, as you've mentioned |1-1| = 0 which is smaller than the maximum element which is equal to 1 so it's not strange ...

Can you download a whole test case on Codeforces? In div1 C the 18th result in the second test is wrong so I have no way to see the input for it.

Found the mistake, I always make a stupid mistake when making larger programs ;((((((

Auto comment: topic has been updated by AliShahali1382 (previous revision, new revision, compare).Which type of round is this? Is it global round or something else?