*Riichi...Tsumo! 2 han 2000 points!*

Hi! Have you ever heard of the game called *Mahjong Soul*? It is a Japanese Mahjong game that is famous for the adorable characters.

We are excited to invite you to take part in Codeforces Round #635, where you can help the characters in trouble. This round will be held on Apr/15/2020 17:35 (Moscow time). Most importantly, it is **rated** for both divisions!

Each division will be given **6 problems** and you will have **2.5 hours** to solve them. An interactive problem may be found in this round. If you are not familiar with interactive problems, you can learn about them here.

The problems were prepared by EternalAlexander, ustze and me Sulfox.

We sincerely thank isaf27 for reviewing and coordinating the round, and MikeMirzayanov for providing such a great contest preparing environment.

Also thanks to the following testers:

ouuan, QAQAutoMaton, PinkRabbitAFO, Dilute for testing at first and polishing the statements.

300iq, HIR180, gisp_zjz, Cekavis, Sad_reacts_only, mmello, firi., StudyingFather, Clever_Jimmy, Polaris_Dane, spookywooky for testing and providing some suggestions.

Good luck to all the participants! Oh, one more thing, you can enjoy *Mahjong Soul* here!

**UPD 1:** The scoring distribution will be:

- Div.1:
**500 — 750 — 1500 — 2250 — (1750 + 1500) — 3250** - Div.2:
**500 — 1000 — 1500 — 1750 — 2500 — 3250**

**UPD 2:** Protips:

- There will be a sticker in each problem statement except 1F. If you are not interested in the story of the characters, you can skip the sentences above the stickers.

**UPD 3:** Congratulations to the winners!

- Div.1

- boboniu (First to solve 1E!)
- maroonrk (First to solve 1F!)
- Golovanov399 (First to solve 1B!)
- Endagorion (First to solve 1D!)
- FizzyDavid
- faebdc
- lzr_010506
- yosupo
- Isonan
- Um_nik

- Div.2

- Bojangles (First to solve 2E!)
- JbopkynyRubkLuxSR
- 01191020csl
- DreamLoIita
- hfyzw
- -45
- soltanmv
- Aquaa
- 18101130I3 (First to solve 2C!)
- DeD_TihoN

**UPD 4:** Thank you all for joining us! Editorial is out!

I've been looking forward to this contest for a long time!

Problems provided by EternalAlexander have never let me down.

Thank you, I'm so glad to hear that!

In such a difficult pandemic, a guy from wuhan is still making contests for us. Thanks writer EternalAlexander

This contest is very great,but it's too late for me. time of Codeforces Round 626 (Div. 1, based on Moscow Open Olympiad in Informatics) and Codeforces Round 625 (Div. 2, based on Technocup 2020 Final Round) is great for me. (I'm a Chinese)

It is also too late for some Russian regions. (I live in UTC+10. I waited for results of system tests till 5 am)

He's the Rick Astley of CodeForces. Never gonna give you up. Never gonna let you down.

>.<GL & HF!

It says there that there will be 2.5 hours to solve 6 problems but on the contest tab it says 2 hours. Which one is correct?

The length on the contest tab will be fixed later.

such a nice round!

Mahjong Soul? Great!

I'm guessing what interesting problems we'll solve. GL & HF !

(づ￣3￣)づ╭❤～

(110 means the phone number of police station，911 in US)

Gravekper can u tell how u added picture in comment section.

![alt](url)

You may have already heard of Mahjong Soul here.

Must I use my e-mail adress to register in Mahjong Soul?

That's the case for the English version. Chinese version is down and Japanese version is too slow. You can try to use Google and Facebook if you like.

Hey,Sulfox.

What is the score of each question?

Forgive me for my rusty English,QAQ

Don't know about other authors, but previous problems by Sulfox were great, looking forward to this round :)

Came here for practicing CP and now end up playing Mahjong Soul :3

Hope you like our problems. GL & HF!

GL & HF. I'm sure you will like the problems lol.

what is GL&HF

Good Luck and Have Fun.

.

.

Hope the problem statements will be as short as the announcement. Thanks.

Hope Strong Pretests .

Riichi & Tsumo with just 2 han 2000 points? Why are you willing to choose Riichi nomi? (?)

btw plz no duliu mahjong problems. (Although I know it's impossible lmao)

Riichi-Tsumo-Yifa-Udora10! yakumann! (How Lucky :D

How did you get Udora 10? That's as unlikely as being snapped by lightning

what is GL & HF???

Good luck and have fun.

Hope it will be a great contest!

No Mahjong Soul over two weeks. :(

I really smiled when I realized that the contest is made by someone from Wuhan , And I hope to see Italians preparing a beautiful round for us very soon <3

EternalAlexander is an intelligent problem provider, hope to enjoy more contests prepared by he together with his team.

All I know about Mahjong is that its terminology is countably infinite.

Don't worry, the problems won't relate to Mahjong basically ;)

Oh godness! Hope the contest will be interesting and not very hard to solve(not AK).

What is AK?

Emmm...It means"All Kill".

If you get all the grades of a contest,you can say that you "AK" the contest.

Thanks. ヾ(^∀^)ﾉ

Wow, another Chinese round!

Chinese round begins at 22:35 in China...

Writers are so considerate for others :)

Not saying this is not OK, as this round is longer than a normal one and prepared by Chinese, why not arrange it to a more convenient time for Chinese while not bringing inconvenience to others? Such as 21:00 in China...I like participating in codeforces round even with the cost of staying up late, but I do hope there are some rounds in a comfortable time for me. And after all, I guess judges also need sleeping.

A lot of rounds at the beginning of the year were at 16:05 UTC or earlier. Even this 14:35 UTC isn't that late in China. Tbh I'd rather have rounds mixed at very inconvenient time and very convenient time for Europe (early morning and evening, e.g. 5:05 UTC and 17:05 UTC) rather than at what's work/schooltime for Europe and late evening for China, like we have here, but it's not like you have it really bad the way it is now.

Sure, I agree with you and I am not commenting to ask for a change but just providing a thought. I remember many Chinese rounds were scheduled to a slightly earlier time.

And I also agree that contests set in work/schooltime are almost the worst thing to be seen...I prefer staying up late...

I think so. This round will end at 1.05 am. It's a little late for me!

Riichi...Tsumo! 2 han 2000 points! what it means?

Both

RiichiandTsumoare kinds ofYakuin mahjong, which are specific combinations of tiles or conditions that yield the value of hands. And2 han 2000 pointsis its score.There is a game called mahjong. These words can be heard when its round ends (and the winning player earns 2000 points).

Only 2000 points is solvable.

Have you mentioned that Before quarantine days there was about 10k participants in Div.2 now it is about 20k?

Will there be an interactive problem in div 2 also?

Although we can't play majsoul in China now, we can take part in a contest of that XD

Use the link provided in the description. It somehow works. Wheeeeeeeeeeeeeeeeee

For that '2 han 2000 points' you just have 20 fu. That's quite a bad hand.

Suddenly realize that this is impossible? cuz 20fu + 2fu (tsumo) is at least 22fu. Or Pinfu + Tsumo can be 20fu, but it should be 3 han if there is Pinfu.

upd : sry, 2 han 30 fu can be 500/1000 when the non-dealer win so it's possible..

[Deleted]

Yes. You cant Richii&Tsumo and get lower than 2000. Only Pinfu&Tsumo is 300/700 (Why Dama ? XD

No. you will get 1300 in 2/20 and its possible. 2/30 is 2000 and when you are rong, it will be 1/40(no Pinfu)

Argh. I'm just a Adept 2 in Majsoul. I'm Garbage

iSulfox

The picture is so cute >.<, i think it's will be a good contest :3

I'm a very simple guy, I see an anime girl, I upvote """D

Japanese Mahjong game-based round? I tried to get 48000pts on this round!

Majsouladmits multiple yakumann :D So we can strive to 288000 in total!大四喜 + 字一色 + 四暗刻単騎 + 天和(or 四槓子) isn't it? (BTW,I like 国士無双(13門)!)

(Sorry for using kanji)

Yup. In

Majsoul`s Old Yaku Room,you can get 7-yakumann. BecauseThree years on stone(?It's called 八連荘 in Japan, achieving 8 straight 和了(done one's hand)

Not this.

There is no

Eight Qinin majsoul :DI meant 石上三年， which is

double riichiwithhaidiOh, I don't know the hand,thanks!(I haven't heard that ever)

BTW Its too hard to get that. Even harder than

Eight Qin.I like these sweet animes with the problems.

[deleted]

I'm a simple man: I see anime — I skip round

I hope it will be a great contest as one of the problem setter Sulfox is a good problem setter. Looking forward to it. CP in quarantine is fun and I am glad that codeforces is good number of contests.

what was the last contest by EternalAlexander??

Oh I didn't make contests on codeforces before but on some Chinese cp platforms.

As a loyal mahjong player (振り込み automaton), I'm really looking forward to this round.

Can Japanese play Japanese Mahjong conveniently?

As a Chinese, If I want to play it,its hard to finding a place which have Japanese Mahjong

Too sad.

I've been curious about where Sooke is these time.

Then I see this!Wow!I really want to take part in this contest!

But I'm not sure if I have enough time.That doesn't matter,lets % Sulfox !

I love how EternalAlexander is from Wuhan, China and has a bat as his pic

That's Azazel, not a bat!

Deleted

More like "sense of pride and accomplishment" is famous for EA.

So just stop saying my avatar is a bat. I really hate this.

Sorry for that.

That's not a bat!

Thanks, I didn't quite catch it when 3 others replied the same thing

From perspective of healthy white heterosexual cisgender patriotic male who doesn't play stupid anime games it looks like a bat.

This is not a bat

That's not a bat!

Jesus I'm sorry I don't recognize Azazel . You guys make me feel like a criminal for not knowing him

I'm sorry that I say it again,because I don't like bat especially at this time.Please forgive me for being impolite

I mean come on, I wasn't being rude. I just thought that was a bat and pointed it out.

If my comment did seem rude to anyone, I'm sorry

.

Who is Azazel? /yiw

A playable character (extreme attack power and agility, but have weak hit points) from the video game series

The Binding of IsaacI hope this will be a good contest

Hope everyone will get a YAKUMAN in this round!

[deleted]

Don't rush my friend, the rating score will be modified one day. Just wait, give them some time.

Ok bro. :'(

Changed!

Yes.see that..thanks both of u.

I am very interested for the problems made by Sulfox.I hope pretests should be strong to avoid hackforces.

Why you hate CF??

His id name is(i_hate_cf) ..so i asked that..but why others are down voting :'(

an announcement without almost copy and paste part......is it even possible to learn this power?

awww, so cute that it'll be 'bout Mahjong Soul characters :3

Have there(in codeforces) more problem about interactive problem without this ( https://codeforces.com/gym/101021/problem/1 ) ??? ..If have..please suggest...I am so poor in interactive problem so need to practice..Thanks in advance. Sorry for my poor english.

You may try it on other platforms also... simply search interactive problems on google and you will get lot of problems to practice.

Thanks bro

You can go to problemset and use interactive tag to find them.

Interactive problems are so boring.. change my mind

They are one of the most interesting on CF

Yeah!

Tell me your mind's password then probable i can change it. I mean nothing can change your mind as long as its locked.

I hope everyone will enjoy this!:)Thanks!

Yet Another Chinese Round?

Yostar round pog Can't wait for Arknights round

I hope I am able to help as many as characters as I can. Sulfox hoping for good and interesting problems. And Thanks for providing Yet Another Chinese Round.

When I load this page, I find this interesting pic flashing but soon I can not see it. After some search I find it again from hidden comment. Why this get so many down vote?

Who would like to play Majsoul before this round?

I wouldn't.

I was graduated from WFLS in 2018 and ... so glad to see awesome OIer from WFLS ...

Why is this contest longer than the usual ones?

Well 6 questions and 2.5 hours probably hinting that the difficulty is more than usual.

Thanks.

I guess the last problem must be data structure.

We will upsolve the Div.2 round live on Youtube after the contest is over: https://youtu.be/FgrB_bvZJ_A

Live Majsoul is better XD

Dude, seriously thank you. That explanation of div2E was awesome. Great content.

Thanks a lot bro, I'll keep them coming!

Contest... End! A FST! B Hacked! -500 rating points! Become Newbie!

Is it not a remarkable coincidence that our question makers are from Wuhan, China?

No,coders have no identity,they are just coders.

I have registered for the contest, if i do not participate in it. Will my ratings get affected?

no, but be responsibility. avoid register contests that u won't participate

I don't know, but I think you'd better not try it. If you have registered in a contest, you can cancel your registeration here, and click the red button after your name.

If you have no submissions, then no.

If you dont submit, your ratings wont be affected. You may even read the question and not submit, you will receive no rating changes.

Hello HNV.

will this round contian a problem like 'give you a set of 14 Mahjong cards,output whether you can ron' XD

if so,that will be really interesting XD

You can't ron because you can only tsumo with 14 mahjong cards

on your hand.so output 'no'.Waiting for a Mahjong problem :)

dude i literally thought there is gonna be a mahjong problem and went to wikipedia to read about it

haha, same

I am a fan of Sooke！

Thanks for the site :D, quite complicated game, hope statements are not complicated as that.

great Sook orz

There will be a hackforces whenever one of the problems maker is Sook.

A nice fact about Sulfox, he has only one blog and couple of comments, his comments are upvoted about 30-50 in average, and his only blog is upvoted 1000 times, also his contributors are more than 120, and its truly too much for one blog and 5 comments.

So what does that mean?, it actually means people loves him and he is a good problemsetter(probable).

Also hope everyone will learn something from this contest and enjoy it(iam not hoping high rate for everyone as it's impossible :])

"There will be a sticker in each problem statement except 1F. If you are not interested in the story of the characters, you can skip the sentences above the stickers."

Not all heroes wear capes.

a

Another day in Div1.

a

Yeah we can see that.

My solution to C gives 9 in 3rd test case on my PC and gfg IDE but giving 7 when submitting on Codeforces

hadn't populated the array with 0.

I just downvoted your contest.

FAQ What does this mean? The amount of contribution (points) on your comment and codeforces account has decreased by one.

Why did you do this? There are several reasons I may deem a comment to be unworthy of positive contribution. These include, but are not limited to:

Rudeness towards me,

Making bad contest,

Being a weeb,

Sarcasm not correctly flagged with italic font.

Am I banned from the Codeforces? No — not yet. But you should refrain from making contests like this in the future. Otherwise I will be forced to issue an additional downvote, which may put your commenting and posting privileges in jeopardy.

I don't believe my comment deserved a downvote. Can you un-downvote it? Sure, mistakes happen. But only in exceedingly rare circumstances will I undo a downvote. If you would like to issue an appeal, shoot me a private message explaining what I got wrong. I tend to respond to Codeforces PMs within several minutes. Do note, however, that over 99.9% of downvote appeals are rejected, and yours is likely no exception.

How can I prevent this from happening in the future? Accept the downvote and move on. But learn from this mistake: your behavior will not be tolerated on CodeForces.com. I will continue to issue downvotes until you improve your conduct. Remember: Codeforces is privilege, not a right.

Being a weeb is the worst reason to downvote a contest

Everytime I take part in contest with 2.5 hours, I become aimless in the last 0.5 hours, just refreshing the dashboard again and again, watching others solve some problems and my rank get lower and lower :(((

Lol I forgot this contest was 2.5 hours so I stopped after 2 hours and just checked again and saw it oops

@fried-chicken hahaha ! True !

Not gonna lie I ignored everything before stickers but couldn't ignore those cool stickers! Thanks for strong pretest in Div2D. Nice round.

I guess i've wasted about minutes watching that things =P

https://codeforces.com/contest/1337/submission/76887917

Can anyone help me find out why the last testcase wasn't printed? (locally it does)

Why are you make complicated ? just print b , c , c .All done.

I haven't read your full solution, but I can see you are using

`pow`

, which is a floating point function, subject to floating point precision errors (especially since you are then checking equality of two floating point numbers with`==`

).Always avoid using floating point numbers unless absolutely necessary.

So what is the recommended method for checking say a number is equal to a^k.

Compute $$$a^k$$$ (without using the

`pow`

function) and check if they are equal :)For powers of 2 in particular (as in the example above), you can compute $$$2^k$$$ by noting that its binary representation is 1 followed by $$$k$$$ zeroes. As such, you can compute it by simply

`1 << k`

.How to solve Div 2 C and D?

Div 2 C:**** Considering that all paths end at the root(node 1) you may want to put the other end of the path as far as possible from the root. Therefore, starting with the highest depth node of the tree, place the nodes to get the highest possible answer using the following equation:`answer[node] = depth[node] - sizeofsubtree[node].`

The size-of-subtree part is because adding an industrial city in the node will decrease the number of tourism cities of all its children. Using a priority queue with a key of the equation above, choose the highest k.

During the contest, I wrote the following algorithm.

Create a max priority queue of leaf node vertices as keys and value as dist[v] — size_subtree[v](distance of v from root and size of subtree rooted at v).

while k > 0, pop from v(we push v to final answer) max heap and push v's parent to max heap(again with same values as dist[p] — size_subtree[p]).

This gave me wrong answer on pretest 6. However, when I just created a vector of these vertices and sorted it with key dist[v] — subtree[v], it got accepted. Here is my submission for priority queue method https://codeforces.com/contest/1337/submission/76925284

CodeYour code can visit a single node multiple times, mark the node as visited when pushing it in the priority queue.

Edit: forgot to mention that you should just add the depth[node] — subtree[node] to the answer with no need for another DFS which I believe isn't working properly..

I agree the other dfs code is redundant but it is logically sound. I didn't realise that my code might have a node being visited mutliple times if it has multiple children. Thanks for your help!

for D, you can sort and use binary search to find the best tuple of three numbers.

my submission

I could not solve C :(

same here!

That's ternary search not binary search.

actually it's binary search only, just check r->b, r->g or r->b, b->g choose min one , and repeat it for each one as 1st element (b, r, g) and (g, r, b) and that's enough.

my sol — 76867720

I'm referring to codersanjeev solution. It's just modified ternary search to compress the two mid points so that they are adjacent. That doesn't make it magically "binary search".

Div 2C : Here's what I did — Perform simple dfs to find the depth of each node(d[i]) and to find the number of nodes lying below each node(n[i]). Now, choose k nodes with largest value of d[i] — n[i] and mark them as industries. Then, simply find the answer with another dfs.

C: compute the size of the subtree of each vertex as well as its depth (distance from root). Now for each vertex compute

`depth[i] - subtree_size[i] + 1`

. Sort the values in non-increasing order. The sum of the first k values is the answer.Proof: notice that when you add a vertex, you gain its depth, but lose 1 from each vertex in its subtree (apart from the one in question). So, it's optimal to add vertices so that vertex

`i`

is only added when all subtree of`i`

except`i`

itself has been added already, and the described order is optimal.The +1 doesn't matter, because it's constant.

It doesn't matter in terms of which vertices we choose, but it matters when we compute the answer.

Hi, tell me if I sort according to subtree size and take smallest subtree size node first is that correct?

It is level of node minus subtree size.

Hi, if I understood you correctly, it isn't. Consider a tree consisting of two paths: one big and one small. Your algorithm will take the vertices from two paths in turns, but it might be optimal to take several vertices from the longer path first before even starting the second.

Oh so are we subtracting depth[i] because of taking into account the happiness of newly added industry? And subtree_size[i] is acting as loss due to newly added industry?

Yes, depth is what's added, and subtree_size — 1 is the combined loss.

https://codeforces.com/contest/1337/submission/76866312

My code failed on test 25 and I am not understanding what can be the issue.

Nice solution. Short and easy.

Thank you very much for the proof! I finally understood it.

Refer here

For D, I had iterated over red and select green by using lower_bound over red and then select blue with lower_bound of (r+g)/2. I make all combinations possible.

here is link to code.

i also want to solve in your solving way..but in the middle feel that code will be big.then stop to write..

I was not able to submit for C question because of this. When I am about to submit solution, contest was just over.

If you make the code modular, its much easier to read / write the code. [submission:76880283]

Div2 C This could be helpful i guess See Here

So I think 1337C - Linova and Kingdom is one of the nice problems. We need to find where to start the pathes in a way that the number of unused vertices on the paths is maximized.

It is obvious that we should choose the cities most far away from the root. But after these are used, which ones continue? It turns out that the number of happiness a city contributes is equal to the level of that city minus the number of children of that city. This is because we add children before the parents, so if we choose a city with children, that envoy will increase happines by the level of the city, but all envoys from child cities will get one happiness less.

To implement we do a DFS collecting the levels and number of children of all cities, and sort that list of numbers, then building the sum of the $$$k$$$ biggest ones.

but I got tle on pretest 7 I implemented the exact same solution!

i did same just used a comparator for sorting. gives me a TLE!! Anyone who can tell me how comparator works ! or why my solution was TLE? thanks in advance!

https://codeforces.com/blog/entry/75996?#comment-605586 see this

I came up with this idea too, and as i thought i implemented it correctly, but my solution failed test #7, can you help me to find the problem in the code? 76890614

It seems you are counting the direct children only, but you should count all nodes in the subtree, ie, children of children, too.

Yeah, looks like, thank you

I implemented this solution i get wrong answer on test 9

Try changing to long long

I got 5 wrong answers in this problem, all because i used set instead of vector, also it took about 10-15 minutes from me to try changing set to vector, in Div. 1 contests its very important if you lose 50 points and i actually lost 200 points for truly no reason :(, problems were very nice but sadly i got too many wrong answers in the contest.

`number of happiness a city contributes is equal to the level of that city minus the number of children of that city.`

. Shouldn't it be muliplied instead of minus or did i miss something?The children (and grandchildren, recursive) get choosen before any parent. So, at the time a parent is choosen it contributes its pathlen to root, and makes the contribution of all children (and grandchildren, recursive) each one one less.

One less because the parent is in the path of each child.

what was pretest 5 in C I thought for 2 hours but couldnt get it!

Did you simply apply DFS to find depth and were returning the maximum K values?

I think n=k

`1 <= k < n`

in the statements 1337C - Linova and Kingdom11 7

1 2

2 3

2 4

3 5

5 7

5 8

7 10

7 11

4 6

6 9

try this, you should get ouput as 22

thanks I got 21

I am getting 22 as output. Even I got W.A on pretest 5.

me too

can u give the labels for 7 nodes that you are making industry for answer to be 22.

labels are: 4, 6, 9, 7, 8, 10, 11

this might work I think waiting for sys tests to over! linksolution

If you are using set, then try using vector, and then sort it, i got 5 wrong answers in the same verdict.

did it using arrays

AT the end did it with vector the same thing but still got WA

Great Round! Solved A and B pretty quickly. Started doing D and ended up wasting an hour. I thought it was pretty easy to do D, and I believe it is, but I didn't get the right idea going. Managed to solve C in the last 30 mins (PS: this is the first graph question I've solved in a contest). Cheers! Looking forward to participating in more contests by Sooke and the other problemsetters.

Pretest 5 of C?

I was selecting cities based on greater depth and if the depth is the same, I choose the one which has fewer cities in its subtree. Is this approach wrong? Submission

I think n=k

Oh my bad sorry

actually, k<n is a condition of the problem, read statements carefully

I was getting pretest 5 wrong when I was applying only DFS to get all the depth of each nodes and getting the maximum K values, which was a wrong approach. Don't know about you though.

Me too. You can have a look at my pretest AC solution. I used DFS too at start and failed pretest 5. Then implemented a BFS solution. I think you can solve it with DFS, just that I couldn't think of how I'd implement it number of leaf nodes is lesser than K.

CodeYes, I did it with DFS only. The thing was I was just using depth of the node instead of depth — number of children in subtree

I dont understand your approach :<

Can you hint me ?

Hey! Sorry for the late reply. After dinner, I wasn't in the mood to type and explain my solution and I slept for a few hours. Here's my explanation: (note that this is similar to most solutions shared using DFS. I just thought BFS approach would be a lot cleaner compared to my DFS solution which failed pretest 5, well because I didn't accomodate for all cases and implemented it wrongly).

Firstly, this is what each important variable means.

$$$G$$$: The adjacency list. Takes $$$2*(N-1)$$$ memory.

$$$L$$$: Stores which nodes are present at some depth

CurDepthwhen built. Takes $$$N$$$ memory.$$$D$$$: Stores the depth of each node. Takes $$$N$$$ memory.

$$$C$$$: Stores the number of children for every node when processing the nodes. Later, I subtract the depth $$$D [i]$$$ of the

i'thnode from each $$$C [i]$$$. Takes $$$N$$$ memory.$$$V$$$: Stores if a node has been visited or not in bfs. Takes $$$N$$$ bits of memory.

Now, the relevant part of the explanation. Once I have the tree built, I process the nodes of the tree layer by layer and store the depth of each node starting from 0 (or 1, if you like 1-based indexing more). This is simple implementation with BFS and saves you from recursion of DFS. After finding the depth of all nodes, I also happen to know the max depth (which is what

CurDepthholds after processing). While processing, I also store all nodes in $$$L$$$ that belong to the same depth level. Then, I iterate over all depths starting from the maximum depth using $$$i$$$ as the loop variable, and for each node $$$j$$$ at depth $$$i$$$, which is what $$$L$$$ tracks, I iterate thorugh all its neighbours $$$k$$$ and check which ones are it children. Basically, the three nested loops you see are finding the number of children of each node. Finding the number of children doesn't suffice because there might be a case when the number of leaves are lesser than K. So, the optimal strategy of picking nodes for industries is to greedily choose those nodes that are the deepest as well as have the least number of children. Or in other words, choose tourism cities to be those that have the highest number of children and have the least depth. This is why there's another loop subtracting the depth from number of children of each node. You need to remember to sort this data for having all best cities ending up towards the end (or beginning, doesn't matter as long as you know what you need to do). Once, all of this is done, you can calculate the answer by summing up the $$$NumberOfChildren - Depth$$$ for the best $$$N-K$$$ nodes.I'm not sure about the time complexity but I think it is $$$O ((N^2)*MaxDepth)$$$, which I thought would be hackable or will fail Systests, but it passed to my surprise. I might even be wrong about the time complexity, don't really know because I'm new to this stuff.

Good luck for future contests!

another way getting depth

bfsYes it is wrong, you should calculate $$$x = depth - subtree$$$ and sort vertices by x and greedily choose the first $$$k$$$ vertices with maximum $$$x$$$ and increase $$$ans$$$ by $$$x$$$.

Yeah got my mistake. Thank you :)

Hey, I was sorting only on the basis of the subtreesize since the depth of a particular level is the same. Can you please tell me where i am going wrong. https://codeforces.com/contest/1337/submission/76880891

You did the same mistake as mine.

Can you explain, I am going level by level, so for a particular level sorting only on the basis of subtree size won't suffice?

Think about what would happen if the number of leaves is lesser than K.

I have handled that in my code I guess, https://codeforces.com/contest/1337/submission/76880891 Can you find where i was making the mistake?

Sometimes its not like a level will fully fill then the last level will, for example the test case below 11 8 1 2 1 3 2 4 2 5 4 6 4 7 6 8 6 9 8 10 8 11

Consider you are at level 4 and has subtree size of 3 and another node is at level 3 and subtree size of 1.

It would be optimal to take the node at level-3 rather than at level-4.

This was the same mistake in my code.

12 6

1 2

1 3

1 4

1 5

1 6

2 7

3 8

4 9

7 10

10 11

8 12

ans should be 13

This will fail .

First let`s establish a relation

Consider a city

Xwith depthD+1and sub-tree size ( excluding itself )S(containing all industries ).Now if we choose to place industry at this city X then the number of industries will be increased hence

S=S+ 1 and the potential tourism cities on path fromXto 1 will be nowD.So total answer for this case =

D* (S+ 1 )Now if we choose to place tourism at this city X then the number of industries will be same and the potential tourism cities on path from

Xto 1 will be nowD + 1and subtree size wont increase.So total answer for this case = (

D+1) *SNow to decide if we should place industry at city

Xwe should have a profit thus first case should yield higher value than second .So profit =

D* (S+ 1 ) — (D+1) *Sprofit =

D—SThis is the profit we obtain if we decide to place industry at city

XNow to we have to maximize the total profit so we have to choose cities with highest profit , thus we can calculate

DandSfor every city and sort them according toD—SNow coming at your logic :

Consider 2 cities A , B .

Depth(A) = 10

Subtree_size(A) = 10

Profit(A) = 0

Depth(B) = 9

Subtree_size(B) = 8

Profit(B) = 1

Ideally we should choose the city B as it has highest profit but your algorithm will choose A.

Hope this helps.

Now choosing city A will imply that the

Wow, thats beautifully explained. The part where you derived D — S formula.

Thanks a lot Lord_Pilo!

F seems hard for div 2. Nobody in div.2 solved it :))

I found a way to find all a_i>=2 in the beginning, but didn't know how to use the straight count to figure out if a_i = 0 or a_i = 1

Hopefully correct, but I found that for given index u < n,

if

`a[u - 2] > 0, a[u - 1] > 0`

, then if you increase`a[u]`

by 1, the number of increment for consecutive tile is x =`a[u - 2] * a[u - 1] + a[u - 1] * a[u + 1] + a[u + 1] * a[u + 2]`

, which is`a[u - 2] * a[u - 1]`

iff`a[u + 1] = 0`

, then you can judge between 0 and 1 somehowI think it's just cause there were a bunch of problems, so there wasn't enough time to finish E. I don't think it was actually that hard, in div1 E just looked like a really good way to get points so maybe that's why there aren't that many solves. I finished it about a minute after the contest ended (maybe) but it took a while to think of a good strategy. Like others have mentioned, if you query each from 1 to n you can determine the exact value by checking the difference of the triplet counts, except you can't distinguish 0 and 1.

To solve this, query the first thing twice, then query from 2 to n-1. Since you query 1 twice, you can then figure out it's true value. Suppose that when you query the ith stone, you know the number of stones in i-1 and i-2th position. Call the number of stones in the i-2,i-1,i,i+1,i+2 position a,b,c,d,e respectively, for the straight count we have c(ab+bd+de). If we just look at the difference before and after adding the stone, this is ab+ d(b+e). Since we know b and a, and b>0 (since we've already queried previous stones at least once), we can determine if d=0. Continuing this we can get all the stones up till the last. For the last query (on the second to last stone), e=0 so we can easily determine the number of stones in that pile.

It seems hard for div.1, too. Only few of them solved the problem.

Is Div1 E1 just finding a basis and then just going through all linear combinations of the basis (or, if basis is too big, go through the complement and subtract these)?

That certainly works, I also found this solution. Too bad it took too long so I didn't have time to write it (rip LGM ;_;)

Yeah I also did try to write something in the last few minutes but WA#4 ...

Do you know if this approach can be extended somehow for E2 ?

I didn't realise that we can do the thing with the complement so I only have E1.

My approach for E1 is meet in the middle with FWHT which works in $$$O(n*m + m*(\frac{m}{2})*2^\frac{m}{2})$$$ and if we use the complement idea this can be easily extended to E2.

How do you use meet-in-the-middle? I did spend some time thinking about this (+FWHT) but in the end, I couldn't really bring these two together.

So first we build the basis in $$$O(n*m)$$$ and after that we split the bits into two groups — the first $$$m-h$$$ and the last $$$h$$$.

Well we will simply do brute force on the first $$$m-h$$$ bits and this way we will generate some pairs of a fixed number $$$k$$$ of set bits that will be in the range $$$[2^{h}, 2^m)$$$ and also some number $$$x < 2^h$$$.

Similarly we do brute force on the lower $$$h$$$ bits and generate some numbers which are again less than $$$2^h$$$. Clearly this will be one of the polynomials for which we will do FWHT.

So here comes the FWHT — we will generate another $$$m$$$ different polynomials for the different $$$k$$$ values mentioned above. Now we simply convolute these newly generated polynomials with the on for the lower $$$h$$$ bits and the final popcount will be $$$k + popcnt(x)$$$, where $$$x$$$ is the index in the convolution.

If we choose $$$h=m/2$$$ we will get the above-mentioned complexity but smth like $$$h=16$$$ was a better choice for E1, as the heavy part of the solution was the convolution after the brute force.

To extend this, we simply use that idea with the complement, but the details seem to be quite annoying.

What is FWHT?

fast Walsh–Hadamard transform, something likes fft.

I don't really get it. What exactly is a complement of a basis?

Ehm you are probably right, it is not the exact term. I meant a basis of the complement of a vector subspace.

Ok, yeah. But why can it be smaller than the basis itself? Like I think that in the first example both basis and basis of a complement have size 3, right?

The thing is, we can handle the two cases differently. If basis itself is at most 20 numbers, then we can almost brute-force.

If the basis is larger than 20 numbers, since m is at most 35 we know that the complement is at most 15 numbers. We can now "brute-force" on that.

Yes, I get what to do if the size of basis is small (including the subtraction thing for complement).

I think I don't understand what a basis of a complement is. So like we want something that can represent all the values original basis can't?

Actually, the more I think about it, the more unsure I am how I intended it all to work together, so my idea is probably wrong (at least the complement part), because a vector space complement is not the set complement (and we want the set complement if we're to subtract, maybe?)

Actually it is a cool idea about the complement of the basis. However, I don't know how to find the complement of basis.

During the contest, I created an easier idea. Let's just honestly find all numbers we can generate by basis. It will cost $$$2^m$$$. However, let's use some DP here. What we can do.

Iterate over first $$$i$$$-bit, and use or not use the vector on the $$$i$$$-th position in Gauss. What we know, that after this all bits on position 1-$$$i$$$ will not change. So we can just store the number of 1s. And for the rest, we need to store all possible masks. Dp [i][cnt][mask] it is easy to show that we can solve it in O($$$ m ^ 2 * 2^{m/2})$$$. If I am not mistaken it will even O($$$ m * 2^{m/2})$$$ However it not passed E2, probably because of big constant. I used hashmap.

76888381

Can you explain please what is the complement of the basis?

After reading your comment, I thought of a slightly different idea.

Generate all the masks using the first $$$m / 2$$$ vectors, and then do the dp using the other ones, with the least significant $$$m / 2$$$ bits of each mask. That way, the dp will not need hashmaps, and the total complexity will be $$$O(m^2 * 2^{m/2})$$$.

77216975

How to solve div2C? It seemed really simple, but I missed the observation and messed up bigtime. Silly me.

Refer here

How to solve problem C? I was thinking to calculate level and no of the child below it for every node and then placing the developer's city in the tree from the bottom and also taking consider no of the child below it and iterating it in increasing order of children.

When you make a city an industry, all of the cities inside its

subtreechange, not just its children. You need to sort your array by depth-subtree size essentially and iterate over that.I have made 3 submissions of problem D, and all of them passed the pretests. Can anybody tell me why am I getting penalty of +2 on D?

There is penalty on resubmission also.

Only the last submission which passes the pretests is considered hence penalty for resubmission.

Thanks, I didn't know that.

For D, if x and z are fixed, then y = (x+z)/2 is optimal. WLOG let x<=y<=z. You can use 3 pointers after sorting the 3 arrays to go through them and find values of x,y,z, where |y-(x+z)/2| is minimal and z is the closest integer greater than x. But there are 6 possible permutations for the order of x,y,z. Simply iterate through all 6 possible array combinations (r,g,b, r,b,g...,g,b,r) and take the minimum of all of these.

I don't know if this works though, I spent the last 20 minutes coding it up and didn't have time to finish because it wasn't working with longs (probably due to me writing integer instead of long somewhere), but it worked for all the other sample test cases.

I did the same thing and got Accepted.

Is there any way of coding the 3! permutations apart from the one in which you have to write 3 loops?

YESSame here, feels bad :(

It can be proved using calculus by finding the zero of a derivative. Note that the derivative at

`0`

is when you have a slope of`0`

(i.e. horizontal line) and this is where the minimum (or maximum but we know maximum doesn't exist) occurs. When you fix two numbers`x`

and`z`

(but can also work for any two pair), the resulting function is`y^2 - (x +z)y + C`

. where`C`

is every term that becomes constant when you fix`x`

and`z`

. The derivative of this function is`2y - (x + z)`

, not that`x+z`

is constant. Now, finding zero of function means doing this:`2y - (x + z) = 0`

->`y = (x + z) / 2`

. So`y`

must be the minimum.Because we are not working in the set of real numbers, we probably need to find the two closest integers (from left and right) if`(x + z) / 2`

doesn't exist.But I think it can also be solved using two nested ternary search if you don't know this.

Code: 76902407

I think calculus is a bit overkill for this. If you know about quadratics, you will known that the tip of the parabola $$$ax^2 + bx + c$$$ is $$$(\frac{-b}{2a}, \frac{-\Delta}{4a})$$$, (where $$$\Delta^2 = b^2 + 4ac$$$, of course).

Maybe... I guess it depends on what you are familiar with. 'Cause personally I forgot about the quadratic minimum derivation.

Video Editorial for 1A/2C