Hello, Codeforces!

We (Denisov, Karavaev1101, perekopskiy) are glad to introduce to you Codeforces Round #660 (Div. 2), which will happen on Jul/30/2020 17:35 (Moscow time). **The round will be rated for the participants with rating lower than 2100**, although higher rated users are more than welcome to take part out of competition.

Huge thanks to those who helped make this round possible:

- adedalic for amazing coordinating and reviewing the round, as well as helping with many different things.
- Karavaev1101, perekopskiy for creating problems and preparation.
- dorijanlendvaj, Tlatoani, BOGDAN_, gachiBASS, pikmike, antontrygubO_o, Osama_Alkhodairy, growup974, errorgorn, ilyasiv, Tathagat_shah, ermakov22 for testing and providing invaluable feedback.
- MikeMirzayanov for Codeforces and Polygon platforms.

There will be **5 problems** and **2 hours** to solve them.

We really hope you enjoy our first contest!

**UPD:** Here is the score distribution:

**750—1000—1500—2000—2750**

**UPD:** Editorial is published!

**UPD: Congratulations to the winners!**

Div. 1:

Div. 2:

!

translation ...

"Round #660 July 30th 2020. Everyone should participate!"

ok

MikeMirzayanov

Cheating caught in today's contest https://codeforces.com/contest/1388/submission/88511065 https://codeforces.com/contest/1388/submission/88510232

They have same codes only function names and variables are cleverly changed.

And the second guy deliberately submitted late when he was sure he will get 3 AC's.

Dont know why people show such teamwork in CF just to increase their ratings.

Their other submissions from the contest are also same.

Shame.

There should be new category of CF user — Plagiarism Police ....XD

With increase of participation in recent days, this was bound to happen, we could just wish to stop it.

if(count(red color) >= 5) {Contest go pew pew over the brain}

Фанаты папича оценят этот раунд

И Капитана Флинта ))

Солью задачи, оплата киви/сбербанк, за пробником в лс

Солью адресс человека выше, пишите в лс

Я уже виллу в Монако купил на вырученные деньги, поменялась прописочка(

Не прошло и 5 минут после оплаты, а задачи уже на почте! Всем рекомендую, отличный продавец

You just want contribution

go awaygo away !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Удачи с вашим первым раундом!

The contest with 5 problems are terrible but nice.

Wow... 2 contests back to back... I am entering heaven. Thank You Codeforces. I was really sad because there was no contest for the past three days.

All the Best Everyone.

And you haven't take part in any of them

I think this looks much better (the presence of many colors in the contest announcement). I hope this will help and the contest will be balanced this time.

I went for upvoting but accidentally did the opposite, now I'm not able to revert. It feels I committed a sin for which there is no redemption.

You were right this contest was much more balanced. Also I finally could solve div2B for once.

Five grandmasters as testers... hahahahhahahaha what a great opportunity for another -100 rating drop

There's many different testers from many different ratings this time, so I think this comp won't be as hard as previous ones...?

everything is fine but i am afraid from the name after pikmike.

I give contests in 3 months approximately. Prepare D and E questions. I had 1600+ rating earlier. Now I have 1500+ rating.

I don't give contests in 3 months now.

i am not able to understand what you are trying to convey,may you clarify more

Can you give your english tutor's contact number ....

Yeah I re-read my statement and it didn't make any fucking sense. Then I couldn't think of anything less embarrassing so :/

TL;DR: The last round's B fucked me. And this time, 'twas a brain fuck.

5 questions means another DIV 1.5 contest.

next round after 11 days :(

Back to back DIV 2 contests, perfect for my rating to enter hell. ;--;

2 rounds back to back. And the next round (#661) is 10 days away. Wow. Sure 10 days will be useful for gaining back the trust in self.

they have started making difficult contests to ease the system

modern problems require modern solutions

luv_yo_name

luv_is_all_i_want

Thank you CodeForces.Love this site very much.....

5 red testers ===>>> great pretests and less heartbreak at system tests.

7 testers of lower levels ===>>> balanced round.

===>>>

5 red testers && 7 testers of lower levels ===>>> balanced round with great pretests and less heartbreak at system tests.

I hope you're right.

I was right ^_^

I miss the days when we discussed if we want more Div3 or more Div4 contests :/

Why aren't we having anymore Div 3 Rounds?

I need div 4 contest to recover from last 2 div 2 rounds.

lol Turns out in next Div3 most likely I will be rated.

what if ur rating increases , "we dont do that here "

hope to give an official submission this time . Left round 657 and 659 cuz can't do anything bout that xD

There will be 5 problems and 2 hours to solve them.Surely the difficulty of contest will be high. After 2 back to back unrated contests due to long queue codeforces has come up with brilliant idea.

Less questions --> higher the difficulty --> less submissions--> no queue --> no unrated round --> No questioning on CODEFORCES PLATFORM.Forget about Division-3 Rounds Codeforces is trying to make even Division 2, a new

DIVISION 1.5.I wonder if that's true

This contest was easier than #657 (Div. 2). The difficulty of the problem A is significantly lower.

wait , no div3 or div4 ?

Hoping for a good contest for me so that i can get out from the bad time i am spending right now.

Hope for a good and balanced round. Testers from specialists to grandmasters provide hope for low to high rated participants for being round of their type. Hoping for good positive in this round.

()

I wonder why does it take time to reveal the scoring distribution ?

In order to unlock the perfect score distribution, the authors must complete a series of quests. They range from solving rejected problems from the previous round to fixing bugs in Codeforces API, and it ends with the most challenging of them all: debug geometry code.

Thats a lot of work ! Delay Accepted :)

Hello problem setters, please try to balance the round. I am suffering from continuous imbalanced rounds. Have mercy on me.

Now, we have score 750 for problem A. Let's wait for another challenging disaster.

750 points for A? Solid A I guess

Удачки с первым контестом, жду с нетерпением чтобы начать писать!

I am getting a runtime error with the message "Exit code is -1073741819" again and again for educational round 92, problem B. I tried my best but does not understand its causes. Please tell me what could be reason for this error. my submission is https://codeforces.com/contest/1389/submission/88348466

For case k=3, your vector v will be empty.

Here, z=-1 and hence in step

you are accessing v[-1]

I am quite concerned about the score distribution of this contest. Is it too difficult for members with a rating below 1500?

And over 1500, I think it's okay

Every div. 2 is too difficult for below 1500.

I agree with you, but that was before the rating point was changed, because now, some coders with a rating below 1500 can successfully complete the contest div.2

I don't know if it's the clone accounts of coders with rating points above 2500 or really a new coder

New coders do have a true rating. Dont judge simply on the basis of displayed/fake rating.

Check here how to find true rating https://codeforces.com/blog/entry/77890

https://codeforces.com/profile/team_scarlet true rating is 2200 and he will participate in div2 again. :/

I participated in a div3 contest that increased my true rating from 2038 to 2195 :/

Does anyone know why atcoder is not having ABC for last couple of weeks?

There is an ABC this sunday.

ABC means ?

AtCoder Beginner Contest

That last sunday was abc, they just replaced the name for some reason.

Why this score distribution? A starts from 750. Seems to be somewhat difficult. Anyways, surely gonna enjoy the problems :D

Yes, but unfortunately, there won't be any rounds for the next week. I think if MikeMirzayanov add a conditional div3 in 4 days, it will be perfect :)

Testers kindly comment on the kind of the problems. I hope there is something for a newbie like me to solve. This is my first rated contest and I hope to solve at least one problem today. Nevertheless, I am going to enjoy them :) Thank you , Codeforces!!

Is it like people aren’t interested in this round? By seeing only 484 upvotes which is less than the normal upvotes for a Div2 round. Can anybody please share the reason for the same ?

Check the number of registered participants, that would be a better assessment of interest of people.

So what according to you is the number of upvotes for ?

Okay got your point :)

How is it decided that an

incorrect submissionwill result in -50 (in the score) or -10 minutes? Or is it just up to the creators?-50 points, as the problems each have points, if they didn't like in educational round or div3 then -10 minutes

Ohh, so educational rounds have this -10 minutes thing. Great Thanks Cheers!!!

From my knowledge they are different rules. Usually educational rounds and div3 rounds are held on extended ICPC rules, which result -10 minutes. Others result in -50 in score.

Thanks, Got It!

As far as I know, Educational rounds and Div.3 rounds are not score based, i.e., in those rounds only the number of solved problems matter. Since those rounds do not involve score, hence they have Time penalty for incorrect submission. On the other hand, Div.1 and 2 are scores based, i.e. each problem has a score associated to it, hence they have Score penalty for incorrect submission.

Thanks for the explanation man

we are friends now! hello, friend! [user:Mike!Mirzayanov,2020-07-30]

too op :-(

All the best friends!!

Do you have editorial for this contest?

Could codeforces consider weekend contest? I have conflicts on the weekdays (unless it's before 6 AM PST or after 5 PM PST). I'm pretty sure other people have more time on weekends too.

Как же смешно читать условия задач, если знаеш рофлы про Пукича и его детство. Спасибо тем, кто писал условия. RoflanSpasibo

Не за что )

Speedforces!

Not being able to solve a problem doesn't make the round bad (or shit as you said).

At least on Topcoder or AtCoder you can leave your submission up there until hacked, here you just get a demanding test case. These problems are getting harder to solve, even on the easiest levels. :(

I submitted my solution of problem A at 0:15. And after submitted my solution to problem B I tried Problem C but couldn't do it.. Then I was feeling kind of bored so I resubmitted my Problem A and B but in Kotlin as I am learning this new language from the last couple of days.. As soon as I submitted it, it added time penalty and my rank decreased.. Why did this happen ?? MikeMirzayanov please can u help ..

Just read the rules. Actually, you explicitly confirmed that you read them while registration on the round.

You cannot submit a problem again if it has been accepted

I did the same thing yesterday.. But it didn't add penalty.. Please check my solution of yesterday's educational round

This is because it is an educational round, and the rules are different. Read the blogpost: link

Thanks MikeMirzayanov hakuna_potato .. I will be careful next time

Who is uncle bogdan I wish I am as talented as him

Well even if the contest ended up being speedforces but I still really enjoyed thinking about C,D & E. They seemed really interesting and I can't wait for the editorial. Thanks for the interesting round!

gradient(difficulty change from D to E) = +infinite

Is topological sort wrong for problem D? I am adding edge from x->b[x] if a[x] is positive, and b[x]!=-1, and the opposite edge if a[x] is negative! Please tell me whats wrong in this?

Try this test

Yes. Almost there :P

Say that you have this case

10000 -1 0

2 3 -1

Now can you find the issue?

P.S. I also made this mistake during the contest.

Start from nodes with in degree 0, remove edges from it, and it should work.

Thx I got it :)

The negative a[x] can be positive during the operation, which is good for maximizing ans

thx

help me with C after the contest

Explanation to problem CIt's a pure DFS problem.

How many people pass through city i? Sum of the people passing through his children, plus p[i]. Call this total_people. What could be the value of h[i]?

First, it has to have the same parity as total_people.

Second (which made me lose score and time), it must hold that h[i] <=total_people and h[i] >=-total_people.

Third, it has to be >= the sum of h[child] for all its children (because mood can only decrease), minus p[i] (It could always be the case that all of the residents of city i changed to negative mood when they reached the city). So, h[i] >= sum(h[children]) -p[i].

That's it.

First reaction after reading C: bruh

is it rated?

Yes it is rated for you<(")Just read the blog before asking it :)

After making silly implementation mistake in B I thought myself as cabin boy Kostya.

Irritating B, as all B's should be

Thanks for the round!

Nice contest, great problems. I really liked how you put the background stories at the top so we could simply ignore them. Although there was a huge difficulty gap between B and C (causing speedsolvers of A and B to get a high ranking), I think the problems were good overall. Hopefully pretests are strong (don't let me down), and I'm still very thankful that you included 36 in the samples for A (I'd probably get WA on pretest 2 and be scratching my head if you didn't)

36, 40, 44

Your text to link here...

whats wrong with my code of second problem please help??

Its mentioned in the question that binary form of(0-9) is concatenated without leading zeros.

ya thanks for help

Wasn't a lot of problem B solution given by the example input.

Seems like yet another rushed contest to maintain schedule.

(With all due gratitude to the the writers)Urging organizers to please be patient for decent enough problems for a well balanced set before arranging a contest.lol, C is more difficult than D and E

For me C and D have same difficulty

I got the solution of D but don't have enough time to implement it.

I got the idea but couldn't understand how to implement. Maybe D is slightly harder.

i could calculate the maximum score correctly but still struggling with the order of operations

I think D was easier than C.If C and D were swapped then it might have more number of submission.

Whay is the solution for C?

Try to solve a simpler version in which all cities are in a straight line. The intuition there carries forward to the general tree case. Its a good dfs question.

Do a bottom-up dfs and for each node i, 2 conditions must be satisfied

You should be able to divide totalPassing = (p[i]+(sum of p[u] in whole subtree)) to "happy" and "unhappy" to get the required h[i]. This can be done be simply checking if totalPassing%2==abs(h[i])%2.

Total number of happy in all of its children should be greater than the "happy" calculated from h[i]. This is because, if a node as n happy people then everyone in the subtree can have at max only n happy people.

If one of them is false for any node, answer is "NO"

also will this hold : |mood[leaf]|=p[leaf] ? (|x| is absolute val)

Yes, why not? |mood[leaf]|=p[leaf] means that for any leaf, all its inhabitants are either all unhappy or happy

Yes got it, so help me here alright, We get the node(call it v) which is parent of some leaves and leaves only (let it be parent of x leaves), so this must hold :

people[v]=| mood[v]- (mood[leaf1]+mood[leaf2]+...+mood[leafx]) |

Now my question is what about the other nodes, I am getting confused here on writing the people equation for other nodes.

That equation isn't correct. Just look at it this way, Total number of people passing a node is the total number of people living in the whole subtree.

First of all you need to be able to divide this total into the required h[i]. Ex- If the total people is 10 and and h[i]=4 then happy people is 7 and unhappy people is 3.

NOw some of these people are gonna pass down into the subtree. Here is the important part, The 7 happy people

canbe converted to unhappy people but the 3 unhappy people cannot be converted back to happy. So here is where we need to check the condition thatTotal number of happy passing a node >= sum of happy people of all everyone in the subtreeAhh I get it,..Thanks for the quick response!

Submission 88531594 I understood the logic but still can't understand what is wrong in my code. Can you please go through and suggest changes? Thanks again!

Is 3rd pretest for problem D — 3rd sample test case?

yeah should be.

I need 100% true answer

Check your submissions, your recent code fails on the sample test — 3. So yeah i'm 100% true.

yes it is, you can see the that test case along with your submission

If you got stuck on it you could literally just look... pretests are shown when you click on submission.

Yeah, thanks, I was so confused and have not noticed that.

Nice problems. I think this round is somewhat similar in difficulty to old div 2s and I like it. Been a while since I struggled with C.

indeed

You are lost when you get WA on C. Like how can you debug this. How can you generate test cases and see which works. I am lucky to debug C

How to solve problem D? Would topological sort fail?

Make a graph using array b and topologically sort it. Start from the beginning and if value of a[i] is non- negative, choose vertex i and update a[b[i]]. After that, sum up the array a.

Can you point out the mistake. I did a topological sorting with graph made using Array a. Link — https://codeforces.com/contest/1388/submission/88522224

yeah, i came up with the same idea, but dunno how to print the order correctly

For positive a[i] just push them in order that you encounter them and for negative, put them in reverse order in which they are present in topo sort.

yeah, did you mean to put all the negative a[i] to another vector then print the positive a[i] as topological order and then print the negative vector in reversed order?

No. Let topologically sorted vertices be in array topo. Then, traverse the array topo from start and if you encounter any vertex i with a[i] >= 0, push it into the answer array and update the corresponding value of a[j] where j = b[i]. Mark vertex i as visited. After that, traverse the topo array in reverse order and add to answer any unvisited vertex.

lol, that is basically the same as my comment, anyway thank you, i got it.

Is E to binary search for most vertical vector?

No

Does problem D involve topological sorting?

Yes. Do you want more information?

Can you point out the mistake in logic or code i did. Link — https://codeforces.com/contest/1388/submission/88522224

It's because you printed all the values in their topological order. Some need to go last. Suppose the graph has the following path -1000,10,10,10. You don't want to use -1000 more than once, right? If you use it before the second element, it will be used at least twice, so you want to use it last.

My python code of the important loop (top is in the order the topological sort):

Code:i did the same by checking if the current value A is less than 0. Then instead of doing A->B i did B->A

Well that's wrong for other reasons. Suppose you have the edges 1,3 and 2,3. supposed 1 is very big, so you reverse it to 3,1. Now you have the graph 2,3,1 and you make a path that was never there (the value of 2 will be added to 1, which shouldn't happen).

I dont have idea how to solve D, but guide me if what im thinking might be correct,(if it's related to graphs? )

So, we have a directed edges with few nodes with negetive value and other with positive.

My approach:

1) find cycles in graph.

2) for remaining nodes(which are not in cycle),

2.a) if it's positive, add the value to the node it is pointing to,

2.b) if it's negetive, add it in the final sum, i.e only once?.

3) for every cycle begin counting the sum, from node which has highest value of Pi

There is written that graph does not have cycles.

Hey can someone explain me, why this sample test case is NO??

Test Case1

3 3 7

13 1 4

1 2

1 3

The mood in city 3 is 4, but it has 7 people in it.

4 % 2 == 0 but 7 % 2 == 1.

Edit: The mood of a city must have the same parity as the number of people who

pass through it.the parity of the happiness index and the number of people going through that city should be same. for example if there are 4 people going through a city the happiness index can be 4,2,0,-2 or -4 but it cant be 3,1,-1 or -3. In your case for city 3 number of people is 7 that is odd therefore happiness index cant be 4 as its even

Problem C was quite interesting .. Overall, nice contest.. Thanks :)

There was a huge gap between B and C.Indeed it was a speedforces type.Ig they are making nowadays tougher one to reduce the load to the system.

I was not able to submit working B. Is there a "funny" trick or something?

Its all 9s except the last ceil(n/4) digits which are 8. I guess you could say its a funny trick.

Ah... I kind of mixed up the logic :/

If somebody is interested in how one can make this one still wrong after aprox and hour of thinking: 88513867

It's always some number of 9's followed by some number of 8's. Number of 8's increases when length of string is 1,5,9,13... as you would be erasing the 1 that separates 8 / 9.

Ans was of form count(9) + count(8), at first I was replacing trailing 8 with 0, and did 2 wrong submission

SpoilerYes, just consider 9 and 8 in the answer.

all the last (n+3)/4 numbers must be 8 and the remaining ones shall be 9. because you want to get maximum r which is only possible if numbers in x have higher length in binary format. since 9 and 8 are only possible 4 bit binary digits , thus we always require to choose numbers as 9 and 8. first of all place n 9s in a string then if last bit of any 9 is getting chucked off then replace that 9 with 8 continue doing so until you replaces last n bits.

It was amazing, thanks!!

What a great contest with fun problems!!!

Sorry my bad :((

Nice Contest Able to solve Only 2 .. but real Div2 is Back How to solve D ?

Build the graph from array 'b', since it's acyclic topo sort exists, process the numbers in that order...watch out when the value is negative, in that case process it's child first....

Man I was first officially right after solving A and B but then got bogged down by debugging on C and didn't get much time to implement D so pretty much no delta change.

Speedforces

Question B ** Why output for 4 is 9998? shouldn't it be 9990? There I can't see any restriction for using 0 in the end.

$$$0$$$ is represented as $$$0$$$ and not $$$0000$$$

binary representation of 9990 is 1001 1001 1001 0 , note that you don't add 4 bits for 0

if you use 0 at end then string k will be(no leading zeros so 0 will be 0 and not 0000) 1001100110010 removing 4 digits from end we get : r1=100110011. Now consider taking 9998, we get 1001100110011000 removing 4 again, r2=100110011001 clearly r2>r1 so 9998 is optimal.

9998 then 1001100110011001 --> 100110011001 9990 then 1001100110010 --> 100110011

since 100110011001 > 100110011 hence 9998

your reasoning will be true if we represent 0 also as 0000, but as can be seen in question they representing it without leading zeros.

You should have every digit 8 or 9, else R will have less digits and won't be the maximum.

no because for x=9990, k=1001100110010 and for x=9998,k=1001100110011000 after removing 4 digits from back in both cses obviously 2nd no. will greater

Oh got it now. Thanks everyone. Happy coding.

Uncle Bogdan is OP!

For problem c, we had to topologically sort elements of a using b, and then do a forward pass picking only non-negative elements and another backward pass picking the remaining elements right?

My solution for problem D outputs "1 3 4 2 5 6 8 9 10 7" as an order of operation for 3rd sample test case. I checked this order by simulating it on computer and it is giving answer -9 which is correct. Can someone help me to find out what is wrong?

the 8 should go after the 7. you should reverse the ones that have neg at the end.

Q2.Can any one help me where I am wrong in question

`#include<bits/stdc++.h> using namespace std; const int a=1e5; int main() { int t; cin>>t; while(t--) { long long n; string s; cin>>n; if(n==1) cout<<8<<"\n"; if(n==2) cout<<98<<"\n"; if(n==3) cout<<998<<"\n"; if(n>=4) { int a=n/4; if(n%4==0) { for(int i=0;i<a;++i) { s.push_back('0'); } for(int i=0;i<n-a;++i) { s.push_back('9'); }

} `

You should be using 8's instead of 0's, as 0 will be read as 0 and not 0000

Problem B. n = 4 k = 1001 1001 1001 0000, r = 1001 1001 1001, x = 9990. But answer is x = 9998. Why??

if you use 0 at end then string k will be(no leading zeros so 0 will be 0 and not 0000) 1001100110010 removing 4 digits from end we get : r1=100110011. Now consider taking 9998, we get 1001100110011000 removing 4 again, r2=100110011001 clearly r2>r1 so 9998 is optimal.

Oh, My mistake. zero is just 0, not 0000.

K for x = 9990 is 1001 1001 1001 0, it is not 1001 1001 1001 0000 because it is given in the problem statement that we must write the digits of x in binary representation without leading zeros.

Problem C: Can anyone explain why is the answer for test case 2 in example 2 of Problem C is "NO".

The P for city 3 is 4, when H is 7. We can't make P even, when H is odd and the city don't has subtree.

In city 3 try to figure out happiness 4 among 7 people. I think it's impossible.

In C, I checked at each node if the happiness count at each node is not smaller than its subtree, I calculated the count of happiness at i th node by fetching total people living at that node and its subtree. Now I have happy-sad at that node (given) and happy+sad = total, I checked if the number of happy people in its subtree is not greater than this value, apart from adding other small conditions. Is it the right approach? Although I could not implement it :(

Did you also check if the total number of people living in node and subtree can be divided into the required happy and sad for that node

Yes, that is what I meant when I said, additional constraints.

I think writing $$$2000$$$ is better than $$$2\cdot 10^3$$$ in case of small constraints. I spent an hour trying to solve E for $$$2\cdot 10^5$$$ before I noticed the constraint. Of course, it was my fault (I even read maximum in place of minimum) but it still is more convenient.

https://codeforces.com/contest/1388/submission/88470430

For problem B. Can anyone explain why this gives TLE? This is in PyPy, but same submission in Python gives AC.

Maybe python interpreter not doing loop, but just s+="8"*smth

When you add a character python creates a whole new string, so your complexity is n^2.

Get used to making lists of characters and then ''.join(L) at the end.

If i understand correctly then s=s+'8' is O(n) and s+='8' is O(1) and i have used s+='8' in my code. Even the same code run in python gives AC so definitely it does not create a new string in Python. Maybe in PyPy it does.

Could be, but that why I:

1) Only use pypy if there is lots of calcualtions involved (for strings,graphs — always python)

2) Use str.join so I won't fail even if I choose pypy by mistake (and it happened many times before).

3) Use str.join because I am human and I can always do s = s + c by mistake...

Well. Optimized appending to the string is pretty much CPython implementation detail and is not guaranteed by the python language per se.

So you're right. In CPython this operation is heavily optimized and I believe is O(1), i.e. it does not allocate a new string every time (there are exceptions actually to maintain external immutability and this immutability is exactly the reason why most people believe that it should always be O(n)) and in PyPy is O(n).

Actualy, I knew that this is not guaranteed in python, but never thought it would become a problem in cp when switching to pypy.

Appreciate the very strong pretests in the contest. Also interesting problems, C was a graph problem after many contests and it was pretty nice imo.

Problem C is quite annoying. I look through some PP code after contest and get nothing about why I'm wrong or even whether I failed to judge something.

Why did you divide by 2?

Edit:

I'll tell you what I don't see. I don't see a condition: h[i] >= sum(h[children_of_i])-p[i]

I don't see the minus p[i] part.

I got what's wrong. My DFS is incorrect. If I wrote correct DFS I would get AC with 50 points penalty.

Was it because "sz[to[i]]" of the parent could be not 0?

I'm sad for you :<

A very good Problemset after decades.

MikeMirzayanov

Cheating caught in today's contest https://codeforces.com/contest/1388/submission/88511065 https://codeforces.com/contest/1388/submission/88510232

They have same codes only function names and variables are cleverly changed.

And the second guy deliberately submitted late when he was sure he will get 3 AC's.

Dont know why people show such teamwork in CF just to increase their ratings.

Their other submissions from the contest are also same.

Shame.

In second question, if n=4 then the answer should be 9990, right?. Similarly if n=5, it should be 99980?

No, 0 is only 0, not 0000. however, 8 is 1000, so 8 should always be used because it's 4 bits instead of 1.

It is better if without leading zero was

bold.For 0 it is represented by 1 bit only so in that case it will not give the required answer. Check problem description for clarification.

No...n=4 -> ans=9998 and n=5 -> ans=99988

88492256 Why my submission of question B got TLE in system testing.... help

Same thing happened to me.

s += '9' will work in time. s = s+'9' will not...

I think the += operator knows to add on to an existing string (like push_back on a vector) while the + operation treats it as creating a new string.

Very annoying (and to me, something I thought would have been caught in a pretest at least).

While adding new charachters to the string you wrote: s = s + '8' which has complexity O(n) and as you do this operation n times the overall complexity of your code becomes O(n^2)

You should have written s+='8' for appending charachters as it has compexity O(1)

It's interesting that if I don't hack Geothermal's problem E code, my first submission of problem E will pass all tests.

:(

Hack人终hack己

:((

I don't understand why your solutions fail this test. Is it because all projections should touch or because of the (0; -1) optimal vector? Can you explain, please.

It's because there are many points pairs are parallel and we don't consider the case well when sorting them. In our method, it will make the program don't catch the case projection is parallel to these points pairs.

Can someone please help me out with C. Here is my approach :

1. Do DFS and store the total population of current city and the cities below it.

2. Then do DFS and find the number of people having mood in the children cities.

3. Find if the current population of city is good enough to satisfy all conditions .

Link to code : https://codeforces.com/contest/1388/submission/88521378

Thanks in advance !

Make sure the number of positive and negative people at each node are greater than zero

NA

I think I still have to practice a lot.

Problems of this round were truly amazing. Loved it <3.

The contest was great! And many many thanks for the good samples.

In the contest i wrote a solution for problem C which exceeded time limit on test 11: According to me it is an O(N) solution. Can somebody figure out the issue.

CodeTry passing your vectors by reference as your current solution makes copies on every function call.

Thanks. It got accepted.

can someone please help as to why this code gives tle on test case 7 in problem B??88487950

instead of s=s+'9' and s=s+'8' try s+='9',s+='8'.i too faced the same problem

thanks! it worked!

Used the s += "8"("9") rather than s = s + "8"("9")

+= take O(1) time

thank you!!I'm surely never forgetting this!

When you do s = s+'9' you are basically creating a new string s+'9' and then assigning it ot s. So for one step its complexity is O(s.length()). So the overall complexity becomes O(n^2).

thank you!! I got it!!

solved the problem D during the contest.forgot to declare ans as long long. result:-suffer.

So, for problem E, if I assume at least two projections touch each other in the optimal answer, I have 2*n possible vectors to choose. And if I check all these cases, the answer should be the minimum of all the answers, right? Is this approach correct ? If so, why is the number of submissions so less ( because of geometry? )?

Don't you have n^2 possible vectors to check? And then you have to check in O(1) instead of O(n)

There won't be n2 possible vectors because we only need to consider the adjacent pairs when sorted by the start point or end points. If we choose two random pairs, the segments whose starting point lie between these pairs cannot be projected without them intersecting.

If we consider the effect of the projection $$$(a, b)$$$ on an interval $$$(l, r)$$$ at height $$$y$$$, then our new interval is projected onto $$$(l - y\cdot\frac ab, r - y\cdot\frac ab)$$$. If we let $$$t = -\frac ab$$$, then this is simply $$$(l+yt, r+yt)$$$. Thus, view the intervals as buses with speed $$$y$$$ and location at time $$$0$$$ equal to $$$(l, r)$$$. We want to find a time $$$t$$$ where no two buses overlap, and we minimize the length of any interval containing all the buses.

If we go to time $$$t = -\infty$$$, the fastest bus should be all the way on the left, while the slowest bus is all the way on the right. At time $$$t = +\infty$$$, the fastest bus in on the right, and the slowest bus is on the left. It's also clear that the time with the minimal interval is when two bus endpoints are touching. To reverse the order, there are $$$\mathcal O(N^2)$$$ crossings. We can process them by time, and keep track of how many buses are crossing at each point in time. Whenever there are $$$0$$$ crossings, then we simply want to find the maximal value of the endpoint of a bus minus the minimal value of the endpoint of a bus. This can be done with Convex Hull Trick. Overall, this will take $$$\mathcal O(N^2 \log N)$$$ where this comes both from sorting the $$$\mathcal O(N^2)$$$ events and the Convex Hull Trick.

88530891

Thanks, that's a nice perspective to look things at.

C: Could anyone please tell why is it giving TLE?

https://codeforces.com/contest/1388/submission/88528267

My approach for C was:

Stored the dfs from 1 in an array. Then Reversed it.

Now we iterate the array. (starting from the leaf nodes.)

For each node we maintain good and bad people passed through it. Then we traverse this reversed dfs, and for each node all it's children's good and bad can be good for our current node. Also the people who live in the present city can be good as well , thereby we try to maximize the good cases. Then we fix current node's good and bad such that we maximize good and have just enough bad to make happiness[i].

For any node if we don't have enough good or bad , it's simply a NO.

Please help in my Code. A failing test case please.

Edit: Sorry I thought its bad instead of sad.

Hey, I feel that you didn't accommodate the fact that bad (sad) people can actually be residents of the given node and children nodes may have less sad nodes than parent because their home was inside parent.

I created a class with intuitive variable names if you want to check my code out! 88503365

what a great contest ! Thank You all .. Denisov make more contest with your team <3

A very good match, but the difficulty gradient is not good. Question b can be a little harder, and question c should be easier. I passed c after the match. If I have two and a half hours in the match, I can solve c. To summarize my code, I feel that my code is too long and not concise. And typing speed is too slow. Many areas need to be improved.

I appreciated the convexity in Problem E. I could best picture the projections as the shadows from an infinitely distant light source.

Respected Sir,

I would request for a recheck on my solution submitted to Problem B from Codeforces Round #660. I have been shown Time_Limit_Exceeded on my solution, but that same solution when I executed it through the custom invocation in codeforces on the same test case it was executed in 658 milliseconds. I would create a great impact on my ratings. I would request you to give a recheck if possible.

I have attached my solution to the problem.

88478273 Codeforces Round #660 (Div. 2) standings

Interesting, it is your s=s+"9" that is slow: it is probably copying everything into a buffer, i.e. it results in a O(n²) solution.

I am still wondering why is this code accepted then[submission:88534456]

n² is not that large. When I tried, I also got accepted sometimes, but with times close to the limit. It means the conditions were not set tight enough.

So I guess for a fair judgment they should allow everyone's solution whether it be s=s+"9" or s+="9" because otherwise, the time limit should have been 1 second

Surprisingly, the same solution is accepted now-Accepted Code. I think, its because of s=s+'8' which take $$$O(n^2)$$$. Replace it with s+='8', it will take significantly less memory

Thank you for very nice problems! They was a good balance of diffculty (at least A, B, C that I tried).

When will the ratings update?

https://codeforces.com/contest/1388/submission/88494957 (Contest Submission) https://codeforces.com/contest/1388/submission/88531246 (Unofficial Submission)

EXACTLY IDENTICAL FILESFirst file gave TLE in System Testing, second file got Accepted. May I know why?You should never do "s = s + '8'". This takes O(length(s)) operations. So your code is actually O(n^2). Replace all the " s = s + x" with "s += x" and you will notice a sharp time difference. See my submission below where I made that change in your code and it took only 30ms compared to the previous 1700 ms.

Usually there is heavy load on the servers during the contest and the code may take longer to run

https://codeforces.com/contest/1388/submission/88533226

I had a same doubt while solving a problem recently.So i used push_back function on string which is O(1). Still , I have a doubt how to push a character in front of a string in O(1)?

You can't, except a trick in the case you only need to push to the front: push to the back and then reverse at the end.

I am not aware of an STL functions which would prepend a character to a string. However, you can store the string as a deque<char> where insertion and deletion at both ends can be done in O(1).

Yea,thanks!

Time difference is real sharp man! Thanks, didn't knew this.

this is my first contest. i solved the first 2 problems but i am still unrated... waws this contest unrated??

NO, the ratings aren't updated yet it will be updated in some time.

how much time does it take to get updated?

It should have been updated by now.but i think they will most probably be updated within 30 minutes.

Why in the problem B this solution gave TLE in the contest [submission:https://codeforces.com/contest/1388/submission/88476304] while in practice the same solution passed [submission:https://codeforces.com/contest/1388/submission/88533918]? Please can you clarify this doubt Denisov MikeMirzayanov

Look at my comment above. The TLE is due to the expression "s = s + '9'"