Hello Codeforces!

gyh20 and I are glad to invite you to Codeforces Round #670 (Div. 2) which will start on Sep/12/2020 16:45 (Moscow time). **Note the unusual start time of the round.**

The contest will last for **two hours**, and you will have **five tasks** to solve. The tasks are prepared by me and gyh20. This round is rated for participants whose **rating is not higher than 2099**. You can see that my current rating is exactly **2099** :)

There might be an interactive problem. You can learn about them here.

We would like to thank:

- isaf27 for coordinating and helping us with the problems.
- namanbansal013, postscript, morzer, Osama_Alkhodairy, csani, growup974, physics0523, bizhouziwei20 for testing the problems and providing useful feedback. Special thanks to csani for helping us with writing the editorials, and physics0523 for improving one problem's solution!
- c2020hxw, fisherking, fangzhijian2020 for discussing the problems with us.
- finally, MikeMirzayanov for great platforms Codeforces and Polygon!

We tried our best to make the statements short and clear, pretests strong and problems interesting. We hope you like the problems!

Score distribution will be announced shortly before the round.

Good luck and have fun!

**Upd:** Score distribution is **500-750-1250-1750-2500**.

**Upd:** For problem reasons, the contest is delayed for 10 minutes. We are very sorry to keep you waiting, sorry again.

**Upd:** Score distribution is changed to **500-1000-1500-2000-2750**.

**Upd:** The round is finished. We're really sorry for B being well-known (none of the testers knew the harder version of this problem in ABC173E). Still, congrarulations to the winners!

Div1 (unofficial):

Div2:

**Upd:** Editorial is out here.

As a tester,I want some contribution!

Have a strong heart, dear.

So why are we downvoting the tester >_<

A tester leaves a rated contest for us, Have some decency please

I won't mind leaving a rated contest just to see my name in the blog and then getting easy contributions later.

You see you're expecting contributions too and so does bizhouziwei20

what we get if we have some contribution.i often see people are asking for contribution ??

nothing,in general but it is an indication that you have been useful to the community.Also there is a list for top contributors

okay Thanks :P

As a coauthor,I am so glad to hold the contest, and i want you-know-what.

I want positive contribution instead of negative contribution!

Don't forget to give the tester bizhouziwei20 you-know-what as well. :)

sad life :(

This happens to me when contest is delayed it hurts alot as i am preparing my self to encounter the problem and just 10 mins more added....

Thanks a lot for replying, I had forgotten about the contest. You reminded me. :)

tianbu Can you highlight the start time of the contest.

I think it is already highlighted.

I mean it start one hour earlier than regular time.

OK, added a sentence

Really looking forward to the contest!I think it will be very good!

Are you son of fisherking ? :v

Definitely not! You can see that he created the account a lot later than I did and there is nothing I can do.

Sorry,it's just a typo.

Sorry for telling that..But when i read your id name , i think the tester fisherking and your id name can be related . Again sorry sir !

Certainly correct

As a problem improver, have fun!

What I improved is following:

In the task E, there was one more constraint "It is guaranteed, that an integer $$$p>1$$$ such that $$$x$$$ is divisible by $$$p^2$$$ does not exist. You want to find $$$x$$$."

I noticed that we can erase this constraint. If you couldn't solve E, it's good that thinking this version first!

Off Topic : What is the cutoff for candidate master.. 1800 ? or 1900 ?

Your rating is exactly 2099 but unfortunately the round is not rated for you.

tianbu, Will there be any subtasks? I really think having subtasks makes contest more interesting.

The contest is ICPC style not IOI style,so there will be no subtasks.

First of all thanks VoidHead for letting me know, but sadly I disagree with your explanation why this contest can't have subtasks. You can see many Div-2 contests had subtasks for eg:Codeforces Round #658 (Div. 2). Also each problems are scored equally in ICPC( like we have in CF educational round ) which is not the case for CF Div-2 rounds.

We used to have subtasks, but we found it meaningless to have subtasks on our problems. So we deleted them in the end.

Every Round is rated for me, except div 1 :3

Div1 people: Are we a joke to you?

forget to mention div 1, Yes excluding div 1.

As if there were any div1 only rounds

Changing profile picture?? kdjonty31

Are the authors middle school students?

`Special thanks to csani for helping us with writing the editorials`

Editorials are already written?

They are written, but not published.

then why not publish them?

So you cant look at the solutions in contest

He will publish as soon as you completely read all the problem statements and try to solve it for 2 hours :v

What if he did not read all the problems ?

He should be :v

Can we have them now? By the way, thanks for the contest, loved it:)

Please Publish Editorials

csani Can you publish editorial now. Thanks!!

It has been published.

As a tester, I would recommend you to participate in this. The problems are really interesting. Good Luck and have fun.

every round is interesting according to testers!!

Not all testers comment on this!!! And everyone has their own opinion on what's interesting.

Not really, it's just that testers who think that the round is bad usually don't say "this round is bad, don't participate" in the comments.

That day is my birthday. It will be a memorable contest lmao :))

hpbd

:3Happy Birthday!

"Has a female profile pic" hpbd

brohappy Birthday. good luck.

Today is my birthday :) I was born on 9/11

belated happy birthday dude[orz pro max]

Many Many Happy Returns!!!! duock28cbg

$$$Yoo$$$ now it's becoming trend of $$$make \,the\, statements \,short \,and \,clear$$$

Its useful too. Isn't it?desp__beginner

As far as there is no story telling(irrelevant statements i.e. not related to that Task) then even complex scenario is fine because converting some complex statements into Mathematical equations/logic is a skill and one should try to learn it.

Hi,desp__beginner I support your thoughts. But I think a clear statement refers to a clear mention of the requirements. Like "if multiple solutions, print any" or this kind of thing... not making a problem easy or spoon-fed.

Is it a trend or...something else?

No.In fact most of the CF contests a few years ago had 5 problems.

(I don’t mean to offend anyone, I actually think testers deserve contribution for the hard work they put in)

I have a question for the authors. Do you know Jiangly? Are you friends?（Sorry. I now feel that this question is not so appropriate. Thank you very much for the author's answer.）

We know Jiangly, because he is the best one in the city,but we are not so good as him so he definitely doesn't know us.

Excited to take part in my first contest!

I have known that the author is very excellent,bless all and hope everyone get higher rating.

After two years without competing here, it is time to come back! :)

your rating graph is quite interesting!

Hope that the interactive one will be nice just like the one in the last contest...

Thank GOD for no scary pictures this time...

Great!! Problems with short and clear statement are really good. Hope this will be a good round.

with strong pretest.

Yeah,this is also very good since my problem will not hack after pretest passed.This hacked really sad me.

Why do all the Div.2 rounds consist of 5 problems nowadays? Why not 6/7?

How many times have you solved more than 3-4 questions lol?

On a serious note, you and I can't even imagine the amount of work it goes into setting a problem and preparing the statements, test cases, etc. So, simply saying things is very easy, when you don't know what is going on behind the scenes

emmm, recent div.2 rounds are really easier than that before. just like me, before Round 658 , I only solved 2-3 questions. But after Round 658, I always solve 4 questions in Round which had 5 problems.

To some degree, it can be named speedforces.

Maybe you just progress

Nope. Earlier I used to solve 2 problems in Div.2 and can't come with a solution for C but in the recent contests I still solve till B but have some logic for C although there is some corner case which I miss out but still. And honestly I don't think this is because of improvement cause i am not practising hard lately.

That means you are improving! Do you want to stay at the level of only being able to solve 2-3 problems?

And, let's just assume for a minute, that the problems have become easier. Do you still know how much effort goes into making those problems, setting up everything, co-ordinating with so many people, getting people to test your problems etc.?

To be very honest, I have tried to sit down for hours by myself to think about making a problem, but I never made a single problem which wasn't already popularly known.

You should have played Round 669.

is this really necessary to reply someone with this first line!? is it really matter how many problems he solved to tell his opinion and ask his question!? these kind of replies will force them to create alt acc(and probably ruin the cf community). dont bully anyone because of their rating PLS!

Your point is valid, but he did ask in a curious way(Not implying or demanding anything). In no way was it appropriate to reply with that first line.

I hope it have strong pretest!

I don't want to FST any more.

So do you have the strong pretests?

This is gonna be my first ever contest. So excited about it. Not very experienced in Competitive Programmer but my journey starts from now. I hope this be a hell of a journey!

UPD : For those who Upvoted me and wished me luck, thank you so much! I think your wishes came true and i got Rank : 652 and Rating got +671

T̶u̶r̶n̶ ̶b̶a̶c̶k̶ ̶w̶h̶i̶l̶e̶ ̶y̶o̶u̶ ̶s̶t̶i̶l̶l̶ ̶c̶a̶n

Good luck on your first contest!

Is there any Chinese description

No.

Happy Programmers' Day!!

Whats Polygon guys?

It's a platform that helps make problems, created by Mike

Socre distribution <- typo

fixed, thanks for pointing out

Still, congrarulations <- another one

The comment is hidden because of too negative feedback, click here to view it

As far as I've seen, this is your first contest arrangement in CF. Best of luck to you too.

(Hope everything will be fine)

Thanks!

you just raised a flag, 10 mins delayed

Interactive problem will be there or not ?

YES

Thanks :)

`There might be an interactive problem.`

I guess that codeforces is trying to make interactive problems a part of the contests, which according to me, is a good thing to do.

haa its really good

start the contest!! im feeling horny

wtf dude, this is codeforces not a nrop website.

STFU you unrated GAY!!

Ok, you idiot. Look what you wrote. Now, try to find any comments like this on this website. See there is no one. I think a better place for you is a nrop website. Also, where did you got enough information to deduce I am gay? Because I'm not.

umm... okay

STFU you bald ass!!

It is not nice to talk to yourself like that.

Hey, we are not same. Thanks.

I know that, but now find a way to prove it. People will start thinking that we are the same.

Go for virtual . :p

Yeah sure with your mother.

Don't you know about virtual contest.? It seems you've started painting your depressions here.

Delayed by 10 min :(

yup!!!

umm , we actually know..

That's why_you_are_here :)

that's enlightening,thanks!

Auto comment: topic has been updated by tianbu (previous revision, new revision, compare).Very sorry,something we didn't expect happened just now ,the contest is delayed by 10 minutes, hope there won't be another delay.

Thanks for the info :)

I think no longer , it should be good this time.

Just curious to know, what was the last minute change because of which score distribution was changed? I think present scoring distribution was accurate.

wish me good luck.

why though? why to you specifically?

because I am loosing my rating. can't you see?

Then what you need is practice and not good luck!

yes bro I do.. But CP is fun and exciting. Why you taking it as a fight? if someone ask wish me good luck if you can just wish them.. What is harm in it? Spread love and have fun bro. This is what life is.

Fighting? Hehe chill. good luck

more practice and then make progress. good luck and high rating!

yes. I do.

goodluck fam

thank you:)

now contribution too, good luck though!

exactly. It is -17. LOl. I think people should not behave like this.

Good Luck Native!!

good luck!

Delay it as much as you want and solve the problem PLEASE DON'T MAKE IT UNRATED due to those issues

you want some rating so badly.

delayforces again lol, but still great thanks for the preparation!

We're very sorry. But please, still enjoy the contest.

Thank you for your effort！！

always thankful. Thanks for the preparation anyway

So you people make quality problems for us, make quality test cases so that only good solutions pass and respond to our queries during the contest.

I don't think you need to apologize for anything. You're already doing more than we deserve.

delay making me more anxious.

Score distribution changes.

Hoping for strong pretests!

How to solve D?

Didn't passed system test yet but my solution idea is :-

Hints:-

ans is

tot = a[0]+d where d = sigma( max( 0, a[i+1]-a[i] ) );

ans = (tot+(tot>=0))/2;

so only edges L and R in queries can change the value of d.

I guess now you can figure out the solution

My solution has not yet passed sys tests but here is how I solved it. Suppose we let initial value of non decreasing array be c so initial value of non increasing array will be a[1]-c.

Now for each next element if is greater than previous value we can keep the value in non increasing array constant therefore the other array remains non decreasing. We can similarly keep the value in non decreasing array constant if the next element is less than the previous one.

Therefore we get the final value of non decreasing array as c+(sum of all positive a[i]-a[i-1]). Now we just have to minimise max(a[1]-c, c+sum of +ve dif). We can see that it will be minimum if both elements are equal.

After each query only 2 differences are changed at max so we can store values of each element by Fenwick Tree and recalculate the optimal c. Upd: passed sys tests.

Observe that since $$$b_{i}$$$ is non-decreasing and $$$c_{i}$$$ is non-increasing, the max is just $$$max$$$$$$(c_{0}, b_{n - 1})$$$.

We can notice that $$$b_{i} \gt b_{i - 1}$$$ will be required when $$$a_{i} \gt a_{i - 1}$$$ and $$$c_{i} \lt c_{i - 1}$$$ will be required when $$$a_{i} \lt a_{i - 1}$$$.

Clearly $$$b_{0} + c_{0} = a_{0}$$$ is a required condition. Let $$$b_{0} = u$$$ and $$$c_{0} = v$$$.

If $$$sum_{inc}$$$ is the sum of $$$a_{i} - a_{i - 1}$$$ whenever $$$a_{i} \gt a_{i - 1}$$$, $$$b_{n - 1}$$$ is clearly $$$u + sum_{inc}$$$. So we want to choose $$$u$$$ such that $$$max$$$($$$a_{0} - u$$$, $$$u + sum_{inc}$$$) is the minimum possible. This is clearly $$$\frac{a_{0} - sum_{inc}}{2}$$$ which gives an answer of $$$\lceil \frac{a_{0} + sum_{inc}}{2} \rceil$$$.

As for the queries, we can notice that only the contributions from $$$a_{l} - a_{l - 1}$$$ and $$$a_{r + 1} - a_{r}$$$ to $$$sum_{inc}$$$ can change. Maybe there is some elegant observation that makes this easy to calculate, but without that also there is an easy way to figure out the change. Subtract the contribution of those two endpoints, update the range (standard lazy prop) and re-add the contribution of them.

BE CAREFULthat you perform ceil correctly for negative values.Damn. That's why my 4th TC got WA.

Wait how, if you ceil incorrectly for negative values it should fail sample 3 as you will do $$$\frac{-2 + 1}{2} = 0$$$ instead of $$$\frac{-2}{2} = -1$$$.

Hmmm. You've got a point. I didn't mention that

I thoughtthat, that might be the issue.Now that I can check the test cases, I'll check what exactly went down.

EDIT: TL;DR: I'm dumb. So I just updated the l and r values of the array, and somehow thought that it would work, without updating the middle values. Will implement the lazy propagation part now.

It should be (a0-sum)/2

Fixed, thanks. I accidentally wrote the minimum obtained instead of the value to be value which when substituted gives the minimum.

We can also do away with the Segment/Fenwick Tree. Since only the contribution of $$$a_l - a_{l-1}$$$ and $$$a_{r+1} - a_r$$$ affect our answer for the current query, keep an array of all differences between consecutive elements and change these two values in that array. Change $$$sum_{inc}$$$ accordingly.

P.S. Don't forget to change the value of $$$a_0$$$ if the current query starts at index 1. Code

Oh yeah since the Fenwick Tree effectively acts like a dynamic difference array here which is sufficient for our needs. Thanks, somehow that didn't strike at that moment during the contest.

How to solve C

Easy AK!

Talking about this ?

Yup!

How to solve E?

I think ask for all prime numbers ( ~9600 ) and make some Inclusion–exclusion principle (ex if you have n = 6 ask for x=2 => s = {1,3,5,6} and for 3 you have to see that 6 is still there (because after the first operation just 1 element %3=0 had to remain in the set) ) .

First we use the second operation to all primes < 1000, all the remaining numbers can either be a prime or $$$x$$$.

We can query

`A 1`

and check if $$$x$$$ is a prime.If $$$x$$$ is not a prime, we query every prime number $$$\leq n$$$, and obtain x by finding all prime divisors of $$$x$$$.

If $$$x$$$ is a prime, we can do the following: delete S prime numbers, and query

`A 1`

to see if $$$x$$$ is in the $$$S$$$ primes numbers you just deleted. When you find the $$$S$$$ numbers that contain $$$x$$$ you do $$$S$$$ extra queries to find $$$x$$$. Let the number of prime numbers be $$$N$$$, The number of operations is $$$N + S + N/S$$$, and we can set $$$S$$$ to $$$\sqrt n$$$code: https://codeforces.com/contest/1406/submission/92637463. It passed the pretests anyway.

перебрать все простые делители до N начиная с таких, что их квадрат больше N, проверять можно по группам, например первые 100 такие делители с помощью функции 2 типа, и проверить есть ли среди данных 100 простых чисел нужное с помощью типа 1 и так далее по сто, а в последнем оставшиеся. Затем перебрать все меньшие простые делители и его степени. Например, k — простое, k * k < n, тогда если н делиться на к, с помощью бинпоиска найдем степень вхождения это числа.(iterate through all Prime divisors up to N starting with such that their square is greater than N, you can check by group, for example, the first 100 such divisors using a function of type 2, and check whether among the data 100 primes necessary using type 1, and so on one hundred, and in the last remaining. Then iterate over all the smaller Prime divisors and its powers. For example, k is a Prime, k * k < n, then if n is divisible by K, we use bin search to find the degree of occurrence of this number.)

Could anyone help me to find why my code for problem C is TLE.92617935

My solution is linear,as far as i know.

memset is the problem. If you have 10000 tests of n=1 then you will do 10000 (number of test) * 1e5 (maxN) operations

You should not clear array sz for each testcase since T can be sufficiently large to give TLE for your implementation. Don't use memset(sz,0,sizeof(sz)) each time. Instead, clear first n cells.

Great Contest!! B was really original problem!!

how to solve B I was getting WA in test 3

Sort the array, took the first and the last 5 elements, now you have an array of 10 elements, just brute force for every 5 pairs and take maximum value.

Can you help me what's wrong here, i tried a similar approach link

Is this approach correct? Sort the array then take

`i`

elements from front(i goes from [0, 5]) and take`[n-4+i to n-1]`

elements from the back. Just compute the max ans from all i. I get WA on test 3, any clues anyone ?Aah, of course it is correct, got my mistake i use using INT_MIN instead of LONG_LONG_MIN ;-)

Notice that there can only be 3 possible answers in B, 1) all largest positive numbers 2) 2 smallest negative and 3 largest positive numbers 3) 4 smallest negative and 1 largest postive numbers. Find the max of all three possible cases

sort the array,the last 5 elements are definitely the biggest now so their product can be the answer but apart from that if the product of first and second number is greater than fourth last and fifth last ,or product of first four numbers is greater than product of second last ,third last,fourth last and fifth last (this will happen when starting numbers are negative ) they can be the answer as well. So, only three cases arise,pick the best.

Actually.. problem B is a special case of this one ABC173_e where K=5

My comment was meant to be sarcastic

E looked really easy to me, but I guess I was wrong.

That's a nice contest!

Although I'm not able to solve E, I think the questions are very interesting and I really hope I can have the ability to make such greate problems :)

How to solve C? it seems it requires some observations, please anyone give hint.

Basically there can atmax exist only 2 centroids and this happens when there exists an edge u-v such size of subtree of u is equal to size of subtree of v (in opposite directions) or basically s[u]=n-s[u].

Now all you have to do is remove an edge from u and connect it to v and only 1 centroid will remain

How to solve idleness limit exceeded for problem E? Please help in this solution — https://codeforces.com/contest/1406/submission/92639473

It seems that cin>>t; isn't needed

Oh Thanks :)

Did anybody else did o(10^5) solution for b?

Yeah but after 4 wrong submissions ;(

My solution has $$$O(n\cdot k)$$$ time complexity; in this case $$$k = 5$$$, so final time complexity is $$$O(n)$$$.

How to solve E? I got to finding out all prime factors (and their powers) except one. There can only be primes with power more than 1 until 317. Until 317, we check all powers of each primes < 10^5. For primes higher than this, it can be a factor with at most one power. We can iterate through all primes from here and apply operation B. We also keep track of how many more multiples must be remaining at this point. If it differs, we know that this must be a factor (and something else). If we continue iterating all of them, we get all the prime factors (and their power is 1 as explained) except the first prime number.

I couldn't figure out how to get this one. Can anyone solve the puzzle? :)

I tried an approach similar to yours where I calculate how much should a number appear if it is divisible by a prime p,and I kept on using operation B for every prime while updating the amount of numbers divisible by all other primes bigger than p and if sometime they did not match up or if the answer was 1 I check it again and then find x or otherwise x=1.I don't know what I am missing but it did not work.

I think we can group the primes greater than 317 into like groups of 100. After every 100 operations of the form "B p" we can do an "A 1" to check if the number of elements in the set has decreased by less than 100, in that case we know one of these primes is a factor of x and we go back over the group and perform an "A p" operation for each prime p in that group. I realized this in the last 5 minutes but didn't have time to fix my code :(

Hard version of problem B: https://atcoder.jp/contests/abc173/tasks/abc173_e

Sorry for the problem being well-known. None of the testers knew this problem before. (In fact B used to be exactly the same as abc173e, but to make it easier we chose k=5). We're really sorry for it.

thank you for this task, I didn't know how to solve it, and now I know

For D is this correct — $$$result = max(a_1 - x, tot + x)$$$ where $$$x = (a_1 - tot)/2$$$ and $$$tot = sum(a_i - a_{i-1}) $$$ $$$if$$$ $$$positive$$$ ?

My idea that passed pretests was $$$sum = a_{1} + tot$$$ and $$$result =\frac{ (sum + (sum \ge 0)) }{2}$$$ (basically ceil division of $$$sum$$$ by $$$2$$$. What is $$$t$$$ in your notation?

Sorry t is tot only

I didn't solve it in-contest but I think you can compute the minimum gain of sequence B and minimum decrease of sequence C by computing the sum of all rising edges (for B) and the sum of all falling edges (for C).

The max of the two sequences is the start point of C or the end point of B. So we can shift up or down the entirety of B and C to find the point that equalizes those two points as closely as possible, which should be the optimum.

CopyPasteForces

Really sorry for problem B. Still we hope you have enjoyed this contest.

Out of curiosity — is there some nice observation about the changes on both ends that allows D to be solved without using lazy prop for the range updates in the queries?

Can anyone tell what is wrong in this code. for B. Link:- https://codeforces.com/contest/1406/submission/92640232

almost same logic, but wrong answer at pretest2... got no idea

Can anyone tell me what these numbers mean? I have never seen them before.

Probably number of tests

Is this a new feature?

Probably YES, but it was previously available on m1.codeforces.com(or m2,m3).

This number tells us the number of pretests in this contest

The number of pretests. I think you could use it to measure if your solution was good enough. A good feature to possibly prevent system test failing imo.

Am I correct in my logic for E?

I first find all primes upto n, then print each of them with a 'B'. After this round of operations only two integers, 1 and x if x != 1 should remain or only 1 if x = 1.

Then I print out each of the primes with 'A'. If I get 1, then i try its powers till i exceed n or get a 0. At each of these steps, if i get 1, I multiply the ans with x.

I think that this should be enough but this fails pretest 2. Can someone please point out why logic falters?

Number of request is about 2*PrimesUptoN which exceeds the given limit for N = 10^5.

My bad, I googled number of primes less than 10000 instead of 100000. I feel so silly right now

This solution will fail since we can't make more than 10000 operations. This will work only if we are allowed to make ~20000 operations.

There are 9592 primes <= 100000. Thus, doing two questions per each one exceeds the limit.

When E seems to be easy than D, but that constraint

You can perform the following operations no more than 10000 timesmade it quite tough to solve.https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/ just put 5 instead of K

Bruh...

YUP. This was insane.

Sorry for this, we had $$$k=3$$$ and find this in the last hour and we don't have other problems, so we just changed $$$k$$$ to $$$5$$$ so you can't google it, and that's the reason for delaying. Sorry again.

And you guys still managed to create strong pretests for B!

For problem A, did the

textchange in between the contest? I think initially the problem was to split the set into twoequalhalves!The problem was worded very oddly, but no, there was B = Ø in one of the examples.

Can anyone please tell why this solution for problem B fails on pretest 3? link

Thanks to KokiYmgch for "The way to find the centroids of a tree"

https://codeforces.com/blog/entry/57593

For Problem E I think we can group the primes greater than 317 into like groups of 100. After every 100 operations of the form "B p" we can do an "A 1" to check if the number of elements in the set has decreased by less than 100, in that case we know one of these primes is a factor of x and we go back over the group and perform an "A p" operation for each prime p in that group. I realized this in the last 5 minutes but didn't have time to fix my code :(

Is problem B solved by sorting the array and then using it like circular array by doubling its size by copying first half to 2nd half and use a sliding window of width 5 and find the max product? I Couldn't submit in time.

https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/

Please help me to figure it out why this solution 92602531 for

problem Bis giving thewrong answerforcase 1and exactly the same solution runs correctly on other IDE (CodeChef and my own local IDE)I thought there may be issues with my current compiler version but after trying it with different versions, it stills gives wrong here but it runs correctly in other IDE's

Div 2 Bwas fromGeeksforGeekshttps://www.geeksforgeeks.org/maximum-product-subsequence-size-k/#:~:text=Answer%20%3D%200.&text=CASE%20III%3A%20if%20maximum%20element,positive%20integer%20in%20the%20subsequence.

please make it

unratedbruh it is not a reason

Many of my friends and other participants solved B in 3 minutes as it is a very famous interview problem.

It was kind of unfair with the participants who haven't solved the question before.

Even if you know, there is a solution on gfg, you should try it yourself during the contest. Forget about those who copied from gfg.

That is the case for all problems, and the reason why it is a good idea to practice a lot.

"pLeAsE MaKE iT uNrAtED"

E was cool! Thanks

But i am actually not understand why we need answer on type B query...

How to solve it?

There are abut 9600 primes <= n, and about 70 numbers <= sqrt(n).

Let me name prime number which <= sqrt(n) as

smalland >sqrt(n) asbigAnd actually there is 0 or 1

bignumber in X factorization.We can understand small primes in X factorization using about 200 queries (asking A queries about p, p*p, p*p*p, ...)

And how to understand is there a

bigprime? We can arrange all big numbers on 100 groups of 100 numbers. Lets check every group separately using B queries and after checking it ask A(1) to understand was bg prime found or not. If it was found check this group again.It will be like 200+9550+200=9950 queries

Sorry for my bad English ;(

Best channel for editorial of this contest . https://www.youtube.com/channel/UCBStHvqSDEF751f0CWd3-Pg/ Do view and subscribe

How to run interactor & solution for an interactive problem in terminal/cmd?

I tried solving B with multiset can any one tell me why am I facing TLE. Worst complexity is O(n * 2logN). Submission

Why? You iterate from 2 to $$$n-1$$$ and inside you iterate over all positive numbers. It's $$$O(n^2)$$$

My approach was to for every i from 2 to n — 2, find the max, min product of two numbers before ith and max, min product after ith of two number. Then multiply them accordingly based on the value of element i. I did not understand how it is O(n^2).

Because at $$$i = 2$$$ you iterate all positive numbers ($$$solvemin(muls)$$$). At $$$i = 3$$$ you iterate all positive numbers except maybe one. At $$$i = 4$$$ you iterate all positive number except maybe two and so on.

And besides you pass multisets by value. So you at each step copy multisets.

P.S. Yes I see now that you iterate at max over two numbers. So copying multisets is you main problem.

In solvemin(muls) I am breaking after I get two numbers from the multiset.

For problem C — if there are two centroids then we disconnect the leaf of 1 centroid and connect it directly to the other centroid. How does this not works ?

you always print 1 2 if there is just one centroid and that s the mistake. The edge 1 2 may not exist

there might be three centroids.

nope , always are 1 or 2

why though ? (im a beginner)

i don't know exactly how to explain. Tree doesn't contains any cycle so you can't have the 1-2 2-3 3-1 case when all nodes are centroids. Wait for the tutorial i m sure that will be a proof.

Whhhhhaaattttttttttt, whyyyyy,come on codeforces (ノ｀Д´)ノ彡┻━┻

yeah..i made the same mistake but i fixed it after 30 minutes and 2 WA :|

I'm so mad at my stupidity.

Deleted.

I did same and it worked for me.

I did the same and it worked for me. https://codeforces.com/contest/1406/submission/92632511

https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/

My approach to problem E:

$$$Ans=p_1^{k_1}\cdot p_2^{k_2} \cdots$$$.

For $$$p \le \sqrt{n}$$$, just ask their exponents directly. This costs about $$$180$$$ queries.

For $$$p \ge \sqrt{n}$$$, there is at most one prime factor. So remove $$$\sqrt{cnt_{primes}}$$$ primes in one time and check if the number of remaining numbers is wrong. If wrong, the factor is in the last $$$\sqrt{cnt_{primes}}$$$ numbers you remove.

There are $$$cnt_{prime}+2\sqrt{cnt_{prime}}+41+66\times 2$$$ queries at most. The time complexity is $$$O(n \ln n)$$$. Since the system test isn't finished, I don't know if it's correct.

edit: passed system test.So you would ask the exponents in a descending manner? Like for example $$$2^{19}$$$, $$$2^{18}$$$, ..., $$$2^1$$$, and see if one of them matches?

Asking in increasing manner is better (i guess). A number $$$n$$$ can be written as the product of at most $$$log(2, n)$$$ prime factors ($$$<20$$$ for $$$n=100000$$$), which means you will stop at the first query for most of the primes.

You can also binary search on exponent. But it isn't necessary here.

Ascending or descending anything is fine.

Ive done the exact same thing but ive get WA(i checked it it was because of the number of queries) and cntprime is around 9700 right!?

There are 9592 primes exactly. I think 408 queries are enough to do other things.

can you please explain with an example . I didn't get for p>=sqrt(n) . What is cnt_primes ?

make this round unrated as B problem's code can be easily found,here are the links-> https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/ https://atcoder.jp/contests/abc173/tasks/abc173_e https://www.codechef.com/problems/MMPROD

all those who have copied these code or have taken the logic from there are at an edge ahead of those who are solving problem honestly

Unrate this contest

Ok. Now shut up

Is LONG_MIN not defined in codeforces?

$$$#include <climits>$$$

I had included this header

So you can use LONG_MIN. Show submission where you cannot use it?

https://codeforces.com/contest/1406/submission/92634664

Its compiling, its just WA

$$$LONG\_MIN$$$ is minimum for $$$long$$$ type

$$$long$$$ type is signed type that at least same size as $$$int$$$. $$$int$$$ type is signed type that at least 2 bytes.

Sizes of that types are different between compilers. So maybe you want use fixed size types.

If you want minimum of 8-byte signed type. You should use i64 and $$$numeric\_limits<int64\_t>::min()$$$ or $$$INT64\_MIN$$$

Try

`LLONG_MIN`

I think LONG_MIN is same as INT_MIN on codeforces, because one time I was getting wrong answer using LONG_MAX and had to change to LONG_LONG_MAX

Yeah because of this I kept getting WA on test 3 for problem B. I think this sould be fixed by codeforces

am i the only one who printed "1 2\n1 2" if there is only one centroid in C?

that doesn't necessarily work because edge 1 — 2 may not exist in the input tree

a solution is just to save an edge from the input and output that one instead

yeah, that was my only mistake

same , cost me 4 wrong answers.

Is it possible to solve prob/B with dp ? Thank you for your responses but is there a recurssive dp solution ?

Definitions:

$$$mx[i][r]:$$$ max product of $$$r$$$ numbers in the prefix $$$1...i$$$.

$$$mn[i][r]:$$$ min product of $$$r$$$ numbers in the prefix $$$1...i$$$.

Now, the transitions:

$$$mn[i][r] = \min(mn[i-1][r], \ mn[i-1][r-1]\cdot a[i], \ mx[i-1][r-1]\cdot a[i])$$$.

$$$mx[i][r] = \max(mx[i-1][r], \ mn[i-1][r-1]\cdot a[i], \ mx[i-1][r-1]\cdot a[i])$$$.

Now, base cases:

$$$mn[0][0] = mx[0][0] = 1$$$;

$$$mn[0][r] = INF$$$ and $$$mx[0][r] = -INF$$$ for $$$r \neq 0$$$;

mn[0][0] and mx[0][0] need to be initialized with 1, not 0

Yes, my bad.

yes

https://codeforces.com/blog/entry/57593 By just applying this links code to tree C problem can be solved

What's next after finding the centroids of the tree using the code in the tree? The problem isn't solved yet!

Finding the centroids are just a helping tool and not the full solution

If $$$N$$$ is odd, is it always the case that the answer is printing a random edge twice (unique centroid)? Can anyone help me in proving that or hacking my submission?

I have tried to find a sub-tree of size exactly $$$N / 2$$$ if $$$N$$$ is even.

My submission: 92627903

Indeed if N is odd, there can be only 1 centroid. Thus printing a random edge twice is valid.

Then finding a subtree of size N / 2 if N is even, is also correct. Since those are the only nodes that satisfy sz[node] >= N / 2 and N — sz[node] >= N / 2 -> node is a centroid.

Congratulations! Your solution is correct!

(I might be wrong though feel free to correct me)

IMO the tricky part of the solution is to find the centroid as after finding centroids we can remove any subtree from one centroid and connect it to the another centroid

And do you have proof of that? Does it always work?

You may have found that as intuitive, but many others don't

The problem tests you if you found that property, not testing you how to find centroids

Yeah applying this code gives you AC but only "JUST" applying wont get you anywhere ,this is the basic difference between having a particular tool and knowing where and how to use it.

In problem C, how can we prove that there can at max 2 centroids? I took this as an assumption (based on observation) and fortunately it turned out be correct.

This is my personal proof for this. Correct me if I'm wrong.

Proof: assume that there's k > 2 centroids. Then, those k vertices should be connected by (k-1) edges (since they should form a tree). But because by Pigeonhole Principle, at least 1 vertex will certainly get 2 or more subtrees (from the other centroids). Then the centroid would be that vertex. Contradiction.

Hence, the number of centroids should be at max 2.

Sorry to ask, in your proof why do the k vertices needed to be connected themselves (i.e. connected by k-1 edges)

If two centroids vertices are connected via a non-centroid vertex, then it would be more optimal to take the non-centroid vertex as the centroid rather than the initial ones.

Probably it would be easier to visualize it by drawing some (possibly random) trees.

So thanks for the nice constest tianbu, and now you may come up with the turorial.

I love problem D so much and thanks for this wonderful contest!

By the way,E is really hard...maybe I will try it later.

Difficulty gap: A<B<C<<D<<<<<E