Hello, Codeforces!

Artyom123, Kotehok3 and I are happy to invite you to Codeforces Round #671 (Div. 2), which will be held on Sep/19/2020 17:35 (Moscow time). **This round will be rated for all participants, whose rating is lower than 2100** (We really hope so).

We would like to thank:

- isaf27 for coordinating the round and helping us <3
- dorijanlendvaj, physics0523, kdjonty31, -____-, postscript, Osama_Alkhodairy, Mohamed.Sobhy, morzer, saurabhshadow, thenymphsofdelphi, MagentaCobra, PM11, MarwanNabil, hugopm, Kirill22 for testing the round and giving us useful feedback
- MikeMirzayanov for great Polygon and Codeforces platforms

You will be given **6 problems** and **2 hours** to solve them (or to play Valorant, because the contest is related to it for some reason). **One problem has an easy version and a hard version**.

You will have to help the agents from the game to grind that rating. Try your best as if you are playing ranked and they are your teammates!

gl hf

Here is the score distribution:

**500** — **750** — **1250** — **(750 + 1000)** — **2250** — **3000**

**UPD:** Editorial

I hope that statements are short with strong pretests.

As a tester, I would say that...Are the problem statements long because of the stories?

Thanks for being honest. I won't participate :)

wowwww

Ameen

Guess what,i can see a lot of solutiions getting rejected in main tests. :((

Wrong answer on test 68 for problem A. Strong pretests? I think not...

"This round will be rated for all participants, whose rating is lower than 2100

(We really hope so).":(

He tries to tell that if it get unrated like the previous round then no rating change

You must remember — peaker has an advantage!

The first 6 problem round in such a long time!

That day will be my birthday, I hope i can do well on that contest and make the day more memorable...

Happy Birthday!

Happy birthday!!

I hope that I can do well along with you...

Happy Birthday! Have Fun!

happy birthday man

happy birthday friend. Even if contest goes bad, still enjoy ur day!

[happy birthday] :)

A true programmer gives every contest ;)

but.. sys fail on A...

Happy birthday bud!!

Happy Birthday!!!

I hope that Dindic`s algorithm will be in the round

Tired of posts about short statements and strong pretests

So what if statements are long? So what if pretests are weak? Everybody is in equal position and competitive spirit does not waver. For all I care you can remove all pretests and I will gladly participate anyways.

What I really need is short queues and strong/correct checkers. And if I can ask for more, then give me an appropriate editorial

Removing pretests will definately make queues shorter! :D (Ofcourse we won't do that)

As for the tutorial, we will try our best to make it as understandable as possible.

some old Codeforces problems have weak pretest causing high amount of hack which cause some lucky people can get free 1500 point by spamming "0" zero in his room

https://codeforces.com/blog/entry/59431#comment-430573

I always forget that people get some score for hacks... Yeah, it does sound bad

So what if pretests are weak? Everybody is in equal positionI really disagree with this. Yes, everybody gets the same pretests. However, people are hurt differently. Some people (me for example) rely on strong pretests for problems like A and B. These problems usually have multitests and generating strong pretests isn't that hard. Others are careful and don't trust the Pretests Passed verdict. They will still check their code for logical/implementational errors and corner cases. Clearly, strong/weak pretests makes a huge difference for these two types of people.

Removing pretests will definately make queues shorter! :D (Ofcourse we won't do that)shishin this seems like exactly what you did. 1 pretest for A (not including samples) and around 70 systests for A (causing such a long queue that systests still hasn't finished). I FSTed A, and I didn't want to wait for all the systests to finish so I decided to stress test myself (by copying an AC submission). Turns out that my program messed up the parity of each position when n is even. However, to my disappointment, such a submission (surprisingly) passes all the pretests. I am interested as to why the for pretests for A are so weak. An easy solution would be allowing 1000 tests and have the first 100 be max cases (to check for tle) and the rest be randomly generated binary strings of length from 1 — 10. Another option is to just generate all binary strings of lengths 1 — 9, in addition to max cases. Both seem really easy to implement and I'm curious why you didn't do so.

I had to make a single pretest in A (except the sample) to avoid long queues. The pretests were made randomly, and I thought that it was strong enough. We didn't find any problems with A's tests during the testing phase.

Even with 1 single pretest in A, you can still easily make strong pretests. Allow there to be 1000 cases, guarantee that the sum of n doesn't exceed 2e5, and make each case small to better account for various corner cases. The issue with problems like A is that special cases only appear when there are no even/odd numbers in all even/odd positions. Thus, randomly generated large tests are very bad at "hacking" solutions. Also, shishin what is the thinking behind having 70 systests for each problem? This is the slowest systest I've ever seen and it seems that a lot of the tests are unecessary.

I would've argued that I'm don't mind exchanging strong pretests for more algorithmic questions, but it seems there is not much to exchange.

I understand not wanting to waste time on A&B. I myself managed to pass pretests with an incorrect checker for that very reason

`^_^`

Of course it would be a different game without pretests, but I would play it anyway. People would work on their solutions more rigorously and really try to prove it before submitting

What I'm trying to say is that weak pretests affects everyone differently, not that

Everybody is in equal positionas you've suggested.

I did not care about weather the statements were long but the pretests on C should have been stronger. I failed it on test 61. Also from what the test case constraints were clarifying the ratings of an individual could only be in the range [-4000,4000] to which I based my solution on and that is why it did not pass. You would tell me "well why did you not ask for a clarification" well the answer is because my submission passed all the pretests thus I imagined that that was the solution.

I have no problem with the problem statements being long. 0 problem. But the pretests must be strong.

Pretty much excited :-D

VALORANT!!!!!!

Clashes with ioi (

Also clashes with IPL first match MI Vs CSK ......:(

Who cares IPL :v

I think IPL is slightly less relevant than IOI to a competitive programming website.

Also, leetcode biweekly contest is clashing with this

Gamers joining Codeforces because of Valorant .

Normal Div 2 Rounds: Score distribution will be announced shortly before the round. Div 2 Round 671: Here is the score distribution Meanwhile other testers: Are we joke to you?

Damn...

As a tester I recommend you to read all the problems.

Thank you! I will first to check the last problem!

I will try !

Thank you problem setters and testers for taking the pain to arrange the contest.

I can assure you. ;-)Hello from the other side.

At least I can say that I've tried.

To tell you I'm sorry

One other thing I couldn't care less about is the score distribution. Seriously, why people always talk about it?

The score distribution tells you roughly how hard the problems are so you can plan your strategy in advance. For example, if you have solved A and B and are stuck at C, you can try working on D1 because it should be easier (at least to the testers) compared to C.

Ok, but except for compound problems, there should never be later problems with lower difficulty rating. It contradicts the very idea of sorting problems by difficulty and most rounds don't even involve compound problems

Conflict in opinion is common to happen. Maybe, according to the problem setters, some problem is harder, but when the testing round occurs, sometimes that problem is found to be easier to the testers.

That's the significance of the scoring distribution. It maps to approximate difficulties of the problem to help the contestants.

:)I just don't see how it can help. If you want to know approximate difficulty, you just look at the number of solutions.

Well the no of accepted solutions also depends on several other factors . Like last time the round went unrated and very few people solved the problems than they would have in normal. Other factors include whether the contest is Div 1,2 or 3.

I use it to determine if I'm participating in codeforces or speedforces.

Again, during the contest you have much more reliable source of information about difficulty — number of successful solutions

got adhocforces instead

For me it was readingforces. I had to read C again and again until enlightenment

Btw the contest was a good example of lying score distribution.

Click That's why short and clear statements are always good.

It certainly affected my rating, but I don't think the round was bad because of that. It was for quite different reasons though.

Hoping to become expert in this contest.

Hey guys i have 20cm cock, wanna twerk? Ladys suck my dhildho

"

A problem has an easy version and a hard version." This hasn't happened in a long time, I'm waiting.I guess It hasn't happened this time either. Looks like you'll have to wait more

Oh, score distribution is cool. Hope the problems will also cool and there'll be no issue with the problems like the last round.

A very unique and interesting rating graph shishin xD

Hope postscript will change some problem in the last few minute, and some guy will solve it in 1 minute.

As a tester, I can confirm that problem statements contain words.

Please, no spoilers.

As a participant, I can confirm that I will participate.

As a participant, I can confirm that problem statements contains

only words but not meaning:(weak pretest! really disappointed.

Participants after knowing that A is also not easy:

That's JETT REBIBE ME!!!

Wasn't scoring distribution supposed to updated only 5 hours or so before the contest?

Why there is this rule?

There is not any rule like that khaliyar. contigo

we want a contest that is related to pubg!!!

I don't want it's so crazy.

I'm waiting for this contest. _

Why is the fourth one 750+1000 instead of 1750? Is that intentional?

thanks

CS:GO themed contest when? :p

Subtasks! After a really long time.

but those subtasks were not really subtasks, it was a joke just change 1 line of code and hard version is solved

Did not see that coming lol.

Super excited :D

When are we planning to have another Codeforces Round #657 (Div. 2) type of round, which was a nightmare more me. Also are we expecting more mathematical problems in this round?

intresting.

bol fy fca.

IPL starts today. Which team are you supporting today? #MIvsCSK

Mumbai Indians

do you watch IPL??

Super Kings

It's your local game, no one outside India watches it, so please never post it here.

Are you sure that no one outside India watches it, if you doesn't watch it?

Even if you post something like champion league or world cup. You will still get downvote because its not related to the blog. And not everyone interested with football

It is not football, it is golf.

I agree this wasn't exactly the right place to post a pic about IPL, but yeah, IPL isn't limited to Indians only. Even you are most welcome to watch it, in case you don't have other engagements and are simply looking for something entertaining! :)

Its wrong to put it in those words, even though he should not have posted it here

IPL succ!

As a novice I always thought, that when the comment has been downrated it has exactly disrespectful words or smth like that, but now I wonder who is just clicking downwote always

I'm looking forward to see "Score distribution is changed to * * * * * *

Sorry IPL, Not today.

Indians be like : Contest or IPL. XD

()

IPL will last for atleast a month so are we expecting these type of memes in every contest? I hope not.

Well, actually today was the first match amidst this lockdown and that also between two teams who meet in the finals many time. But yeah, this is not the right place for such memes

The round is also clashing with

Leetcode biweekly contest 35. Can there be possibly any change in contest time?Who else is just waiting rn for contest to start?

As an author, I`m waiting rn for the contest to start too.

Great, I am sure its going to be fun and exciting :)

I think Gamers are joining Codeforces because of Valorant :P

This will be the first contest I have ever done. I hope all participants have fun and challenging problems to solve and learn.

I thank all of you; authors for your time and energy.

I hope will not be like the last contest, because it was disappointment.

forget the past, lets live the present :D

as a tester you will enjoy this round

zer nazan Tester ke nisti. Translation: Your are not tester

doost bashin. sohrabi ro ham azyat nakonin. translate : be friends.

As a participant, I will not trust your opinion anymore.

as a participant i had not enjoyed this round at all :(

I wasnt tester :)

but i was participant :)

I'm not sure about problem statements, after they say about the "Valorant" part. Hope to have good statements. :3

Wishing everyone higher rating

god damn it I fu**ed up in A and B it could've been way better

i didn't participate but i think this contest could be greater if the problem statement in A & B was better than this

Can you please tell me how to solve D2 and E? Thank You very much.

$$$D2$$$

$$$D2$$$ can be solved in the same way as $$$D1$$$ — just place $$$a_{n/2+1}$$$ ~ $$$a_n$$$ and $$$a_1$$$ ~ $$$a_{n/2}$$$ alternatively.

$$$E$$$

$$$E$$$ can be solved like this: If $$$n$$$ has a divisor which is square number $$$k^p (k,p>1)$$$ greater than $$$1$$$, find all divisor of $$$n$$$ (let's define these as $$$a_i$$$.) which is not the multiple of $$$k$$$, and order $$$a_i \times k^p, a_i, a_i \times k, a_i \times k^2 , \ldots a_i \times k^3$$$. Repeat this to get an answer, and the minimum number of moves is $$$0$$$.

If $$$n$$$ has no such divisor, it means $$$n$$$ is equal to the product of $$$m$$$ ($$$2 \le m \le 9$$$) distinct prime numbers. And there will be $$$2^m-1$$$ divisor (except $$$1$$$), and we can consider those as combination of different prime factors — $$$b_i$$$ is multiple of $$$j_{th}$$$ prime factor of $$$n$$$ if $$$i \And (2^{j-1})$$$ is nonzero. Now we just need to rearrange the numbers from $$$1$$$ to $$$2^m-1$$$ so that the bitwise and values of all two adjacent enlements are nonzero. And for each number $$$i$$$, print $$$b_i$$$. The minimum number of moves is $$$1$$$ if $$$m = 2$$$, $$$0$$$ if $$$m \ge 3$$$. The rearrangement method can be found by considering the following sequence: $$$\left[1,3,2,0,6,4,5,7\right]$$$

Thank You, Have a Good Night!

Approach for C??

First, check if the average distance from x for each element = 0. If it is, then the answer is 0 if all elements already equal x, else 1. Otherwise, if x is in a, then the answer is 1. Otherwise, answer is 2.

ohh shit i didnt check if x is in a thing otherwise same logic. :/

you do it in atmost 2. the case for 0 is easy. 2 can be done by infecting (n — 1) elements while using the last one as offset and then infecting the last one while using any element as offset. The case for 1 was tricky and I did some guess work.

Div 2B

XD

This round is too unbalanced. In A — D you just need to figure out the idea, and don't have to know any algorithms, and around 3500 people solved all of them, but E is probably too hard for div2.

For A-D, I agree with you but E is definitely not hard. It's basic number theory and observations(constructive algo).

Hint for E : Just find prime factors of the number and try to put the product of then any two primes in between those 2 primes. Obviously, you need to handle some corner cases like prime square or number is product of only 2 primes.

Well, the gap between D and E this time was surely overwhelming...

See, it is very difficult to make a balanced contest. Actually in D2, simple greedy algorithm worked(place the smallest (n/2) elements and then remaining in the gaps between them). That's why D2 had so many submissions. I think no one in the testing team thought that such algorithm would worked, else problem D would have been definitely replaced.

This was basically speedforces

YES YES YES!!!

Thanks for a "nice" B:)

I hope Codeforces employs some proof readers for the statements. The statements should be better.

Waste so much time in B that could not focus on C even when I kind of had the idea that max 2 answer is possible. Definitely my fault but the statements could have been better.

can explain what is meant by nice stairs in the statement thanks for help

Can you explain why stairs with n = 5 is not nice stairs? We can form 3 squares of size 2, and remaining of size 1

In the shape 6 squares are used, but with n=5 (for 5 stairs) 5 squares should be used

You should for only n squares

because we have to use n numbers of squares.

The nice stairs can be covered by n disjoint squares made of cells, not n types of squares. There are 6 squares in your picture, not 5.

by mistake I downrated your comment, and now CF is not allowing me to change rating. sorry.

That's OK

thank you all for your replies. I'm really stupid

i don't see the right reply

it was given in statement every cube should have one of it's corner as stair. orange cube doesn't have any corner as stair, thus not nice.

Edit here is statement

`All squares should fully consist of cells of a staircase.`

sirearsh hmmm...nice observation. From editorial : "the minimal amount of squares needed to cover the staircase is not less than n, where n is the height of a staircase."

so if you try to cover any constraints, (count, or making cell of square as staircase), you are actually covering other itself.

For problem C when can the answer be 1 ??

When average of all arr[n] (arr[n] != x) == x

If all elements already sum to zero (after making sure that all elements are not equal to x) or if x is in a

when mean is x or there is at least one rating already equal to kjoll's rating

When the average of the elements is equal to x (fix each one so and make it x, the net rating change is 0), or when there is at least one other element equal to x (this on is already infected, so all rating changes will go to this one, and change all other values to x)

Why is it not 2 in test case 3 2 1 1 2 so before the round 1 the 3rd account is infected but how can every account be infected in 1 round?

it is 0 if everyone has rating x. Otherwise, it is 1 if either at least one other person has rating x or if the sum of the ratings of the n people is n*x. Otherwise, it is 2.

can explain what is meant by nice stairs in the statement thanks for help

I knew that when sum equal to 0 then it will be 1.But why when x is present in a[] then also result is 1?

When x is present in A[], then what we can do is make all others as X and add up the rating changes and add it to A[i] where A[i]==X;

for example 4 5 5 3 6 1

5(-2+1-4) (3+2) (6-1) (1+4) are the rating changes and all are infected

IS following approach correct for E!

find all factors then find all prime factors of it let all prime factor sequence be like a1<a2<a3. then first club all factors divisible by a1 together that is put them side by side then remaining factors divisible by a2 together then by a3 and so on I think then finding minimum answer required is easy!

Please correct me if I am wrong!

Brute force will TLE

I guarantee my complexity to be sqrt(n)*log(n) for each case and I think it will pass lets implement later on when sys tests pass!

Your approach will definitely get TLE

sqrt(n)*2e5 not sqrt(n)*log(n)

Do some paperwork you will understand how

WHy 2e5! Cant we store prime factors of that number in a vector rather than iterating overall primes again and again! I think you didn't get the idea completely! and just iterating over the prime factors and factors in a single go !. By the way correct approach is in comment below!

what you are doing now basically is sieve of 1e9

The approach will not TLE.

It is possible that i am misunderstood and. thinking of a different thing than what you both are trying to say.

So here is what we do.

Given an input n, we find its prime factors in O(sqrt(n)) time.

Once this is done, we also find the divisors of n in O(sqrt(n)) time, and classify it in accordance with the smallest prime factor dividing it.

Then, we simply print these divisors in a nice fashion.

This algorithm takes time O(sqrt(n) + tau(n)). If we sum this across all test cases, we have an upper bound of

100*[sqrt(10^9) + (2*10^5)], which is good.

Nice explanation, thanks

That is the idea I used. The first thing to note is that the answer is either 0 or 1, because gcd(lcm(a,b), lcm(b,c)) > 1 for a,b,c>1.

Now, if n has only one prime factor then the problem is easy. Otherwise, suppose the prime factors of n are p_1, p_2, ..., p_k.

Write n, then write all factors of n that are divisible by p_1, but keep p_1*p_2 for the end. Then, write all factors of n divisible by p_2 but not p_1, but keep p_2*p_3 for the end. Keep doing this.

For example, if n = 60, we can do

60,2,4,10,12,20,30,6,3,15,5

where 60 is written first, followed by multiples of 2 with 2*3 = 6 written in the end, followed by multiples of 3 that are not multiples of 2 but keeping 3*5 = 15 in the end, followed by multiples of 5 that are not multiples of 2 and 3.

The only case where an issue occurs is when n=p*q for distinct primes p,q.

So the answer is 0 for n not of the form pq for distinct primes pq, and 1 otherwise.

Thanks buddy I was on the verge !

I don't understand why I fkd up D1 ;'(

what was your approach?

I thought of putting smaller numbers first in alternate places then the remaining in the places left in case reordering is required. like if n=6 and arr[]=1 2 3 4 5 6 so answer is 2 and order is 4 1 5 2 6 3 if n=5 and arr[]=1 2 3 4 5 so ans=2 and order is 3 1 4 2 5 I still couldn't find anycase giving maximum answer other than this.

I think, you did right. I did the same and it passed both D1 and D2. Sort the list. Put first n/2 elements in one vector and last n/2 in another. Now print alternatively.

I hope you printed a[i], and not i. That is what I did the first time :P

I just saw your sol; where did you sort the array?

I didn't sort , just printed alternatively straight away. Is this wrong?

Yeah, I don't get why that would give you the optimal solution.

I did the same,I sorted the array and then took one answer arrray and kept filling it (one big one small),so that the pattern is bsbsbsbs

b:-big

s:-small

so is your solution right? then why is my failing ?

It is still in pending queue.

You can check submission.

username:-namanbhardwaj.official submission:-93254332

Enjoyed the problemset. Kept me challenged till the last minute. Thank you!

How to solve B

If you look closely a stair is valid only when the height of the longest staircase is

`2^x - 1`

,`x >= 1`

how do you even prove it? forget that how to even verify it ? it is inhumane to expect us to draw 15 staircase

same

I did it out using OEIS :p but there was no time left

True that! Very inhuman of me!!!!

I did till 7. next in line had only two possibily 13 or 15 just tried the general terms of both the cases. They made me draw it :(

Adding a proof to (B):

Proving that 2^x-1 works is obvious: note that there is a big 2^(x-1) times 2^(x-1) square with one corner on the lower right; use this, and now we are left with two smaller stairs of parameter 2^(x-1)-1.

Now suppose n is not of the form 2^x-1 and n is nice. Note that the minimum number of squares needed to cover the stairs with parameter n is at least n, because no two of the top-most cells of the n steps can belong to the same square. Therefore, each square in the decomposition of this stair should cover one top-cell in some stair. Furthermore, note that the bottom right cell and a top-most cell of any step do not belong to a common square except if n is odd, in which case bottom-most cell and the top-most cell of the middle step can belong to the same square. But this square splits the stairs into two stairs of parameter (n-1)/2. Now induct, and we are done.

I really don't get it, can you please explain what is big ? " Moreover, if n is not of the form 2^x-1, then the bottom right cell and a top-most cell of any step do not belong to a common square. " did you mean bottom left cell and why should it be of the form 2^x -1 to fulfill that?

Okay, so each unit square is a cell. Each column is a stair. A square is a union of cells that makes a square shape, as shown in the problem.

Consider a staircase with n steps. We want to partition the n*(n+1)/2 cells into squares. Let f(n) be the least number of squares required to do so. Consider the topmost cell of each stair; there are n of them . Note that no two of these topmost cells can belong to a common square. Therefore, the answer f(n) is at least n, for all n.

Suppose the answer is exactly n. Then, the n squares that partition the staircase must cover the topmost cells of each stair; you cannot have a square that misses all topmost cells of all stairs, otherwise we will need > n squares.

Now, consider the bottom right cell. The square that covers this must also cover the topmost cell of some stair. For even n, this is impossible. For odd n, this is possible only when you take a square that covers the topmost cell of the middle stair and the bottom right cell of the staircase. But upon taking this square, we are left with 2 staircases of sizes (n-1)/2 each. This means

f(n) = n => f((n-1)/2) = n-1/2. Thus n is nice implies n is odd and (n-1)/2 is nice. From this, you can conclude that n is of the form 2^x-1 for some x.

Thanks but I don't get this

" Now, consider the bottom right cell. The square that covers this must also cover the topmost cell of some stair " why so ?

If it doesn't, then you still need at least n squares to cover the topmost cells.

Every square must cover some topmost cell of a stair. Also, no square can cover two different topmost cells.

So if you want n squares to cover the staircase, it MUST be the case that EACH of the n squares in the decomposition of the staircase covers the topmost cell of some stair. Simply consider the one which also covers the bottom right cell.

Thanks for the wonderful explanation,got it now

this is elegant

u need to figure out the pattern of a nice stair.

Spoiler1st: (2^0)^2 * 2^0 2nd: (2^0)^2 * 2^1 + (2^1)^2 * 2^0 3rd: (2^0)^2 * 2^2 + (2^1)^2 * 2^1 + (2^2)^2 * 2^0 ...

Literally AdHocForces :(

B was more difficult than C +D combined for me, how do you even solve it?

same...

can explain what is meant by nice stairs in the statement thanks for help

It took me time to realize "What is a nice stair?"

can you explain what is meant by nice stairs in the statement thanks for help

I can at least say that no even number can form a nice stair. I hope I'm right

No every non-even number can't form a nice stair

may be observation(in my case) . pow(2,i)-1 is the key.

Pure Adhoc bud..

The valid nice patterns are of sequence 1,3,7,15 .... (1 << n) — 1 for the general term. Just brute force out each value until u have the blocks left. Since the no of blocks required are sigma(n) a n nice staircase hence the values in this sequence grow exponentially. Just didnt like the set today :(

for me it was easy . basically what i did is tried every n from 1 to 11. i saw that the answers were 1 3 7 which is basically just multiply with 2 then increase by 1. so i just run a loop until x is small for the next 2 * i + 1.

codeThink like this:

First you have a $$$1 X 1$$$ square only and it is the first nice stair.

So, you have a nice stair of size $$$1$$$.

Now, add a $$$2 X 2$$$ square at the bottom and a nice stair of size 1 at its left. Then you get the 2nd nice stair.

So, you have a nice stair of size $$$3$$$.

Now, add a $$$4 X 4$$$ square at the bottom and a nice stair of size 3 at its left, and you get the 3rd nice stair.

So, you have a nice stair of size $$$7$$$.

and so on...

Notice that, at each step, the size of the nice stair is $$$2 \times previousSize + 1$$$.

So, you get a $$$2^n - 1$$$ sized nice stair after n such iterations.

lol, I think you are just describing the diagram mentioned in the problem, how to prove that squares of 3X3 5X5 aren't important?

I just shared the way I did that, maybe drawing it by following what I said would help.

fine,thanks

https://oeis.org/A006095

Hi, This was my first contest. And I was able to solve only 1 problem and it passed the pretest. Will I get any ratings? My rating is zero as of now!

Thanks and regards.

Yes, you will.

how long does it take to receive a rating?

About

new account's ratings: (http://codeforces.com/blog/entry/77890)only after system testing can you test right ?

My brute force idea for E was working in 3 seconds for number of factors=700. Any idea to optimize it to pass under 1s TL or a more elegant solution?

I have a simple idea:

Say we organize numbers(except N) in blocks of their spf, first number of each block is the prime, pi, followed by it's divisors where it is the spf, and the last number of each block is always pi*(pi+1), and for the number with no next block, we put N. We get answer as 0 here. And we only get 1 boundary case where a number is of the form p1*p2 where answer is always 1.

PS : consider primes in increasing order of values

Cool idea, I did not try any such good construction(there are so many good constructions possible). Thanks!

Anytime!

I did exactly this, hard luck on implementation i guess :(

How do the ratings change in this testcase for problem C?

1 step only,

change rating +1 for first guy and -1 for second guy.

We can change it to

4 3 and the answer will be 1. Note that the first one is already affected , so we can change the rating of second and make it infected

As 3 is already infected we can use this to change 4,

to change 4 to 3(x) we need -1 and to compensate we add 1 to 3(A[0]), rating change will be {1,-1} therefore we required only 1 step (Answer).

wtf was b

can explain what is meant by nice stairs in the statement thanks for help

its basically just 1 + 3 + 7 + 15 + 31 + .... its the problem statement that confusing..

can any one why my solution for F is not working!? here is my submission it mean a lot if you find its bug : ) https://codeforces.com/contest/1419/submission/93269063

Can anyone tell me why I was having TLE in problem D2 ?

My submission: https://codeforces.com/contest/1419/submission/93268164

You wrote i<=v1.size()-2, where v1.size() is size_t type(like unsigned int). If n = 1, v1.size()-2 is -1, but because it is unsigned type, (-1) % (unsigned mod) = (unsigned mod-1), which is very big(around 2e9). So you increase i by 1, while i < 2e9. This is TLE.

P.S. To avoid this, you can simply write something like this: i + 2 <= v1.size()

check: 1 1

v1.size() returns unsigned integer. hence v1.size()-2 becomes 4294967295.

All the questions could be done in 4-5 lines. No data structure/ implementation knowledge was tested. There were essentially two pretests for B-D, you passed one if you got the idea, and failed the other (they made those specific cases where you would fail, like for C, count of x >=1 in the set will give contest max 1 ans, you would print 2).

My point is, having 1-2 questions like these don't hurt. But when all the questions feel like this, something's wrong. In a div2 contest, we need more of algorithm/ data structure based questions, not the puzzle types. And because of the puzzle nature, the difficulty order of the questions did not correspond to the number of solves. I would request the future problem setters to please take a note of this.

Quality of problems has been deteriorating rapidly here.

I was hoping one graph problem among C — E, but there was none, instead it was a puzzle game as OP said.

D1 < A < B < C = D <<< E << F

the order of your submission, lol

Lol I think this order- D1=D2<A<B<C<<<<E<<F. D1 and D2 had were just 4-5 lines solution

can you explain what is meant by nice stairs in the statement thanks for help

Took me so long to understand C right...

can explain what is meant by nice stairs in the statement thanks for help

Read the statement again. They have explained it pretty clear.

See the color coding in the diagram.There is

14x4 box,22x2 box,41x1 box.In totality there are7boxes which is equal to n.Could anyone explain the proof of B ?

can you explain what is meant by "nice stairs" in the statement thanks for help

If I am not wrong then nice stairs are the stairs with rows = 1,3,7,15... They are 'nice' because they can be constructed using 'square' shaped blocks. Also, it is quite logical to use large sized blocks for larger stairs.

"nice" stairs means a stair consists only of

`N x N`

stair cells in this problem.If you look at the highest(rightmost) column of i-th "nice" stair set, it's 1, 3, 7, 15 , ... which is

`2^i-1`

I don't get how D1 and D2 were different. I barely made any changes to pass D2 after D1

Exactly lol. After B I solved D1 in 10 mins and since it was too easy i thought it wouldn't work for D2 and went to C. After C when I thought more about D2 I realized all i had to do was calculate answer seperately

At first, I was surprised to see the submissions to problem D2 and C1 rising rapidly, this actually makes you panic when you have not solved the problem. But later I felt like someone fooled me. The problems were damn easy. Short statements strong pretests.

I couldn't understand A and B's English.It seems that I need to study English, not algorithms ...

What Problem B actually say

i read it many times but cannot understand that

Can any one say me what they actually want

its basically just 1 + 3 + 7 + 15 + 31 + .......

Actually i want to understand problems statement it's quiet unclear to me

list of nice stair is 1, 3, 7, 15, 31, .. so on..

how many different nice stair u can make with n cell you have..

if we take staircase of height 6, then it can be an answer, with 1 square of size 3, 2 squares of size 2 , and rest will be squares of size 1, they will be disjoint , only corners of same colour square will touch.....

can we solve B without searching on oeis. :)

Draw out the stair cases. I did till 7 (1,4,7 were only possible) so next in line would be 13 or 15 just check for both the cases which one matches

if we take staircase of height 6, then it can be an answer, with 1 square of size 3, 2 squares of size 2 , and rest will be squares of size 1, they will be disjoint , only corners of same colour square will touch.....

yes, look at how height of stairs is changing.

starts with 1, and then height = (height * 2) + 1

and they gave us the higher limit of this in sample. look at my submission for more help.

I did. My approach — First I observed the pattern, nice numbers are 1, 3, 7, 15 ..

this is enough to come up with formula f(i) = 2*f(i — 1) + 1. Also, if you observe how nice stairs are made, if number x is nice then you need to combine two stairs of type x with a middle square of length (x + 1) resulting in a nice stair of length 2*x + 1. Rest is simple greedy approach.

Wasted a lot of time in question B, Unable to get the Statement specially "n disjoint" but finally solved it.

Problem $$$A$$$:

~~Digit~~FST GamesGoodbye, master~

im feeling bad for ur C too.

Me too...(cry silently)

Thanks, anyway

Sorry, but what is FST? Some kind of error?

Answering to myself, it must be Failed System Testing...

FailedSystemTestSolutions which were wrong but passed the pretests, will (almost always) receive FST.

If you get FST on a task, you'll lose all scores on it.

Yeah, I know. Couldn't figure out the abbreviation

For some reason I thought it was something like

Memory Limit Exceeded / Time Limit Exceeded and I search those among the error list

Or

StackOverflow / Integeroverflow

looks like we tried same shit for D2 at first attempt lol.

Problem C: 4 38 -21 83 50 -59 can anyone just EXPLAIN me how answer is 2?

See if a rating is infected then basically you don't care about it afterwards. So use it to make every other rating to infected.

suppose I make A[0] infected, then how he will infected others ??

in the statement, they said all changed need to be 0. then I need to decrease/Increase the same amount of rating from another one.

can you please explain to me how it's happening? how he will infect others if I infected only one.

So A[0] is infected, now make all non-infected rating equal to A[0]. For this, you have to increase or decrease other ratings. Calculate the net change and subtract that value from A[0], that's all, other ratings are also infected. See we can do this because A[0] is already infected, and it is mentioned in the problem statement that once a rating is infected, it will remain infected forever, So we change A[0] to any value to satisfy the other condition that, some of all changes in ratings after a contest should be zero. If you still didn't get it I can explain with an example.

thanks brother, got it. i just didn't notice the "New ratings can be any integer", I thought rating is fixed for everyone. :( :( :(

thanks again.

First contest ratings are [38, 38, 38, whatever] so you infect 3 people. And in second contest last person is infected.

how it become -21 = 38, 83 = 38, 50=38 ??

all rating changes must be equal to zero???

The total of 4 people ratings [-21, 83, 50, -59] should stay the same yes. -21 + 83 + 50 + -59 = 53 so after the first contest ratings will be [38, 38, 38, X] and 38 + 38 + 38 + X = 53. So X is -61. [-21, 83, 50, -59] became [38, 38, 38, -61] after first contest.

thanks brother. :)

or I think I didn't get the problem correctly :(

more simply you can get all ratings to x except for the last guy, and compensate it all on the last guy to make whole difference 0. and then do last guy, make him compete with any one and make last guy infected as he is the only one remaining. that is case for 2, (only 2 operations needed).

if the last guy == x, he was already infected, that is case for 1 (only one operation needed) else 2.

more formally

`all elements == x: output 0`

`if (count x in array > 0 || sum == x) output 1`

`else 2`

got it brother, thanks. :)

Test Case 68 on A was nasty.

Also, there wasn't enough of a distinction between D1 and D2. They were both just basically greedy brute forces. I solved D1 without even looking at D2, and I then solved D2 by changing 1 line.

Overall, this was a bad round.

hey how are you able to see the test case?

They are open after the contest.

what brute force?

I did left the contest early because of bad statements A+B.

True. Read B like 4-5 times and finally did what i could understand from the explanation.

Thank you for your effort.

(750 + 1000) for D1 + D2, but each one of them is actually easier than B. In the previous contests, it is used to be counted as: if you solved D1 you deserve 750, then solving D2 gives you the complete degree because you solved the difficulty that deserve 1750. But today, you can get 1750 by solving two problems, each of them is easier than the problem that deserve 750 in this contest.

In other words, my mind used to think of D2 as 1750 not as 1000 or lower.