orz reirugan for VIP testing

Just a tip, asking for upvotes on online forums typically doesn't go well...

Alright, I'm not joking, I will definitely skip this round! Look at what you all did! Now I'm on the bottom of the last contribution page https://codeforces.me/top-contributed/friends/false/page/341 :(

Stop leaving trash comments!

thats rough buddy

orz orz orz orz orz sammyuri for MVP testing orz orz orz orz orz orz orz orz orz orz orz orz orz orz orz orz orz orz orz orz

I am not Mohamed Light twin :wilted_rose:

I believe they will be sent to chromate's basement, where they will be forced to create new problems non-stop (how do you think he pumps out these contests so fast?)

Yeah i also want to break my all time high in this round.

It is your mistake. They had clearly mentioned days in advance the exact from and to time of maintenance, which was not covering the whole day. If you really cared so much, you would have fixed a convenient time to solve at least one problem to maintain your "streak".

But sure, lets blame it on the maintainers for doing their job, and not your ignorance :)

I found it very tricky but then I printed possible values and observed something and then just guessed based on that

my observation was if we break string into chunks of 2.. both chars can't be same in a chunk .. and it passed pretests

for me B >> both C

how did you do B

man it wasnt a guess you simply had to figure out maths in a x can be just 4x+3y+2z and y as -x and z if you will run through example youll realise you need 4 only when y is negative you can make y as zero d wasnt guess again too

nah on B, I did dp :)

A is hard ngl QAO B is fun C1 is a little bit tricky, but i guess it's my problem C2 is easy when you solve C1 D just trick me with the sample testcase lol

can you tell logic for B please ?

realize that using the 1st operation and the 3rd operation is the same than using 2 time the 2nd operation, so that makes you realize that when u use 1st operation u just ignore 3rd and work with 1 and 2 op, and viceversa, after that u just make the 'y' be the one u want and check if u can make the 'x' to be what u want too, then if u can it's 'yes' otherwise 'no'

Let $$$A, B, C$$$ be the number of moves of the mentioned types. Then we must have $$$(x, y) = A\cdot (2, 1) + B\cdot (3, 0) + D\cdot (4, -1),$$$ which is a system of equations over non-negative values $$$A, B, C$$$. Working out the system, you will get

$$$B = (x - 2y) / 3 - 2C$$$ and $$$A = y + C$$$.

Note that knowing the $$$C$$$ we will know $$$B, A$$$ using the above relations, in other words all solutions can be parameterized by $$$C$$$. We must ensure that

• $$$C$$$ is non-negative

• $$$(x - 2y)$$$ is divisible by $$$3$$$

• $$$B$$$, $$$A$$$ are non-negative, which gives lower and upper bound on $$$C$$$ as $$$(x - 2y) / 6 \geq C$$$ and $$$C \leq -y$$$

checking that all above conditions hold, gives the answer

It was quite simple first bring y to 0 and change x accordingly and then check if x%3=0 ie if you can reach x with jumps of 3 which is 2nd operation.

Think of it as a 1D problem instead of 2D. First try to reac y from 0. y can be reached from 1st operation (if y>0) or 3rd operation (if y<0). after reaching y check if the x we reached after reaching y <= target x or not (if not then return NO). now we just need to check if we can reach to target x or not. there is two type of horizontal movements we can do.

use operation 2 to move 3 steps

use operation 1 and 2 both and move 6 steps

as 6 is a multiple of 3 itself, there is no point of using operation 1 and 2.

so simply just check if the remaining distance in x direction is a multiple of 3 or not!

$$$O(n^2)$$$ does not pass

I kind of wanted to make them realize that sum of $$$d(n)$$$ is $$$\mathcal{O}(n \log \log n)$$$ and therefore the time complexity is fine, and then there's bitset solutions that pass within the TL unfortunately

That being said all of our intended (including tester's) solutions pass in around $$$1.5$$$ s or less