Yeah we understand and acknowledge your concern, CodeAgon Schedule was scheduled around on 30 Jan but then it changed. In the short period of notice, it's very difficult to change the time and inform the same every registrant.

Hope it helps :)

can you explain your sol thoroughly? because i think two pointer will not work in this i guess.

Thanks for self-explanatory code !!

I also have linux, but use Vscode with this extension to run test cases (https://marketplace.visualstudio.com/items?itemName=DivyanshuAgrawal.competitive-programming-helper). Which will automatically terminate your program if its run infinitely with returning the error of SIGTERM.

Friend++ :)

nice idea, btw how do you have practised so u r CM today :) pls share ur tips bhai, sometimes it like ki mera saturation aa gya hai :(

Hope it will become true some day ;--;

nice !

In case odd we need to place one row of vert dominoes

I think there should be horizontal dominoes instead of vert..as in one row only horizontal dominoes can come.

for me UNO reverse :(

thanks man, got it.

the thing is as i read question in contest, i start a mathematical approach and don't try many cases, which makes a easier prob hard for me. If u have any thing to guide then pls share. :)

thanks once again :}

no probs bro, i got the editorial now, cool

can u pls explain the solution for E...pls

can u pls explain E...pls

https://mirror.codeforces.com/contest/1315

if anyone wants to take reference this is last vk cup.

can u pls refer what technique/method/algo is this? I didn't got this wholly, so want to read on this.

Firstly I thought doing this will stuck in infinite loop, but now this seems magical !!

which is root cause of so much cheating in these days.

Much waited for this but Clashing with Leetcode :( ,anyways preferred cf over all others :)

MikeMirzayanov Please ban these type of user immediately to set an example, humble request, Please :\

tnks for the round :)

tenks man

thanks man, after reading it looks simple logic but why i can't think of these logics in contest :(

noob me :{

can you please elaborate your thought about LCM..

The one who is always ONLINE XD...

synonym for error_404 :)

jk :}

u messed up some where in counting CC, just check it once.

can you please explain your approach, i didnt get it..

https://atcoder.jp/contests/arc121/submissions/22994246

This legend really writes the neat code, worth to go through it.

no, i did BS with while loop.

but i think the case u might have missed is, if for a element x[i] u found lower bound as y[j] from other array. then u should check y[j-1] also because at last we have to take min(abs(x[i]-y[j]), x[i]-y[j-1]) as our target is to minimise this difference not to find lower_bound.

I also missed this, i guess my code also failed because of this only.

Hope this help ;)

same i did bruhh.. but with binary search, but dont know where my code is failing :(

see submissions of different peoples'. u might find one. ;)

smallest and lovely solution for I Hate 1111 can be:-

as we know we have to check only for 11, 111 (from editorial),

• if we can represent x = a*11 + b*111 (where b is as low as possible), we return true
• else, we return false

And also 111 = 11*10 + 1, therefore equation reduces to x = (a+10*b)*11 + b

now b = x%11 and a + 10*b = x/11

• so if the above conditions match we output true,
• else false.

The neat sol. can be found at :- https://mirror.codeforces.com/contest/1526/submission/117617666

Hope it helps ;)

implementForces :)

oh okay, will look into it. it will be good if u can point any(link).

Thanks a lot :joined_hands.

yeah thanks but wouldn't cp algorithms website is enough?

lol

i guess by doing tons of questions :)

Ok lets say you took some no. which is not co-prime with n in p then they would have a common factor say m, now we can express p and n as

p = m*a

n = m*b

for arbitrary a,b (which are co-prime to each other)

and now for getting (p mod n) = 1 ,we should have p = k*n + 1 and if we substitute values of p and n from our equations we have.

m*a = k*m*b + 1 but clearly the rhs is not multiple of 'm' which is contradiction.hence p should have all no.'s which are co-prime with n.

hope it helps :)

wouldn't*

thanks a lot

but When its 2d dp we can visualise it by drawing, but how to do same with 3d dp? is there any method to verify that our 3d dp is filling in appropriate order?

thanks in advance :)

thanks buddy

arrange a good setup.

okay thanks, i am not use to these things but will try and report you back soon bro.

How can i check compiling options, I don't know about it? btw i use VS code and "code runner" extension to run them in VS code.

Auto comment: topic has been updated by coder_Aditya_404 (previous revision, new revision, compare).

how in my ide compiler(g++) it shows TLE, can you explain or direct me to sources where i can find more about it? Thanks

Can we solve problem D if we are allowed to swap any two piles once? It could be a variation for this problem.

Suggesions are much appreciated.

@Codeforces please apply the points which are feasible to authors.

please count me in too.

Arnab_riya

okk thanks

bro why you have made sieve that contains the no's which are not prime.

Cn we solve it by making sieve that contains prime no's instead and just checking their presence and return true if present?

thanks in advance

bro why you have made sieve that contains the no's which are not prime.

Cn we solve it by making sieve that contains prime no's instead and just checking their presence and return true if present?

thanks in advance

mast explanation diya brother :)

kis ke :)