You wrote the exact same comment in the announcement of rounds #748 and #749, but didn't participate in both of them. Will you participate in this one? XD

will see !!

why are tou skiped?? you cheated again ?!?!?!

sum of delta = 0 intensifies

At the time of the round the CheReKOSH olympiad will be held, where the problems from this round will be used!

At this point I dont know which is the usual time anymore :))

your comment makes me nervous.

Согласен с тобой, я бы тоже поднял бы рейтинг

It's better to put the meme inside a spoiler so it doesn't take so much space :)

Probably.

Yes, subtask means same problem will have different constrains and one of those can be passed with a easier solution.

Me too lol hood luck

lol i think u will be master nice job ! i had a really bad contest however

lesgooo

Maybe today's contest is early because otherwise it would have clash with CodeChef's Cook-Off !!

No, you won't.

Boruto you are annoying as ever !! First cheating in chunin exams and now this

What part of the sentence, "The round will be rated for all the participants with rating strictly less than 2100," do you not get?

there should be no zeroes in array B

Not you, todays problem A was tricky af. Not hard, just tricky. Seemed trivial at first but then...

I feel that was the whole point of that qn.. There was a way("we had to think that") to get it within the limits. I couldn't pass my code btw..

That made it interesting for odd sized array.

Most probably you must have taken the limit as sqrt(n) but it should be sqrt(2*n)

O(n*sqrt(n)) is well enough.try constant optimization.

dp[k][n], and kmax = 450

Unfortunately, it's true. That problem didn't have a multitest testing so it didn't work out to cut off as much solutions as possible using a little number of pretests. A little number of tests was needed to avoid queuing.

unfortunately yes :(

I don't think that it is possible to solve this problem with gaussian basis. The expected solution uses the fact that you have at most 5000 different values for $$$a_i$$$. You just need to solve $$$dp[k][xor] =$$$ minimum position you can reach value xor, using only $$$a_i \lt = k$$$

for Problem A : the answer is either 0 or 1 proof : let A , B , C be durations ( 1 min each ) of the 3 musics A + 2B + 3C is the total duration now consider 2 groups : ( B + C ) and ( B + C ) so we remain with A + C duration now if A + C is odd => 1 will be extra or else A + C will be equally distributed

1. First of all, it's obvious that answer can't be more than 3.
2. Answer can't be 3, because in that case we can move from the bigger subset 1 or 2 to smaller one (or, if bigger subset consists only from 3's, that means all 1's in 2's in smaller, and we can exchange 3 with 2 or 1 and make answer smaller).
3. Answer can't be 2 using the same logic. Obviously, if bigger subset has at least one 1, we can make the answer 0 easily. If answer is 2, then all 1's in smaller subset. Bigger subset consists only of 3's? Exchange 3's from that subset with 2 and 1 from smaller, afterwards move 1 to smaller. Answer is zero. Bigger subset consists only of 2's? Exchange 2 from bigger with 1 in smaller. Answer is zero. 3's and 2's? Exchange 2 with 1 from smaller.

4.Answer is obviously can be 1, because a + 2b + 3c can be odd.

Any natural (and > 0 depending on definiton) numbers a, b, c, such that a + 2b + 3c is even can be divided to equal groups.

Proof: answer can't be bigger than 1, proved that in 1. — 3. If cannot be divided, then answer is 1, and that means that sum of bigger subset and smaller subset is odd, which contradicts with what we prove. So the answer is (a + 2b + 3c) % 2.

the problem guarantees $$$a_i \ne 0$$$ for all $$$i$$$, so at least one of $$$a_1+a_2$$$, $$$a_1+a_3$$$, $$$a_2+a_3$$$ is not zero. Just find it out and solve $$$a_4$$$ to $$$a_n$$$.

Just guess the solution.

a,b,c (1≤a,b,c≤10^9) — I didn't see this limitation myself at the beginning and went to solve C.

a,b,c >= 1