We're thrilled to invite you to participate in our upcoming **Brain Booster #6** programming contest at **SeriousOJ**.

Details:

Problem setter : Bullet, thakuremon, _MJiH_

Problems: 8 (one problem will have subtasks)

Duration: 2:15 hours

Difficulty: Similar to Codeforces Div 3

Special thanks to CLown1331, owner of **SeriousOJ**.

Registration: Please register for the contest on **SeriousOJ**. If you encounter any issues on the register, be sure to check your spam folder for the confirmation email.

Good luck & Have fun!

**Update: Congrats to the winners**!

Hope all participants enjoy the problemset and have fun :)

Excited for this round !!

excited to participate in this contest

The problem set is thoughtfully designed and well-presented. We hope all participants will thoroughly enjoy this contest!

1 hour left for the contest....

ok

Very Much Excited for the contest & Hopefully for the "Plater"Please consider adding rust to the list of available languages

Hi Egor

Rust is availabe now.

https://judge.eluminatis-of-lu.com/records/66fed389ebfacc0008b79bbf

Thanks

Here is rough editorial incase anyone wants it

A: N/3 + (N%3 > 0)

B: xorsum

C: Say e evens o odds. Let q = min(e,o)//2*2. Game is same if (e-q, o-q) and lasts < 4 turns, so simulate.

D. dp[i] answer from position i. Lets maintain minat[d] = min(dp[i] for i if A[i]%d==0). Now for i=N-1..0 for d dividing A[i], we can update dp[i] by dp[i+1] + K, or (for d dividing A[i]) minat[d] + A[i]/d. Finally we can update minat[d].

E. Let dp[i][s] be the least cost to make s "abc"s from S[i:].

F. prefix xor sums, want P[i]^P[j] = P[j]^P[k], so for every P[i] = P[k] we add k — i — 1. For each P[i]=v lets info[v][0] += 1, info[v][1] += i. So that our answer increases by info[P[k]][0] * (k — 1) — info[P[k]][1].

G. For each position we want to know how many lower or higher on the left and right. We can use a fenwick tree and sort the queries. For the hard version we want mergesort tree.

H. segtree with lazy prop to support toggle. use euler tour to flatten the tree first and convert subtrees to subarrays.

congratulations

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