Innopolis Open 2025-2026. Elimination Round 1

5 problems from Innopolis Open 2025-2026. Elimination Round 1 (contest 106158), difficulty -. 5/5 solutions verified against sample I/O.

5 items

Innopolis Open 2025-2026. Elimination Round 1

Special | 5 problems | 5/5 verified | Difficulty - | 5m 27s

#	Problem	Time	✓
A	Shustrik, Persik, and Eternal Friendship	1m 7s	✓
B	Magic Circle	47s	✓
C	Pleasant Paths	1m 10s	✓
D	Hamster Spectacle	1m 8s	✓
E	Bracket Dance	1m 15s	✓

Practice › Codeforces Solutions › Innopolis Open 2025-2026. Elimination Round 1 ›

CF 106158B - Magic Circle

We have $n$ wizards arranged in a circle, indexed from 1 to $n$. Each wizard starts with a very large identical mana value, so initially all wizards are tied. Then we apply $q$ rituals. Each ritual defines a sequence of affected wizards.

Practice › Codeforces Solutions › Innopolis Open 2025-2026. Elimination Round 1 ›

CF 106158D - Hamster Spectacle

We are given a sequence of positive integers arranged in a line. Each position represents a hamster, and each hamster contributes a set of “skills”, where a skill is simply any divisor of its number.

Practice › Codeforces Solutions › Innopolis Open 2025-2026. Elimination Round 1 ›

CF 106158E - Bracket Dance

We are given a balanced bracket string of length $n = 2^k$. The only allowed transformation is applied at a chosen scale $m = 2^x$. We partition the string into consecutive blocks of size $m$, and inside each block we split it into two halves of equal size.

Practice › Codeforces Solutions › Innopolis Open 2025-2026. Elimination Round 1 ›

CF 106158C - Pleasant Paths

We are working with a rooted tree where the root is fixed at vertex 1. Each edge has a weight that behaves like a “popularity” value. A walk always starts at the root and repeatedly descends until it reaches a leaf.

Practice › Codeforces Solutions › Innopolis Open 2025-2026. Elimination Round 1 ›

CF 106158A - Shustrik, Persik, and Eternal Friendship

We are given a string of digits arranged in a circle, and we are allowed to rotate it any number of times. Each rotation is one cyclic shift either left or right.