Pinely Round 2 (Div. 1 + Div. 2)

Solutions for Pinely Round 2 (Div. 1 + Div. 2) (contest 1863). 3/9 problems verified against sample I/O. Difficulty range: 800-3500.

9 items

Pinely Round 2 (Div. 1 + Div. 2)

Type: Div. 1+2 | Problems: 9 | Verified: 3/9 | Rating range: 800-3500 | Time: 15m 58s

Problem	Name	Rating	Tags	Solve Time	Verified
A	Channel	800	greedy, implementation	1m 54s	✓
B	Split Sort	1100	greedy, math, sortings	1m 36s	✗
C	MEX Repetition	1100	implementation, math	2m 9s	✓
D	Two-Colored Dominoes	1400	constructive-algorithms, greedy	1m 53s	✓
E	Speedrun	2100	brute-force, dfs-and-similar, dp	1m 50s	✗
F	Divide, XOR, and Conquer	2600	bitmasks, dp, math	1m 44s	✗
G	Swaps	2800	combinatorics, dp, graphs	1m 36s	✗
H	Goldberg Machine 3	3500	dp, trees	1m 35s	✗
I	Redundant Routes	3500	constructive-algorithms, dp, trees	1m 41s	✗

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863I - Redundant Routes

We are given a tree, so between any two vertices there is exactly one simple path. From this tree we want to choose several distinct vertex-sets, where each chosen set must itself be the vertex set of some simple path that contains at least two vertices.

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863H - Goldberg Machine 3

We are dealing with a complete rooted binary tree, meaning every non-leaf node has exactly two children. Each leaf has a “hunger value” representing the number of cookies it must ultimately receive. Non-leaf nodes have a selector that decides which child a cookie moves to.

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863G - Swaps

We are given an array where each position points to a value in the same range as indices. You are allowed to repeatedly pick an index i and swap the value stored at position i with the value stored at position a[i].

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863F - Divide, XOR, and Conquer

We are given an array of integers and a strange operation that repeatedly reduces the array by splitting it into two non-empty parts, computing the XOR of each part, and discarding the part with the smaller XOR value.

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863E - Speedrun

We are given a set of quests, each of which can only be completed at a specific hour within a repeating game day. Some quests depend on others, meaning they cannot be completed until certain prior quests are done.

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863D - Two-Colored Dominoes

We are given an $n times m$ board partially covered with dominoes. Each domino occupies exactly two adjacent cells, either horizontally or vertically. Some cells may remain empty.

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863C - MEX Repetition

We are given an array of distinct integers ranging from 0 up to n, inclusive. The task is to repeatedly update the array in a very specific way: for each element from left to right, replace it with the MEX (minimum excluded value) of the current array.

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863A - Channel

A channel has n subscribers. When a new post is published, exactly a subscribers are online, so those a people immediately read the post. After that, we receive a sequence of notifications. A '+' means some subscriber came online, and a '-' means some subscriber went offline.

Practice › Codeforces Solutions › Pinely Round 2 (Div. 1 + Div. 2) ›

CF 1863B - Split Sort

We are given a permutation of numbers from 1 to n, and we want to transform it into the identity permutation where each value sits in its matching index position.