Codeforces Round 1091 (Div. 2) and CodeCraft 26

Solutions for Codeforces Round 1091 (Div. 2) and CodeCraft 26 (contest 2217). 0/4 problems verified against sample I/O. Difficulty range: 1000-2600.

4 items

Codeforces Round 1091 (Div. 2) and CodeCraft 26

Type: Div. 2 | Problems: 4 | Verified: 0/4 | Rating range: 1000-2600 | Time: 7m 34s

Problem	Name	Rating	Tags	Solve Time	Verified
B	Flip the Bit (Easy Version)	1000	greedy, implementation	2m 15s	✗
D	Flip the Bit (Hard Version)	1900	greedy, implementation, math	1m 50s	✗
E	Definitely Larger	2000	binary-search, constructive-algorithms, data-structures	1m 39s	✗
G	Down the Pivot	2600	combinatorics, dp, math	1m 50s	✗

Practice › Codeforces Solutions › Codeforces Round 1091 (Div. 2) and CodeCraft 26 ›

CF 2217G - Down the Pivot

We are working with rooted binary trees where each node carries a binary label, either zero or one. The tree structure is arbitrary as long as every node has at most two children. On top of that structure, we assign labels independently.

Practice › Codeforces Solutions › Codeforces Round 1091 (Div. 2) and CodeCraft 26 ›

CF 2217E - Definitely Larger

We are given a fixed permutation p of size n. Alongside it, we must construct another permutation q of the same size.

Practice › Codeforces Solutions › Codeforces Round 1091 (Div. 2) and CodeCraft 26 ›

CF 2217D - Flip the Bit (Hard Version)

We are given a binary array where each position contains either 0 or 1. A subset of positions is marked as special, and all special positions initially share the same value, call it $x$.

Practice › Codeforces Solutions › Codeforces Round 1091 (Div. 2) and CodeCraft 26 ›

CF 2217B - Flip the Bit (Easy Version)

We are given a binary array of length n and a single special index p (since k=1 in this easy version). The bit at this index is considered "correct," and the goal is to make the entire array equal to this bit by performing the fewest number of flip operations.