Codeforces Round 663 (Div. 2)

4 problems from Codeforces Round 663 (Div. 2) (contest 1391), difficulty 800-2600. 2/4 solutions verified against sample I/O.

4 items

Codeforces Round 663 (Div. 2)

Div. 2 | 4 problems | 2/4 verified | Difficulty 800-2600 | 16m 7s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Suborrays	800	constructive-algorithms, math	25,062	4m 35s
C	Cyclic Permutations	1500	combinatorics, dp, graphs	16,025	1m 29s	✓
D	505	2000	bitmasks, brute-force, constructive-algorithms	5,762	7m 33s	✓
E	Pairs of Pairs	2600	constructive-algorithms, dfs-and-similar, graphs	1,926	2m 30s

Practice › Codeforces Solutions › Codeforces Round 663 (Div. 2) ›

CF 1391D - 505

We are given a binary grid and we are allowed to flip cells from 0 to 1 or 1 to 0. The goal is to transform the grid so that every square submatrix whose side length is even contains an odd number of ones.

Practice › Codeforces Solutions › Codeforces Round 663 (Div. 2) ›

CF 1391A - Suborrays

We are asked to construct a permutation of the numbers from 1 to n such that every contiguous segment behaves in a very specific way under the bitwise OR operation. For any subarray, we take all values inside it and compute their bitwise OR.

Practice › Codeforces Solutions › Codeforces Round 663 (Div. 2) ›

CF 1391E - Pairs of Pairs

We are given a connected undirected graph, and the task is not to compute a classical graph property, but to construct one of two global structures that are guaranteed to exist. The first possible output is a simple path that visits at least half of the vertices.

Practice › Codeforces Solutions › Codeforces Round 663 (Div. 2) ›

CF 1391C - Cyclic Permutations

We are given a permutation of numbers 1...n. From that permutation, a graph is constructed on the indices of the array. For every position i, we connect it to the nearest larger element on the left and the nearest larger element on the right, whenever such elements exist.