Codeforces Round 143 (Div. 2)

5 problems from Codeforces Round 143 (Div. 2) (contest 231), difficulty 800-2100. 3/5 solutions verified against sample I/O.

5 items

Codeforces Round 143 (Div. 2)

Div. 2 | 5 problems | 3/5 verified | Difficulty 800-2100 | 13m 50s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Team	800	brute-force, greedy	429,994	5m 22s
B	Magic, Wizardry and Wonders	1500	constructive-algorithms, greedy	3,915	1m 26s	✓
C	To Add or Not to Add	1600	binary-search, sortings, two-pointers	9,696	2m 58s	✓
D	Magic Box	1600	brute-force, geometry	4,332	1m 29s	✓
E	Cactus	2100	data-structures, dfs-and-similar, dp	2,320	2m 35s

Practice › Codeforces Solutions › Codeforces Round 143 (Div. 2) ›

CF 231A - Team

Three friends evaluate each contest problem independently. For every problem, we are given three values, each either 0 or 1. A value of 1 means that friend is confident they know how to solve the problem. A value of 0 means they are not confident.

Practice › Codeforces Solutions › Codeforces Round 143 (Div. 2) ›

CF 231E - Cactus

We are given a connected undirected graph that is guaranteed to be a vertex cactus. In a vertex cactus, every vertex belongs to at most one simple cycle. Cycles may touch the rest of the graph through articulation points, but two different cycles can never share a vertex.

Practice › Codeforces Solutions › Codeforces Round 143 (Div. 2) ›

CF 231D - Magic Box

We are asked to compute the sum of numbers visible on a box from a given viewpoint in three-dimensional space. The box is axis-aligned, meaning all edges run along the X, Y, and Z axes. Its minimal corner is at the origin, and the opposite corner is at coordinates $(x1, y1, z1)$.

Practice › Codeforces Solutions › Codeforces Round 143 (Div. 2) ›

CF 231B - Magic, Wizardry and Wonders

We are asked to reconstruct an initial sequence of integers given the final result of a repeated transformation. Vasya has n cards, each containing an integer between 1 and l.

Practice › Codeforces Solutions › Codeforces Round 143 (Div. 2) ›

CF 231C - To Add or Not to Add

We are given an array of integers and a limited number of increment operations. Each operation increases a single element by one, and the same element can be increased multiple times.