Codeforces Round 173 (Div. 2)

5 problems from Codeforces Round 173 (Div. 2) (contest 282), difficulty 800-2200. 4/5 solutions verified against sample I/O.

5 items

Codeforces Round 173 (Div. 2)

Div. 2 | 5 problems | 4/5 verified | Difficulty 800-2200 | 8m 47s

# Problem Rating Tags Accepted Time ✓
A Bit++ 800 implementation 358,005 1m 16s ✓
B Painting Eggs 1500 greedy, math 13,971 2m 15s ✓
C XOR and OR 1500 constructive-algorithms, implementation, math 10,178 1m 34s ✓
D Yet Another Number Game 2100 dp, games 1,614 2m 14s
E Sausage Maximization 2200 bitmasks, data-structures, trees 4,323 1m 28s ✓
Practice › Codeforces Solutions › Codeforces Round 173 (Div. 2) ›

CF 282A - Bit++

The problem presents a toy programming language with only one variable, x, initially set to zero. Each line of the program is a statement that either increments or decrements this variable.
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Practice › Codeforces Solutions › Codeforces Round 173 (Div. 2) ›

CF 282B - Painting Eggs

We are asked to distribute a sequence of eggs between two children, A and G, where each egg has an individual cost for each child. The key constraints are that the total paid to A and the total paid to G must not differ by more than 500.
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Practice › Codeforces Solutions › Codeforces Round 173 (Div. 2) ›

CF 282E - Sausage Maximization

We are given a single array of integers, which represents a sausage in Bitland. The goal is to cut this sausage into two non-overlapping segments: a prefix for BitHaval and a suffix for BitAryo. Either segment can be empty.
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Practice › Codeforces Solutions › Codeforces Round 173 (Div. 2) ›

CF 282D - Yet Another Number Game

We are asked to determine the winner in a turn-based game with a very small sequence of integers, where two players alternate moves. On each turn, a player can either reduce a single element by any positive amount or reduce all elements by the same positive amount.
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Practice › Codeforces Solutions › Codeforces Round 173 (Div. 2) ›

CF 282C - XOR and OR

We are given two binary strings. An operation chooses any adjacent pair of bits. If the pair is (x, y), we compute: - p = x xor y - q = x or y We then write p into one position and q into the other position, in either order. The length of the string never changes.
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