Codeforces Round 671 (Div. 2)

Solutions for Codeforces Round 671 (Div. 2) (contest 1419). 4/7 problems verified against sample I/O. Difficulty range: 900-2800.

7 items

Codeforces Round 671 (Div. 2)

Type: Div. 2 | Problems: 7 | Verified: 4/7 | Rating range: 900-2800 | Time: 14m 47s

Problem Name Rating Tags Solve Time Verified
A Digit Game 900 games, greedy, implementation 1m 13s ✓
B Stairs 1200 brute-force, constructive-algorithms, greedy 1m 37s ✓
C Killjoy 1500 greedy, implementation, math 6m 23s ✗
D1 Sage's Birthday (easy version) 1000 binary-search, constructive-algorithms, greedy 1m 16s ✓
D2 Sage's Birthday (hard version) 1500 binary-search, brute-force, constructive-algorithms 59s ✓
E Decryption 2100 constructive-algorithms, implementation, math 1m 37s ✗
F Rain of Fire 2800 binary-search, data-structures, dfs-and-similar 1m 42s ✗
Practice › Codeforces Solutions › Codeforces Round 671 (Div. 2) ›

CF 1419C - Killjoy

Got it. Let’s carefully reason through this failure. The input: The expected output: The actual output from the previous solution: Clearly, the previous solution underestimates the counts.
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CF 1419F - Rain of Fire

Fix a value t. Brimstone may wait at any detachment for an arbitrary amount of time. When he decides to move, he chooses one of the four axis directions and keeps moving until he reaches the first detachment lying on that ray.
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CF 1419E - Decryption

The problem revolves around arranging the divisors of a given composite number in a circle such that no two adjacent numbers are coprime.
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CF 1419D2 - Sage's Birthday (hard version)

We have a row of ice spheres, each with a positive integer price, and Sage will pick out spheres that are cheaper than both of their immediate neighbors. The spheres at the ends are never cheap because they lack two neighbors.
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CF 1419D1 - Sage's Birthday (easy version)

We are given distinct prices of ice spheres and may rearrange them in any order. A sphere is considered cheap only if it has both a left neighbor and a right neighbor, and its price is strictly smaller than both of them. The first and last positions can never contribute.
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CF 1419B - Stairs

A staircase with $n$ columns contains columns of heights $1,2,dots,n$. The total number of cells in such a staircase is $$1+2+cdots+n=frac{n(n+1)}2.$$ The problem only cares about nice staircases.
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CF 1419A - Digit Game

We are asked to simulate a simple turn-based digit game between two agents, Raze and Breach. Each game starts with a positive integer consisting of $n$ digits. The digits are numbered from left to right, starting with 1.
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