Codeforces Global Round 14

Solutions for Codeforces Global Round 14 (contest 1515). 3/9 problems verified against sample I/O. Difficulty range: 800-3500.

9 items

Codeforces Global Round 14

Type: Global | Problems: 9 | Verified: 3/9 | Rating range: 800-3500 | Time: 24m 27s

Problem Name Rating Tags Solve Time Verified
A Phoenix and Gold 800 constructive-algorithms, greedy, math 3m 9s ✗
B Phoenix and Puzzle 1000 brute-force, geometry, math 3m 39s ✓
C Phoenix and Towers 1400 constructive-algorithms, data-structures, greedy 3m 46s ✗
D Phoenix and Socks 1500 greedy, sortings, two-pointers 2m 48s ✗
E Phoenix and Computers 2200 combinatorics, dp, math 1m 45s ✓
F Phoenix and Earthquake 2600 constructive-algorithms, dfs-and-similar, dsu 2m 56s ✗
G Phoenix and Odometers 2700 dfs-and-similar, graphs, math 1m 31s ✓
H Phoenix and Bits 3500 bitmasks, brute-force, data-structures 2m 40s ✗
I Phoenix and Diamonds 3400 binary-search, data-structures, sortings 2m 13s ✗
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CF 1515I - Phoenix and Diamonds

We are maintaining a multiset of diamond types where each type has a fixed weight and value per item, but its available count changes over time due to arrivals and sales. On top of this evolving collection, we must answer queries that simulate a greedy packing process.
codeforces competitive-programming binary-search data-structures sortings Jun 10, 2026
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CF 1515H - Phoenix and Bits

Method 1b already provides the key idea. If $(umldots u1u0)3$ is a ternary integer, its decimal value is obtained from the nested form $$(cdots((umcdot3+u{m-1})cdot3+u{m-2})cdots)cdot3+u0.$$ For pencil-and-paper work this means: Start with the leading digit.
codeforces competitive-programming bitmasks brute-force data-structures sortings Jun 10, 2026
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CF 1515G - Phoenix and Odometers

We are given a directed graph representing intersections in a city and one-way roads between them, each with a positive integer length. A set of cars starts at specific intersections, each with an odometer that begins at a number s and wraps around to zero after reaching t.
codeforces competitive-programming dfs-and-similar graphs math number-theory Jun 10, 2026
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CF 1515F - Phoenix and Earthquake

The problem describes a scenario where a country’s road network has been destroyed by an earthquake, leaving n cities and m roads in a disconnected state.
codeforces competitive-programming constructive-algorithms dfs-and-similar dsu graphs greedy trees Jun 10, 2026
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CF 1515D - Phoenix and Socks

Method 1b already provides the key idea. If $(umldots u1u0)3$ is a ternary integer, its decimal value is obtained from the nested form $$(cdots((umcdot3+u{m-1})cdot3+u{m-2})cdots)cdot3+u0.$$ For pencil-and-paper work this means: Start with the leading digit.
codeforces competitive-programming greedy sortings two-pointers Jun 10, 2026
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CF 1515E - Phoenix and Computers

We have a row of n computers, all initially off, and Phoenix wants to turn all of them on. He can manually switch on any computer that is currently off, but with a twist: if a computer has both its neighbors already on, it will turn on automatically without manual intervention.
codeforces competitive-programming combinatorics dp math Jun 10, 2026
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CF 1515C - Phoenix and Towers

We are given a set of n blocks, each with a height hi. Phoenix wants to construct exactly m towers using all the blocks. Each tower's height is the sum of its blocks, and a "beautiful" arrangement requires that no two towers differ in height by more than x.
codeforces competitive-programming constructive-algorithms data-structures greedy Jun 10, 2026
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CF 1515B - Phoenix and Puzzle

We are given a set of identical right isosceles triangular puzzle pieces. Each triangle has two equal sides and a right angle. Phoenix wants to use exactly $n$ of these triangles to form a perfect square, without overlaps or holes.
codeforces competitive-programming brute-force geometry math number-theory Jun 10, 2026
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CF 1515A - Phoenix and Gold

Method 1b already provides the key idea. If $(umldots u1u0)3$ is a ternary integer, its decimal value is obtained from the nested form $$(cdots((umcdot3+u{m-1})cdot3+u{m-2})cdots)cdot3+u0.$$ For pencil-and-paper work this means: Start with the leading digit.
codeforces competitive-programming constructive-algorithms greedy math Jun 10, 2026