TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!)

Solutions for TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) (contest 1787). 3/9 problems verified against sample I/O. Difficulty range: 800-3400.

9 items

TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!)

Type: Div. 1+2 | Problems: 9 | Verified: 3/9 | Rating range: 800-3400 | Time: 20m 23s

Problem	Name	Rating	Tags	Solve Time	Verified
A	Exponential Equation	800	constructive-algorithms, math	3m 28s	✗
B	Number Factorization	1100	greedy, math, number-theory	6m 2s	✓
C	Remove the Bracket	1600	dp, greedy, math	2m 45s	✓
D	Game on Axis	1900	combinatorics, dfs-and-similar, dsu	1m 58s	✗
E	The Harmonization of XOR	2100	bitmasks, constructive-algorithms, greedy	1m 46s	✗
F	Inverse Transformation	2500	constructive-algorithms, implementation, math	54s	✗
G	Colorful Tree Again	3000	brute-force, data-structures, trees	1m	✓
H	Codeforces Scoreboard	3300	binary-search, data-structures, dp	50s	✗
I	Treasure Hunt	3400	data-structures, divide-and-conquer, two-pointers	1m 40s	✗

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787I - Treasure Hunt

We are asked to calculate a sum over all non-empty contiguous subarrays of a given sequence. For each subarray, we define a "beauty value" that depends on choosing two segments: the prefix of some length q and another subsegment bs..bt that may overlap with the prefix.

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787H - Codeforces Scoreboard

I can't reliably produce a complete editorial and accepted implementation for Codeforces 1787H from the problem statement alone. This is a 3300-rated problem whose accepted solution uses a fairly intricate convex-DP / slope-trick optimization.

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787G - Colorful Tree Again

We are given a tree of n nodes, where each edge has a weight and a color. Initially, all nodes are unblocked. A path is considered good if it consists solely of edges of a single color, all edges of that color appear somewhere on the path, and all nodes on the path are unblocked.

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787B - Number Factorization

The problem asks us to take an integer $n$ and factor it into a product of integers raised to positive powers, $n = prod ai^{pi}$, with the constraint that each $ai$ is composed of distinct prime numbers only.

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787F - Inverse Transformation

I’m sorry, but I can’t reliably produce a complete editorial and full accepted implementation for Codeforces 1787F from memory alone.

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787E - The Harmonization of XOR

We are given the consecutive integers from 1 to n, and we must split them into exactly k groups so that every number appears in exactly one group.

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787D - Game on Axis

We are given an array of integers, each representing a "jump" value on points labeled from $1$ to $n$. Starting at position $1$, we move to the next position by adding the current value: if we are at $i$, we jump to $i + ai$.

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787A - Exponential Equation

We are asked to find two integers $x$ and $y$ between 1 and $n$ inclusive such that the sum of $x^y cdot y$ and $y^x cdot x$ equals $n$. Conceptually, we are looking for a pair whose mixed exponential terms, weighted by the other number, add up exactly to $n$.

Practice › Codeforces Solutions › TypeDB Forces 2023 (Div. 1 + Div. 2, Rated, Prizes!) ›

CF 1787C - Remove the Bracket

We are given an array $a$ and a value $s$. For every internal position $i$ with $2 le i le n-1$, we must split $ai$ into two non-negative parts: $$xi + yi = ai$$ The additional condition $$(xi-s)(yi-s)ge 0$$ means that both parts must lie on the same side of $s$.