Codeforces Round 825 (Div. 2)

Solutions for Codeforces Round 825 (Div. 2) (contest 1736). 3/6 problems verified against sample I/O. Difficulty range: 800-2600.

6 items

Codeforces Round 825 (Div. 2)

Type: Div. 2 | Problems: 6 | Verified: 3/6 | Rating range: 800-2600 | Time: 18m 58s

Problem	Name	Rating	Tags	Solve Time	Verified
A	Make A Equal to B	800	brute-force, greedy, sortings	1m 57s	✗
B	Playing with GCD	1200	math, number-theory	4m 30s	✓
C1	Good Subarrays (Easy Version)	1300	binary-search, data-structures, schedules	2m 6s	✓
C2	Good Subarrays (Hard Version)	2400	binary-search, data-structures, dp	5m 4s	✓
D	Equal Binary Subsequences	2200	constructive-algorithms, geometry, greedy	3m 19s	✗
E	Swap and Take	2600	dp, greedy	2m 2s	✗

Practice › Codeforces Solutions › Codeforces Round 825 (Div. 2) ›

CF 1736C2 - Good Subarrays (Hard Version)

We are asked to analyze an array of positive integers and count subarrays that satisfy a specific property: each element in the subarray is at least as large as its 1-based position within that subarray.

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CF 1736E - Swap and Take

Codeforces 1736E: Swap and Take

Practice › Codeforces Solutions › Codeforces Round 825 (Div. 2) ›

CF 1736D - Equal Binary Subsequences

We are given a binary string of length $2n$. The task is to split the indices into two groups of size $n$ so that if we read characters in each group in increasing index order, both groups produce exactly the same binary string.

Practice › Codeforces Solutions › Codeforces Round 825 (Div. 2) ›

CF 1736B - Playing with GCD

We are given an array a. We want to know whether there exists another array b of length n + 1 such that every element of a is the GCD of two neighboring elements of b: [ ai = gcd(bi, b{i+1}) ] for every position.

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CF 1736C1 - Good Subarrays (Easy Version)

We are asked to count all contiguous subarrays of a given array where each element is at least as large as its position in the subarray.

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CF 1736A - Make A Equal to B

This is a Type B (prove) problem. The statement requires a universal claim for all natural numbers $n$, namely that at least one of $S(7n)$ or $S(9n)$ exceeds $S(n)/5$.