Codeforces Round 907 (Div. 2)

Solutions for Codeforces Round 907 (Div. 2) (contest 1891). 1/6 problems verified against sample I/O. Difficulty range: 800-2500.

6 items

Codeforces Round 907 (Div. 2)

Type: Div. 2 | Problems: 6 | Verified: 1/6 | Rating range: 800-2500 | Time: 9m 40s

Problem Name Rating Tags Solve Time Verified
A Sorting with Twos 800 constructive-algorithms, sortings 2m 3s ✓
B Deja Vu 1100 brute-force, math, sortings 41s ✗
C Smilo and Monsters 1500 binary-search, constructive-algorithms, greedy 1m 44s ✗
D Suspicious logarithms 1900 binary-search, brute-force, math 1m 9s ✗
E Brukhovich and Exams 2500 brute-force, greedy, implementation 1m 57s ✗
F A Growing Tree 2000 data-structures, dfs-and-similar, trees 2m 6s ✗
Practice › Codeforces Solutions › Codeforces Round 907 (Div. 2) ›

CF 1891F - A Growing Tree

We are asked to process a sequence of operations on a growing rooted tree. The tree starts with a single node numbered 1, and each node has a numerical value, initially 0.
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CF 1891E - Brukhovich and Exams

We are given a sequence of exams with known difficulties. Smilo considers consecutive exams with coprime difficulties unpleasant. The "sadness" of the year is the total number of consecutive exam pairs whose greatest common divisor is one.
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Practice › Codeforces Solutions › Codeforces Round 907 (Div. 2) ›

CF 1891A - Sorting with Twos

We are given an array of integers, and the only operation allowed is a kind of prefix subtraction. Specifically, you can pick a prefix of length $2^m$ for any non-negative integer $m$ such that $2^m le n$, and subtract 1 from every element in that prefix.
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CF 1891D - Suspicious logarithms

Working
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Practice › Codeforces Solutions › Codeforces Round 907 (Div. 2) ›

CF 1891C - Smilo and Monsters

We are given several independent scenarios. In each scenario, there are multiple groups of enemies, each group having some initial size.
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Practice › Codeforces Solutions › Codeforces Round 907 (Div. 2) ›

CF 1891B - Deja Vu

We have an array of positive integers and a sequence of queries. A query with value x examines every array element. If an element is divisible by 2^x, we add 2^(x-1) to that element.
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