Codeforces Round 842 (Div. 2)

Solutions for Codeforces Round 842 (Div. 2) (contest 1768). 1/6 problems verified against sample I/O. Difficulty range: 800-2900.

6 items

Codeforces Round 842 (Div. 2)

Type: Div. 2 | Problems: 6 | Verified: 1/6 | Rating range: 800-2900 | Time: 17m 52s

Problem	Name	Rating	Tags	Solve Time	Verified
A	Greatest Convex	800	greedy, math, number-theory	1m 45s	✓
B	Quick Sort	900	greedy, math	3m	✗
C	Elemental Decompress	1300	constructive-algorithms, greedy, implementation	1m 35s	✗
D	Lucky Permutation	1800	constructive-algorithms, dfs-and-similar, graphs	5m 23s	✗
E	Partial Sorting	2300	combinatorics, math, number-theory	2m 12s	✗
F	Wonderful Jump	2900	dp, greedy	3m 57s	✗

Practice › Codeforces Solutions › Codeforces Round 842 (Div. 2) ›

CF 1768D - Lucky Permutation

We are given a permutation of integers from 1 to $n$, and we can swap any two elements to change the permutation. The goal is to perform the minimum number of swaps so that the resulting permutation has exactly one inversion.

Practice › Codeforces Solutions › Codeforces Round 842 (Div. 2) ›

CF 1768F - Wonderful Jump

Input: Expected: Two different configurations (two different games), each printed as a pair of 18-card hands. Actual output: Two identical blocks: This means: - Your program does print two cases - But they are not correctly separated as distinct constructions in memory - Or…

Practice › Codeforces Solutions › Codeforces Round 842 (Div. 2) ›

CF 1768E - Partial Sorting

We are given a permutation of size 3n, which is an array containing every integer from 1 to 3n exactly once in some order. We have two allowed operations: we can either sort the first 2n elements or sort the last 2n elements.

Practice › Codeforces Solutions › Codeforces Round 842 (Div. 2) ›

CF 1768B - Quick Sort

We are given a permutation, meaning a rearrangement of numbers from 1 to n. In one move, we pick any k elements, remove them from the array, sort just those chosen elements, and append them back to the end. The rest of the array keeps its relative order.

Practice › Codeforces Solutions › Codeforces Round 842 (Div. 2) ›

CF 1768A - Greatest Convex

The problem asks us to find the largest integer x less than a given integer k such that the sum of x! and (x-1)! is divisible by k. In plain language, we want a number just below k where adding its factorial to the factorial of the previous number gives a multiple of k.

Practice › Codeforces Solutions › Codeforces Round 842 (Div. 2) ›

CF 1768C - Elemental Decompress

This is a Type B (prove inequality) problem, not Type C. The task is to prove that $$frac1{a^3(b+c)}+frac1{b^3(c+a)}+frac1{c^3(a+b)}ge frac32$$ for all positive $a,b,c$ satisfying $abc=1$.