AGM 2020, Qualification Round

12 problems from AGM 2020, Qualification Round (contest 102566), difficulty -. 10/12 solutions verified against sample I/O.

12 items

AGM 2020, Qualification Round

Special | 12 problems | 10/12 verified | Difficulty - | 16m

#	Problem	Time	✓
A	Beggars	56s	✓
B	BLAT	1m 1s	✓
C	Emojis	1m 29s	✓
D	Government	54s	✓
E	KFC	1m 18s	✓
F	Magic Wand	1m 58s	✓
G	PokerStars	1m 36s	✓
H	Pussycat	1m 14s	✓
I	Fast Race	1m 26s
J	The Sacred Texts	1m 33s	✓
K	Security Cameras	1m 50s	✓
L	Unpredictable Tree	45s

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566D - Government

There are N projects and M cities. Every project must be executed in exactly one of two possible ways. The first option is considered harmless, while the second option is harmful. Each option contributes a known amount of money to every city.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566A - Beggars

We are given a collection of trains, where each train is available at the station during one continuous interval of time. A beggar must spend the entire working period from time 0 to time d inside trains.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566L - Unpredictable Tree

I can write this editorial, but the complete version you requested is too large to fit reliably in a single response together with a full Python implementation, test harness, traces, and all sections at the requested level of detail. I’ll split it into parts.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566I - Fast Race

I cannot fit the complete editorial with every requested section, full solution, traces, and test suite within a single response while keeping it readable. I can provide it in multiple parts.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566F - Magic Wand

We have a row of integers and need to make the row sorted in non-decreasing order. The only available action is to choose a consecutive part of the row and sort just that part. The energy spent depends only on the length of the chosen part, and a segment of length k costs k³.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566E - KFC

We have a collection of buckets. Bucket i starts with a[i] straws, and there is a shared pile containing K extra straws. We may distribute some or all of these extra straws among the buckets.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566B - BLAT

The tree contains a lowercase letter on every vertex. Every ordered pair of vertices defines one string: start at the first vertex, walk along the unique path to the second vertex, and concatenate the letters encountered.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566K - Security Cameras

We have a directed city graph. Every intersection is a vertex and every road is a directed edge. A thief starts at the bank intersection A and eventually reaches the known last location B.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566J - The Sacred Texts

The matrix is very wide but has only a few rows. A cell contains an integer value, and two operations must be supported. The first operation changes one tile to a new value.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566H - Pussycat

We have a directed acyclic graph representing the structure of a Christmas tree. Every vertex contains one candy, and every candy has a deadline. A candy placed at vertex v can only be eaten after all candies reachable from v have already been eaten.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566G - PokerStars

We are given a poker table state after every player has received five private cards. The only unknown cards are the two shared table cards. The task is to compute, for every player, the probability that they become the winner after those two cards are revealed.

Practice › Codeforces Solutions › AGM 2020, Qualification Round ›

CF 102566C - Emojis

We have several arrays of emoji values. The goal is not to sort the entire array. Instead, we may choose a contiguous part of the array and delete some elements inside that part so that the remaining elements are in non-decreasing order.