Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2)

9 problems from Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) (contest 2084), difficulty 800-3500. 2/9 solutions verified against sample I/O.

9 items

Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2)

Div. 1+2 | 9 problems | 2/9 verified | Difficulty 800-3500 | 22m 42s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Max and Mod	800	constructive-algorithms, math	17,425	1m 51s
B	MIN = GCD	1100	greedy, math, number-theory	16,610	1m 29s	✓
C	You Soared Afar With Grace	1400	constructive-algorithms, data-structures, greedy	11,704	5m 14s
D	Arcology On Permafrost	1600	binary-search, brute-force, constructive-algorithms	8,839	2m 30s
E	Blossom	2400	binary-search, combinatorics, dp	1,791	3m 21s
F	Skyscape	2900	constructive-algorithms, data-structures, greedy	508	2m 22s
G1	Wish Upon a Satellite (Easy Version)	2600	dp, games	755	1m 44s
G2	Wish Upon a Satellite (Hard Version)	3500	data-structures, dp	108	2m 18s
H	Turtle and Nediam 2	3500	dp, greedy	112	1m 53s	✓

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084H - Turtle and Nediam 2

We start with a binary string and repeatedly apply a deletion rule based on a length-3 window. For a binary triple, the median is simply the majority value.

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084G2 - Wish Upon a Satellite (Hard Version)

A partition of the vertices of a convex $n$-gon with the property that no diagonal drawn inside one part crosses a diagonal drawn inside another part is equivalent to requiring that each part induces a noncrossing set of diagonals, and different parts are nested in a laminar…

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084G1 - Wish Upon a Satellite (Easy Version)

We are given a partially filled permutation of length $n$, where some entries are zero, representing missing numbers. The goal is to fill in the zeros with the remaining numbers from $1$ to $n$ to form a complete permutation.

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084F - Skyscape

We start with a permutation a. An operation chooses a segment [l, r] whose rightmost element is the minimum value inside that segment. Then that minimum element is moved to the front of the segment by performing a cyclic right shift. If the segment is [x1, x2, ...

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084C - You Soared Afar With Grace

We are given two permutations a and b of the same size. In one move, we are allowed to pick two positions and swap them in both arrays simultaneously.

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084E - Blossom

We are given a partially filled permutation of length $n$. Some elements are missing and represented by $-1$. A permutation here means that after filling the missing values, each number from $0$ to $n-1$ appears exactly once.

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084D - Arcology On Permafrost

Working

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084A - Max and Mod

We are asked to construct a permutation of the integers from 1 to $n$ such that, for every index $i$ from 2 to $n$, the maximum of the current and previous element, modulo $i$, equals $i-1$. Concretely, if $p$ is our permutation, the condition is $max(p{i-1}, pi) bmod i = i-1$.

Practice › Codeforces Solutions › Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2) ›

CF 2084B - MIN = GCD

We are given a sequence of positive integers, and we need to determine if it is possible to split the sequence into two contiguous parts after some rearrangement, such that the minimum value in the first part equals the greatest common divisor of the second part.