Codeforces Round 176 (Div. 1)

5 problems from Codeforces Round 176 (Div. 1) (contest 286), difficulty 1400-2800. 4/5 solutions verified against sample I/O.

5 items

Codeforces Round 176 (Div. 1)

Div. 1 | 5 problems | 4/5 verified | Difficulty 1400-2800 | 8m 52s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Lucky Permutation	1400	constructive-algorithms, math	3,450	1m 41s	✓
B	Shifting	2200	implementation	1,172	1m 30s	✓
C	Main Sequence	2100	greedy, implementation	1,602	1m 50s	✓
D	Tourists	2600	data-structures, sortings	457	1m 40s
E	Ladies' Shop	2800	constructive-algorithms, fft, math	1,020	2m 11s	✓

Practice › Codeforces Solutions › Codeforces Round 176 (Div. 1) ›

CF 286B - Shifting

We are asked to generate a permutation of numbers from 1 to n that becomes "beautiful" after a sequence of block-cyclic shifts. The transformation takes a permutation and an integer k and splits the permutation into consecutive blocks of length k.

Practice › Codeforces Solutions › Codeforces Round 176 (Div. 1) ›

CF 286C - Main Sequence

We are given an encrypted version of a “correct bracket sequence” using integers instead of traditional parentheses. Each integer represents a bracket type, and its sign indicates whether it is an opening or closing bracket.

Practice › Codeforces Solutions › Codeforces Round 176 (Div. 1) ›

CF 286E - Ladies' Shop

We are given a set of bags, each with a minimum weight capacity. We need to define a minimal set of item weights such that any total weight that a bag can hold can be formed using these items in unlimited quantities.

Practice › Codeforces Solutions › Codeforces Round 176 (Div. 1) ›

CF 286D - Tourists

We are given a sequence of moments when pairs of tourists start walking, and another sequence of “walls” that appear over time.

Practice › Codeforces Solutions › Codeforces Round 176 (Div. 1) ›

CF 286A - Lucky Permutation

We are asked to construct a permutation of numbers from 1 to n such that applying the permutation twice maps an element at position i to the mirrored position n - i + 1. Formally, if p is the permutation, then for every i between 1 and n, we must have p[p[i]] = n - i + 1.