Codeforces Round 166 (Div. 2)

5 problems from Codeforces Round 166 (Div. 2) (contest 271), difficulty 800-2200. 3/5 solutions verified against sample I/O.

5 items

Codeforces Round 166 (Div. 2)

Div. 2 | 5 problems | 3/5 verified | Difficulty 800-2200 | 7m 18s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Beautiful Year	800	brute-force	177,634	1m 19s	✓
B	Prime Matrix	1300	binary-search, brute-force, math	15,439	1m 21s	✓
C	Secret	1500	constructive-algorithms, implementation	3,135	1m 41s
D	Good Substrings	1800	data-structures, strings	14,002	1m 33s
E	Three Horses	2200	constructive-algorithms, math, number-theory	598	1m 24s	✓

Practice › Codeforces Solutions › Codeforces Round 166 (Div. 2) ›

CF 271C - Secret

We are asked to distribute n sequentially numbered words among k Keepers such that each Keeper receives a non-empty subset of words, the subsets are pairwise disjoint, their union covers all words, and no subset forms an arithmetic progression.

Practice › Codeforces Solutions › Codeforces Round 166 (Div. 2) ›

CF 271A - Beautiful Year

We are given a four-digit year and we want to move forward in time until we reach the next year whose decimal representation does not reuse any digit.

Practice › Codeforces Solutions › Codeforces Round 166 (Div. 2) ›

CF 271E - Three Horses

We start with a set of “cards”, each card is a pair of integers $(a, b)$ with $a < b$. From one initial card $(x, y)$, we can repeatedly apply three transformation rules that behave like operations on this pair.

Practice › Codeforces Solutions › Codeforces Round 166 (Div. 2) ›

CF 271D - Good Substrings

We are given a string made of lowercase English letters. Each letter is labeled as either good or bad using a separate 26-character binary mask. We are also given an integer k, which limits how many bad letters we are allowed to tolerate inside a substring.

Practice › Codeforces Solutions › Codeforces Round 166 (Div. 2) ›

CF 271B - Prime Matrix

We are given a grid of positive integers. From this grid, we are allowed to repeatedly choose any single cell and increment its value by one. Each increment costs one move.