Codeforces Round 260 (Div. 1)

5 problems from Codeforces Round 260 (Div. 1) (contest 455), difficulty 1500-2900. 5/5 solutions verified against sample I/O.

5 items

Codeforces Round 260 (Div. 1)

Div. 1 | 5 problems | 5/5 verified | Difficulty 1500-2900 | 7m 50s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Boredom	1500	dp	71,930	1m 31s	✓
B	A Lot of Games	1900	dfs-and-similar, dp, games	6,839	1m 49s	✓
C	Civilization	2100	dfs-and-similar, dp, dsu	6,464	2m 1s	✓
D	Serega and Fun	2700	data-structures	3,655	1m 27s	✓
E	Function	2900	data-structures	786	1m 2s	✓

Practice › Codeforces Solutions › Codeforces Round 260 (Div. 1) ›

CF 455A - Boredom

We are given an array of integers. In one move, we choose a value x that is currently present, earn x points from deleting one occurrence of it, and all occurrences of x - 1 and x + 1 disappear from the array as a consequence.

Practice › Codeforces Solutions › Codeforces Round 260 (Div. 1) ›

CF 455B - A Lot of Games

The game can be viewed as moving through a trie built from the given strings. We start at the root, which represents the empty string. On each turn a player chooses one outgoing edge, corresponding to appending a character.

Practice › Codeforces Solutions › Codeforces Round 260 (Div. 1) ›

CF 455C - Civilization

The graph consists of cities connected by roads. The condition that between any two connected cities there is exactly one simple path means every connected component is a tree. Since the graph may be disconnected, the whole graph is a forest.

Practice › Codeforces Solutions › Codeforces Round 260 (Div. 1) ›

CF 455D - Serega and Fun

We have an array of integers, each between 1 and n, and we need to support two types of operations efficiently. The first operation rotates a segment of the array one step to the right. The second operation counts how many times a particular value occurs in a segment.

Practice › Codeforces Solutions › Codeforces Round 260 (Div. 1) ›

CF 455E - Function

We are given an array of integers, a[1…n], and a recursive function f(i, j) defined on it. The function essentially accumulates sums along specific paths: the base case is simply f(1, j) = a[j], and for later rows, each f(i, j) is the sum of a[j] plus the minimum of f(i-1…