Codeforces Round 263 (Div. 1)

5 problems from Codeforces Round 263 (Div. 1) (contest 461), difficulty 1200-3000. 3/5 solutions verified against sample I/O.

5 items

Codeforces Round 263 (Div. 1)

Div. 1 | 5 problems | 3/5 verified | Difficulty 1200-3000 | 8m 20s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Appleman and Toastman	1200	greedy, sortings	16,432	2m 14s	✓
B	Appleman and Tree	2000	dfs-and-similar, dp, trees	7,017	1m 26s
C	Appleman and a Sheet of Paper	2200	data-structures, implementation	1,393	1m 22s
D	Appleman and Complicated Task	2800	dsu, math	401	1m 2s	✓
E	Appleman and a Game	3000	binary-search, shortest-paths, strings	516	2m 16s	✓

Practice › Codeforces Solutions › Codeforces Round 263 (Div. 1) ›

CF 461A - Appleman and Toastman

We start with a single group containing all given numbers. Whenever Toastman receives a group, he adds the sum of all numbers in that group to the score. If the group contains more than one number, Appleman may split it into two non-empty groups and send both back to Toastman.

Practice › Codeforces Solutions › Codeforces Round 263 (Div. 1) ›

CF 461B - Appleman and Tree

We are given a tree with n vertices, where some vertices are black and others are white. The tree is rooted implicitly by the way the edges are described: each node i for i 0 has a parent p[i-1].

Practice › Codeforces Solutions › Codeforces Round 263 (Div. 1) ›

CF 461C - Appleman and a Sheet of Paper

We have a one-dimensional sheet of paper represented as an array of width n. Appleman can fold the sheet at a position p, which means the left segment [0, p) flips and lies over the right segment [p, currentwidth).

Practice › Codeforces Solutions › Codeforces Round 263 (Div. 1) ›

CF 461D - Appleman and Complicated Task

We are given an $n times n$ grid that is mostly empty, except for a small number of cells that are already fixed as either x or o. Every other cell must be filled with one of these two characters.

Practice › Codeforces Solutions › Codeforces Round 263 (Div. 1) ›

CF 461E - Appleman and a Game

Appleman starts with an empty string. In one second he may append any substring of t to the end of the string he is building. For a fixed target string s, Appleman chooses an optimal sequence of substrings and finishes s in the minimum possible number of seconds.