Codeforces Round 191 (Div. 2)

5 problems from Codeforces Round 191 (Div. 2) (contest 327), difficulty 1200-2300. 4/5 solutions verified against sample I/O.

5 items

Codeforces Round 191 (Div. 2)

Div. 2 | 5 problems | 4/5 verified | Difficulty 1200-2300 | 8m 20s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Flipping Game	1200	brute-force, dp, implementation	49,874	1m 23s	✓
B	Hungry Sequence	1200	math	21,634	1m 43s
C	Magic Five	1700	combinatorics, math	3,668	1m 37s	✓
D	Block Tower	1900	constructive-algorithms, dfs-and-similar, graphs	2,495	1m 12s	✓
E	Axis Walking	2300	bitmasks, combinatorics, constructive-algorithms	1,583	2m 25s	✓

Practice › Codeforces Solutions › Codeforces Round 191 (Div. 2) ›

CF 327B - Hungry Sequence

We need to construct an increasing sequence of n positive integers such that no later element is divisible by any earlier element. The input contains a single number n. We must output any sequence of length n satisfying two conditions.

Practice › Codeforces Solutions › Codeforces Round 191 (Div. 2) ›

CF 327C - Magic Five

We are given a digit string a and an integer k. The actual plate is not a itself, but the string obtained by concatenating a with itself k times. From this long string, we may delete any subset of positions, except that we are not allowed to delete every digit.

Practice › Codeforces Solutions › Codeforces Round 191 (Div. 2) ›

CF 327E - Axis Walking

We have n positive segment lengths. Their total sum is d, which is also the destination point on the number line. A route is simply an ordering of these lengths. While following a route, we keep a running prefix sum.

Practice › Codeforces Solutions › Codeforces Round 191 (Div. 2) ›

CF 327D - Block Tower

We are given a 2D grid with n rows and m columns. Each cell is either empty, where we can place a tower, or a hole, where no tower can be built.

Practice › Codeforces Solutions › Codeforces Round 191 (Div. 2) ›

CF 327A - Flipping Game

We are given a binary array, every element is either 0 or 1. We must choose exactly one contiguous segment and flip every value inside it. A flip changes 0 to 1 and 1 to 0. After performing this single operation, we want the resulting array to contain as many ones as possible.