Codeforces Round 105 (Div. 2)

5 problems from Codeforces Round 105 (Div. 2) (contest 148), difficulty 800-1900. 3/5 solutions verified against sample I/O.

5 items

Codeforces Round 105 (Div. 2)

Div. 2 | 5 problems | 3/5 verified | Difficulty 800-1900 | 9m 21s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Insomnia cure	800	constructive-algorithms, implementation, math	112,352	1m 34s
B	Escape	1500	implementation, math	6,014	1m 12s	✓
C	Terse princess	1700	constructive-algorithms, greedy	3,381	2m 14s
D	Bag of mice	1800	dp, games, math	8,901	1m 55s	✓
E	Porcelain	1900	dp	4,868	2m 26s	✓

Practice › Codeforces Solutions › Codeforces Round 105 (Div. 2) ›

CF 148A - Insomnia cure

The problem asks us to figure out how many dragons get affected by a princess who has a unique way of defending herself. She targets every k-th, l-th, m-th, and n-th dragon with different actions. The total number of dragons is d.

Practice › Codeforces Solutions › Codeforces Round 105 (Div. 2) ›

CF 148B - Escape

We are asked to simulate a pursuit scenario between a princess and a dragon. The princess runs at a constant speed, and the dragon flies faster but only begins chasing after a delay. Each time the dragon catches up, the princess can drop a bijou to distract him.

Practice › Codeforces Solutions › Codeforces Round 105 (Div. 2) ›

CF 148C - Terse princess

We need to construct an array of groom fortunes so that the princess reacts in exactly the required way. For every groom after the first one, two special situations are possible. If the current fortune is larger than every previous fortune, the princess says Oh....

Practice › Codeforces Solutions › Codeforces Round 105 (Div. 2) ›

CF 148D - Bag of mice

We have a bag containing white and black mice. The princess moves first, then the dragon, then the princess again, and so on. Whoever draws a white mouse immediately wins. If someone draws a black mouse, the game continues. There is one extra rule during the dragon’s turn.

Practice › Codeforces Solutions › Codeforces Round 105 (Div. 2) ›

CF 148E - Porcelain

We have several shelves of porcelain items. Inside one shelf, the items form a line, and at any moment we may only remove the current leftmost item or the current rightmost item. After removing one item, the next item on that side becomes accessible.