Codeforces Round 973 (Div. 2)

6 problems from Codeforces Round 973 (Div. 2) (contest 2013), difficulty 800-2700. 2/6 solutions verified against sample I/O.

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Codeforces Round 973 (Div. 2)

Div. 2 | 6 problems | 2/6 verified | Difficulty 800-2700 | 18m 16s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Zhan's Blender	800	constructive-algorithms, math	31,224	1m 44s	✓
B	Battle for Survive	900	constructive-algorithms, greedy, math	28,236	6m 22s
C	Password Cracking	1400	constructive-algorithms, interactive, strings	14,578	2m 33s
D	Minimize the Difference	1900	binary-search, greedy	7,466	2m 11s
E	Prefix GCD	2200	brute-force, dp, greedy	4,215	3m 50s
F1	Game in Tree (Easy Version)	2700	binary-search, brute-force, data-structures	651	1m 36s	✓

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CF 2013B - Battle for Survive

We are given a group of fighters, each starting with a positive strength value. We repeatedly pick two still-alive fighters, and the one with the smaller index is always removed. The survivor’s strength is updated by subtracting the removed fighter’s strength from theirs.

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CF 2013C - Password Cracking

Codeforces 2013C: Password Cracking

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CF 2013E - Prefix GCD

We are given a list of positive integers, and we are allowed to permute them in any order before processing. After choosing an order, we build prefix GCDs: the first value is just the first element, the second is the gcd of the first two elements, and so on until the full array.

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CF 2013D - Minimize the Difference

We are given a sequence of numbers arranged in a line. In one move, we are allowed to take one unit from some position and push it to the next position on the right. This means mass can only flow to the right, never backwards, and every move preserves the total sum of the array.

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CF 2013F1 - Game in Tree (Easy Version)

We are given a tree. One player, Alice, always starts at vertex 1. The other player, Bob, starts at a vertex that lies on a fixed simple path between two vertices u and v.

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CF 2013A - Zhan's Blender

Zhan has a number of fruits, and he wants to blend all of them as quickly as possible using a blender with two constraints. Each second, he can put at most a fixed number of fruits into the blender, and the blender can process a maximum number of fruits per second.