Codeforces Round 651 (Div. 2)

5 problems from Codeforces Round 651 (Div. 2) (contest 1370), difficulty 800-2700. 3/5 solutions verified against sample I/O.

5 items

Codeforces Round 651 (Div. 2)

Div. 2 | 5 problems | 3/5 verified | Difficulty 800-2700 | 18m 3s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Maximum GCD	800	greedy, implementation, math	64,410	1m 34s	✓
B	GCD Compression	1100	constructive-algorithms, math, number-theory	25,013	4m 11s
D	Odd-Even Subsequence	2000	binary-search, dp, dsu	10,923	2m 25s	✓
E	Binary Subsequence Rotation	2100	binary-search, constructive-algorithms, data-structures	5,220	1m 45s	✓
F2	The Hidden Pair (Hard Version)	2700	binary-search, dfs-and-similar, graphs	1,669	8m 8s

Practice › Codeforces Solutions › Codeforces Round 651 (Div. 2) ›

CF 1370F2 - The Hidden Pair (Hard Version)

We are given a tree and two hidden distinct nodes inside it. We do not know which nodes they are, but we can interactively probe the tree by choosing any subset of nodes.

Practice › Codeforces Solutions › Codeforces Round 651 (Div. 2) ›

CF 1370B - GCD Compression

We are given an array of length $2n$. We are allowed to remove exactly two elements permanently. After that, the remaining $2n-2$ elements must be partitioned into pairs. Each pair is replaced by the sum of its two elements, producing an array $b$ of length $n-1$.

Practice › Codeforces Solutions › Codeforces Round 651 (Div. 2) ›

CF 1370D - Odd-Even Subsequence

We are given a sequence of numbers and asked to choose a subsequence of fixed length $k$. Once we pick this subsequence, we alternate its elements into odd and even positions based on their order inside the subsequence, not their original positions in the array.

Practice › Codeforces Solutions › Codeforces Round 651 (Div. 2) ›

CF 1370E - Binary Subsequence Rotation

We are given two binary strings of equal length, and we want to transform the first string into the second using a very unusual operation.

Practice › Codeforces Solutions › Codeforces Round 651 (Div. 2) ›

CF 1370A - Maximum GCD

We are asked to find the largest possible greatest common divisor among all pairs of distinct integers from 1 to n. For a given n, we need the maximum gcd of any two numbers a and b where 1 ≤ a < b ≤ n.