Problem A Solution

Direct implemantation yields the solution.

Problem B: Maximizing Odd Frequency Counts

Given a sequence of length 2n with elements from [1,2n], define f(a) as the count of distinct elements appearing an odd number of times in sequence a. Partition the sequence into two subsequences p and q, each of length n. The goal is to maximize f(p) + f(q).

Strategy: Analyze each element's frequency modulo 4:

• If remainder is 1: contributes 1
• If remainder is 2: contributes 2 (can distribute odd counts to both sequences)
• If remainder is 0: contributes 2 if both sequences receive odd counts, otherwise 0

When remainder-0 elements are even in count, pair them for mutual benefit. With odd count, use one remainder-0 element with a remainder-1 element to achieve contribution 2.

Problem C: Alternating Game on Sequences

Two players alternate k turns on sequence a of length n. First player adds b_i to any a_i on odd turns; second player subtracts b_i on even turns. First player aims to maximize maximum subarray sum, second player minimizes it.

Key Insight:

• For even k: Sequence remains unchanged as second player can reverse first player's moves
• For odd k: After k-1 even turns, first player can add b_i to one position

Implemantation: Calculate prefix and suffix sums. For position i, maximum subarray sum containing i is:

sum(a) - min(prefix[1..i-1]) - min(suffix[i+1..n]) + (k%2 ? b[i] : 0)

Problem D: Palindrome Flip Operations

Transform binary string s to t using at most 2n operations. Each operation selects a palindromic substring [l,r] and flips all bits.

Solution: Always possible for n ≥ 4. Strategy:

1. Convert s to all-0s or all-1s using adjacent identical bits
2. Convert uniform string to t using reverse process
3. Handle alternating patterns (0101...) by flipping first three bits

Problem E: Minimum Special Points on Grid

Given n×m grid with values, select k special points such that any cell adjacent to special points with value ≥ neighbor becomes special. Find minimum k and handle value decrement queries.

Observation: Minimum k equals count of cells strictly smaller than all neighbors. Use union-find to merge equal-valued connected components. For queries, udpate affected components and neighbors.

Problem F1: Constructing Sequences with Distinct GCDs

Construct sequence of length n where consecutive GCDs are distinct, minimizing distinct values used.

Construction: Use numbers of form 2^p·3^q. For k=37, we get 704 distinct GCD pairs. Find Euler path in complete graph of k+1 vertices. For even k+1, remove (k+1)/2-1 edges to enable Euler path.

Implementation:

// Generate sequence using 2^p * 3^q
vector<long long> sequence;
for(int i = 0; i < n; i++) {
    int p = path[i] - 1;
    int q = k - p;
    sequence.push_back(pow2[p] * pow3[q]);
}