A. Special Characters

This problem involves constructing a string of length n with paired characters. A solution exists only when n is even, as characters must appear in pairs. For odd n, output is "NO". For even n, we output "YES" followed by a string constructed in pairs, for example, "ZZYYXX..."

#include <iostream>
#include <vector>

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
    
    int numTests;
    std::cin >> numTests;
    
    while (numTests--) {
        int len;
        std::cin >> len;
        
        if (len % 2 == 1) {
            std::cout << "NO\n";
            continue;
        }
        
        std::cout << "YES\n";
        std::string result;
        for (int i = 0; i < len / 2; ++i) {
            char baseChar = 'Z' - (i % 3);
            result += baseChar;
            result += baseChar;
        }
        std::cout << result << "\n";
    }
    return 0;
}

B. Array Fix

The goal is to transform an array into a non-decreasing sequence by strategically splitting two-digit numbers. The greedy strategy is to split a two-digit number if its tens digit is not smaller than the previous element and its units digit is not smaller than the tens digit. After processing all elements, we verify if the resulting sequence is sorted.

#include <iostream>
#include <vector>

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
    
    int testCases;
    std::cin >> testCases;
    
    while (testCases--) {
        int size;
        std::cin >> size;
        std::vector<int> original(size);
        std::vector<int> processed;
        
        for (int i = 0; i < size; ++i) {
            std::cin >> original[i];
        }
        
        for (int i = 0; i < size; ++i) {
            int num = original[i];
            if (num < 10) {
                processed.push_back(num);
            } else {
                int tens = num / 10;
                int units = num % 10;
                if (tens <= units && (processed.empty() || tens >= processed.back())) {
                    processed.push_back(tens);
                    processed.push_back(units);
                } else {
                    processed.push_back(num);
                }
            }
        }
        
        bool isSorted = true;
        for (size_t i = 1; i < processed.size(); ++i) {
            if (processed[i] < processed[i - 1]) {
                isSorted = false;
                break;
            }
        }
        std::cout << (isSorted ? "YES" : "NO") << "\n";
    }
    return 0;
}

C. Arrow Path

This problem can be modeled as a graph traversal on a 2xN grid. We use a breadth-first search (BFS) to explore reachable states, where each state is a combination of position (row, colum) and the move type (vertical or horizontal). Starting from (1,1), we alternate between move types. The goal is to determine if the target cell (2,n) is reachable, specifically after a horizontal move.

#include <iostream>
#include <vector>
#include <queue>
#include <utility>

const int ROWS = 2;
const int MAX_N = 200005;

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
    
    int testCases;
    std::cin >> testCases;
    
    while (testCases--) {
        int n;
        std::cin >> n;
        
        std::string grid[3];
        std::cin >> grid[1] >> grid[2];
        
        std::vector<std::vector<bool>> reachableHorizontal(3, std::vector<bool>(n + 2, false));
        std::vector<std::vector<bool>> reachableVertical(3, std::vector<bool>(n + 2, false));
        
        std::queue<std::pair<std::pair<int, int>, bool>> q;
        q.push({{1, 1}, false});
        
        while (!q.empty()) {
            int r = q.front().first.first;
            int c = q.front().first.second;
            bool isHorizontalMove = q.front().second;
            q.pop();
            
            if (!isHorizontalMove) {
                if (r > 1 && !reachableVertical[r - 1][c]) {
                    reachableVertical[r - 1][c] = true;
                    q.push({{r - 1, c}, true});
                }
                if (r < ROWS && !reachableVertical[r + 1][c]) {
                    reachableVertical[r + 1][c] = true;
                    q.push({{r + 1, c}, true});
                }
            } else {
                int nextC = (grid[r][c - 1] == '>') ? c + 1 : c - 1;
                if (nextC >= 1 && nextC <= n && !reachableHorizontal[r][nextC]) {
                    reachableHorizontal[r][nextC] = true;
                    q.push({{r, nextC}, false});
                }
            }
        }
        
        std::cout << (reachableHorizontal[2][n] ? "YES" : "NO") << "\n";
    }
    return 0;
}

D. Tandem Repeats?

This task asks for the maximum length of a tandem repeat that can be formed in a string. A tandem repeat consists of two substrings of equal length, where one can be transformed into the other's reverse by replacing wildcard characters ('?'). The brute-force O(N³) approach iterates through all possible pairs of starting positions and lengths, checking for potential matches by comparing corresponding characters from the end of the substrings inward.

#include <iostream>
#include <string>
#include <algorithm>

const int MAX_STR = 5005;

int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(NULL);
    
    int testCases;
    std::cin >> testCases;
    
    while (testCases--) {
        std::string s;
        std::cin >> s;
        int n = s.length();
        int maxLen = 0;
        
        for (int i = 0; i < n && (n - i) / 2 > maxLen; ++i) {
            for (int len = (n - i) / 2; len > maxLen; --len) {
                int j = i + len;
                bool isValid = true;
                for (int k = 0; k < len; ++k) {
                    char c1 = s[i + len - 1 - k];
                    char c2 = s[j + k];
                    if (c1 != '?' && c2 != '?' && c1 != c2) {
                        isValid = false;
                        break;
                    }
                }
                if (isValid) {
                    maxLen = len;
                    break;
                }
            }
        }
        
        std::cout << 2 * maxLen << "\n";
    }
    return 0;
}

E. Clique Partition

Problem E, "Clique Partition," is a more advanced graph theory challenge that requires partitioning the vertices of a graph into cliques. The problem typically involves complex combinatorial optimization techniques and is beyond the scope of this brief analysis.