To solve this problem, we need to focus on the two key elements:

1. The cost of a point on the grid, which is defined as A[x] * B[y].
2. We need to find the minimum cost path from (1,1) to (P,Q).

Understanding the problem:

• A[i] is defined as max(i, N / i) if i divides N; otherwise A[i] = 0.
• Similarly, B[i] is defined as max(i, M / i) if i divides M; otherwise B[i] = 0.

Given this, the cost to move from (x, y) is A[x] * B[y].

Steps:

1. Calculate the values for A and B:

   • For A[x], calculate the maximum of x and N / x if x divides N. Otherwise, set A[x] = 0.
   • Similarly, for B[y], calculate B[y] based on M.

2. Dynamic Programming Approach: Since the movement is allowed in four directions (up, down, left, right), a simple dynamic programming solution should work where we calculate the minimum cost to reach each point on the grid.

Strategy:

• We start at (1,1) and initialize the cost to get there as A[1] * B[1].
• For each test case, calculate the minimum cost for each point (x, y) using a BFS-like strategy (or dynamic programming) to explore the points. idea is to update the minimum cost by considering the neighbors and the cost to move to them.

Here's how you can write a Java solution for the problem:

java
import java.util.*;

class Codechef {
    
    static int[] A, B;
    
    // Function to compute A[i] based on N
    static void computeA(int N) {
        A = new int[N + 1];
        for (int i = 1; i <= N; i++) {
            if (N % i == 0) {
                A[i] = Math.max(i, N / i);
            } else {
                A[i] = 0;
            }
        }
    }
    
    // Function to compute B[i] based on M
    static void computeB(int M) {
        B = new int[M + 1];
        for (int i = 1; i <= M; i++) {
            if (M % i == 0) {
                B[i] = Math.max(i, M / i);
            } else {
                B[i] = 0;
            }
        }
    }
    
    public static void main (String[] args) throws java.lang.Exception {
        Scanner sc = new Scanner(System.in);
        
        // Number of test cases
        int T = sc.nextInt();
        
        while (T-- > 0) {
            // Reading N, M, P, Q for each test case
            int N = sc.nextInt();
            int M = sc.nextInt();
            int P = sc.nextInt();
            int Q = sc.nextInt();
            
            // Compute A[] and B[]
            computeA(N);
            computeB(M);
            
            // We use a BFS to find the minimum cost path
            Queue<int[]> queue = new LinkedList<>();
            boolean[][] visited = new boolean[P + 1][Q + 1];
            int[][] cost = new int[P + 1][Q + 1];
            
            // Starting point (1, 1)
            queue.add(new int[] {1, 1});
[1][1] = A[1] * B[1];
            visited[1][1] = true;
            
            // Directions for movement (left, right, up, down)
            int[] dx = {-1, 1, 0, 0};
            int[] dy = {0, 0, -1, 1};
            
            while (!queue.isEmpty()) {
                int[] point = queue.poll();
                int x = point[0];
                int y = point[1];
                
                // Explore all four directions
                for (int i = 0; i < 4; i++) {
                    int nx = x + dx[i];
                    int ny = y + dy[i];
                    
                    // Check if the new position is within bounds
                    if (nx >= 1 && nx <= P && ny >= 1 && ny <= Q) {
                        int newCost = A[nx] * B[ny];
                        
                        // If not visited or found a lower cost path
                        if (!visited[nx][ny] || cost[nx][ny] > cost[x][y] + newCost) {
                            visited[nx][ny] = true;
                            cost[nx][ny] = cost[x][y] + newCost;
                            queue.add(new int[] {nx, ny});
                        }
                    }
                }
            }
            
            // Output the minimum cost)
            System.out.println(cost[P][Q]);
        }
        
        sc.close();
    }
}

Explanation of the Code:

1. Helper functions:

   • computeA(int N) calculates the A[i] array where for each index i, it stores max(i, N / i) if i divides N and 0 otherwise.
   • computeB(int M) calculates the B[i] array in the same way for M.

2. Main Functionality:

   • We initialize a BFS-like approach to explore all possible points from (1, 1) to (P, Q).
   • The cost array keeps track of the minimum cost to reach each point, and the visited array ensures we don't revisit points unnecessarily.
   • For each point, we calculate the cost of moving in the four directions and update the cost array if we find a cheaper path.

3. Edge Cases:

   • Since we have constraints as large as 10^8, we only compute values for relevant positions of A[i] and B[i] instead of explicitly constructing large arrays.

Complexity:

• For each test case, we perform a BFS over a grid of size P x Q, so the time complexity per test case is approximately O(P * Q) in the worst case. However, this can be optimized with more advanced techniques like Dijkstra's algorithm or A* if necessary.

This solution should work efficiently for reasonable input sizes within the problem's constraints.