Consider the following method

public static int[][] operation(int[][] mat, int[] vec, int c)
{
int[][] result = new int[mat.length][mat[0].length];

for (int j = 0; j

Question

Consider the following method

public static int[][] operation(int[][] mat, int[] vec, int c) 
{
  int[][] result = new int[mat.length][mat[0].length];

for (int j = 0; j < mat.length; j++) 
  {
    for (int k = 0; k < mat[j].length; k++) 
    {
      if (k == c) 
      {
        result[j][k] = mat[j][k] + vec[j];
      } 
      else 
      {
        result[j][k] = mat[j][k];
      }
    }
  }
  return result;
}

The following code segment appears in another method in the same class.

int[][] m = {{1, 2, 3, 4},
             {4, 3, 2, 1},
             {2, 1, 3, 4},
             {1, 3, 2, 4}};

int[] v = {4, 4, 4, 4};

int[][] grid = operation(m, v, 1);

Which of the following represents the contents of the grid as a result of executing the code segment?

BlackTom AI · Accepted Answer

The operation method adds vec[j] to the element in column c of each row j, leaving all other entries unchanged.

1. Relevant knowledge points and concepts
- Java 2D arrays: mat.length gives number of rows, mat[j].length gives number of columns in row j.
- Loop indices: j iterates rows (0–3 here), k iterates columns (0–3 here).
- Conditional: if (k == c) the code replaces result[j][k] with mat[j][k] + vec[j]; otherwise it copies mat[j][k].

2. Step-by-step reasoning
- We call operation(m, v, 1). So c = 1, meaning only column index 1 in each row is modified.
- For each row j, the code checks every column k. Only when k == 1 does it add vec[j] to mat[j][1]; all other columns are copied as-is.

3. Calculations with intermediate steps
Row 0 (j = 0): original row = {1, 2, 3, 4}
- column 0 (k=0): not c → result[0][0] = 1
- column 1 (k=1): equals c → result[0][1] = mat[0][1] + vec[0] = 2 + 4 = 6
- columns 2,3: copied → 3, 4
Result row 0: {1, 6, 3, 4}

Row 1 (j = 1): original {4, 3, 2, 1}
- column 1: 3 + vec[1] = 3 + 4 = 7
Result row 1: {4, 7, 2, 1}

Row 2 (j = 2): original {2, 1, 3, 4}
- column 1: 1 + vec[2] = 1 + 4 = 5
Result row 2: {2, 5, 3, 4}

Row 3 (j = 3): original {1, 3, 2, 4}
- column 1: 3 + vec[3] = 3 + 4 = 7
Result row 3: {1, 7, 2, 4}

4. Why the correct answer is correct
- The code explicitly modifies only column c of each row by adding the corresponding vec element (vec[j]). Given c = 1 and vec elements = 4, each mat[row][1] increases by 4 while all other entries remain identical.

5. Why other possible options would be incorrect
- Any option that changes columns other than column 1 is wrong because the conditional restricts changes to k == c.
- Any option that adds the wrong vec element (e.g., vec[k] instead of vec[j]) or performs a different arithmetic operation would not match the loop logic.

6. Conclusion
- The resulting grid is:
{{1, 6, 3, 4},
 {4, 7, 2, 1},
 {2, 5, 3, 4},
 {1, 7, 2, 4}}
This matches the provided correct answer.

类似问题

Which statement declares a two-dimensional integer array called myArray with 3 rows and 4 columns?

What is the output of the following code: int arr[4][4] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}; for (int j = 2; j < 3; j++) for(int i = 0; i < 4; i++) cout << arr[i][j];

Consider the following code segment. int[][] mat = new int[4][6]; for (int r = 0; r < 4; r++) { for (int c = 0; c < 6; c++) { if(c < 3 - r || c > 3 + r) { mat[r][c] = 0; } else { mat[r][c] = 1; } } } Which of the following represents mat after this code segment is executed?

Consider the following method intended to swap the first and second rows in a two-dimensional array: public static void swapRow(int[][] a) { /* missing code / } Which of the following correctly replaces / missing code */?

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