构造一个二进制矩阵，其每一行和每一列的和为素数

给定一个整数N ，任务是构造一个大小为N * N的二进制矩阵，以使矩阵的每一行和每一列的和为素数。

例子：

Input: N = 2 
Output: 

Explanation: 
Sum of 0^th row = 1 + 1 = 2 (Prime number) 
Sum of 1^st row = 1 + 1 = 2 (Prime number) 
Sum of 0^th column = 1 + 1 = 2 (Prime number) 
Sum of 1^st column = 1 + 1 = 2 (Prime number)

Input: N = 4 
Output: 

方法：请按照以下步骤解决问题：

• 初始化一个二进制矩阵，例如mat [] [] ，大小为N * N。
• 将mat [i] [i]的所有可能值更新为1 。
• 将mat [i] [N – i -1]的所有可能值更新为1 。
• 如果N为奇数，则将值mat [N / 2] [0]和mat [0] [N / 2]更新为1 。

下面是上述方法的实现。

C++

1 1
1 1

Java

1 0 0 1 
0 1 1 0 
0 1 1 0 
1 0 0 1 

Python3

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to construct
// the required binary matrix
void constructBinaryMat(int N)
{
    // Stores binary value with row
    // and column sum as prime number
    int mat[N][N];
 
    // initialize the binary matrix mat[][]
    memset(mat, 0, sizeof(mat));
 
    // Update all possible values of
    // mat[i][i] to 1
    for (int i = 0; i < N; i++) {
        mat[i][i] = 1;
    }
 
    // Update all possible values of
    // mat[i][N - i -1]
    for (int i = 0; i < N; i++) {
        mat[i][N - i - 1] = 1;
    }
 
    // Check if N is an odd number
    if (N % 2 != 0) {
 
        // Update mat[N / 2][0] to 1
        mat[N / 2][0] = 1;
 
        // Update mat[0][N / 2] to 1
        mat[0][N / 2] = 1;
    }
 
    // Print required binary matrix
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            cout << mat[i][j] << " ";
        }
        cout << endl;
    }
}
 
// Driver Code
int main()
{
    int N = 5;
 
    constructBinaryMat(N);
 
    return 0;
}

C#

// Java program to implement
// the above approach
class GFG{
 
// Function to construct
// the required binary matrix
static void constructBinaryMat(int N)
{
  // Stores binary value with row
  // and column sum as prime number
  int mat[][] = new int[N][N];
 
  // Update all possible values
  // of mat[i][i] to 1
  for (int i = 0; i < N; i++)
  {
    mat[i][i] = 1;
  }
 
  // Update all possible values
  // of mat[i][N - i -1]
  for (int i = 0; i < N; i++)
  {
    mat[i][N - i - 1] = 1;
  }
 
  // Check if N is an odd
  // number
  if (N % 2 != 0)
  {
    // Update mat[N / 2][0]
    // to 1
    mat[N / 2][0] = 1;
 
    // Update mat[0][N / 2]
    // to 1
    mat[0][N / 2] = 1;
  }
 
  // Print required binary matrix
  for (int i = 0; i < N; i++)
  {
    for (int j = 0; j < N; j++)
    {
      System.out.print(mat[i][j] + " ");
    }
    System.out.println();
  }
}
 
// Driver Code
public static void main(String[] args)
{
  int N = 5;
  constructBinaryMat(N);
}
}
 
// This code is contributed by Chitranayal

输出：

# Python3 program to implement
# the above approach
 
# Function to construct
# the required binary matrix
def constructBinaryMat(N):
     
    # Stores binary value with row
    # and column sum as prime number
    mat = [[0 for i in range(N)]
              for i in range(N)]
 
    # Initialize the binary matrix mat[][]
    # memset(mat, 0, sizeof(mat));
 
    # Update all possible values of
    # mat[i][i] to 1
    for i in range(N):
        mat[i][i] = 1
 
    # Update all possible values of
    # mat[i][N - i -1]
    for i in range(N):
        mat[i][N - i - 1] = 1
 
    # Check if N is an odd number
    if (N % 2 != 0):
         
        # Update mat[N / 2][0] to 1
        mat[N // 2][0] = 1
 
        # Update mat[0][N / 2] to 1
        mat[0][N // 2] = 1
 
    # Print required binary matrix
    for i in range(N):
        for j in range(N):
            print(mat[i][j], end = " ")
             
        print()
 
# Driver Code
if __name__ == '__main__':
     
    N = 5
 
    constructBinaryMat(N)
     
# This code is contributed by mohit kumar 29

时间复杂度： O(N ² )
辅助空间： O(N ² )