Construya una Array Binaria cuya suma de cada fila y columna sea un Número Primo

Dado un número entero N , la tarea es construir una array binaria de tamaño N*N tal que la suma de cada fila y cada columna de la array sea un número primo .

Ejemplos:

Entrada: N = 2 
Salida: 

1 1
1 1

Explicación: 
Suma de la fila 0 ⁼ 1 + 1 = 2 (Número primo) 
Suma de la fila 1 ⁼ 1 + 1 = 2 (Número primo) 
Suma de la columna 0 ⁼ 1 + 1 = 2 (Número primo) 
Suma de 1 ^ª columna = 1 + 1 = 2 (Número primo)

Entrada: N = 4 
Salida: 

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

Enfoque: siga los pasos a continuación para resolver el problema:

• Inicialice una array binaria , digamos mat[][] de tamaño N * N.
• Actualice todos los valores posibles de mat[i][i] a 1 .
• Actualice todos los valores posibles de mat[i][N – i -1] a 1 .
• Si N es un número impar , actualice el valor mat[N/2][0] y mat[0][N/2] a 1 .

A continuación se muestra la implementación del enfoque anterior.

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
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;
}

// 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

# 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

// C# program to implement
// the above approach
using System;
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++)
    {
      Console.Write(mat[i, j] + " ");
    }
    Console.WriteLine();
  }
}
 
// Driver Code
public static void Main(String[] args)
{
  int N = 5;
  constructBinaryMat(N);
}
}
 
// This code is contributed by gauravrajput1

<script>
    
// Javascript program to implement
// the above approach
 
// Function to construct
// the required binary matrix
function constructBinaryMat(N)
{
     
    // Stores binary value with row
    // and column sum as prime number
    var mat = Array.from(Array(N), () =>
                         Array(N).fill(0));
 
    // Update all possible values of
    // mat[i][i] to 1
    for(var i = 0; i < N; i++)
    {
        mat[i][i] = 1;
    }
 
    // Update all possible values of
    // mat[i][N - i -1]
    for(var 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[parseInt(N / 2)][0] = 1;
 
        // Update mat[0][N / 2] to 1
        mat[0][parseInt(N / 2)] = 1;
    }
 
    // Print required binary matrix
    for(var i = 0; i < N; i++)
    {
        for(var j = 0; j < N; j++)
        {
            document.write( mat[i][j] + " ");
        }
        document.write("<br>");
    }
}
 
// Driver Code
var N = 5;
 
constructBinaryMat(N);
 
// This code is contributed by rutvik_56
 
</script>

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

Tiempo Complejidad: O(N ² )
Espacio Auxiliar: O(N ² )