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C program to print box number pattern of 1 and 0 with cross center
10001
01010
00100
01010
10001

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Write a C program to print the given number pattern of 1's and 0's with cross in the center using loop. How to print box number pattern of 1, 0 with X in center using loop in C programming. Logic to print box number pattern with cross center using loop in C program.

Program Code

/**
 * C program to print box number pattern with cross center
 * www.atnyla.com
 */

#include <stdio.h>

int main()
{
    int rows, cols, i, j;

    /* Input rows and columns from user */
    printf("Enter number of rows: ");
    scanf("%d", &rows);
    printf("Enter number of columns: ");
    scanf("%d", &cols);

    for(i=1; i<=rows; i++)
    {
        for(j=1; j<=cols; j++)
        {
            if(i == j || (j == (cols+1) - i))
            {
                printf("1");
            }
            else
            {
                printf("0");
            }
        }

        printf("\n");
    }

    return 0;
}

Output

Enter number of rows: 5
Enter number of columns: 5
10001
01010
00100
01010
10001

Explanation

Required knowledge

Basic C programming, Loop

Logic to print box number pattern with cross center

In the given pattern, 1 is printed only when -

  • Current column equals to current row.
  • Current column equals (total columns + 1) - current row.

Below is the step by step descriptive logic to print the given number pattern.

  1. Input number of rows and columns to print from user. Store it in some variable say rows and cols.
  2. To iterate through rows run an outer loop from 1 to rows. The loop structure should look like for(i=1; i<=rows; i++).
  3. To iterate through columns run an inner loop from 1 to cols. The loop structure should look like for(j=1; j<=cols; j++).
  4. Inside the inner loop check if(i == j || (j == (cols+1)-i)) then print 1 otherwise print 0.
  5. Finally, move to the next line after printing all columns of a row.

How to learn from this program

First read the algorithm, then study the program code line by line. After that, compare the code with the output and finally go through the explanation. This approach helps learners understand both the logic and the implementation properly.