Program to print number pattern with 1, 0
11111
00000
11111
00000
11111
C Programming Language Loop control in C Language (Article) Loop control in C Language (Program)
8194Program:
/** * C program to print number pattern of 1, 0 at even/odd rows */ #include 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++) { printf("%d", (i%2)); } printf("\n"); } return 0; }
Output:
Enter number of rows: 5 Enter number of columns: 5 11111 00000 11111 00000 11111
Explanation:
The previous method if fine and easy to understand. However, you can further optimize the above program. i%2 returns 1 in case of odd and 0 in case of even. Hence, you can remove the if else checking condition. The whole if else condition can be transformed to single printf("%d", (i%2));
Required knowledge
Basic C programming, Loop
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Logic to print 1, 0 number pattern at alternate rows
If you look at the pattern carefully you will notice that for all odd rows 1 is printed and for even rows 0 is printed. Hence, before printing numbers inside inner loop, you need to check even odd condition. If the current row is odd then print 1 otherwise 0.
Below is the step by step descriptive logic to print 1, 0 number pattern at alternate rows.
- Input number of rows and columns to print from user. Store it in some variable say rows and cols.
- To iterate through rows run an outer loop from 1 to rows. The loop structure should look like for(i=1; i<=rows; i++).
- Run another inner loop to iterate through columns from 1 to cols. The loop structure should look like for(i=1; i<=cols; i++).
- Inside the inner loop before printing any number, we need to check condition. Which is for every odd rows 1 is printed and for every even rows 0 is printed. Hence, check if(i%2 == 1) then print 1 otherwise 0.
- Finally, after the inner loop when all columns are printed move to next line. Which is pri
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