Program to print number pattern at even/odd columns
10101
10101
10101
10101
10101
C Programming Language / Loop control in C Language
6212Program:
/** * C program to print number pattern with 1/0 at even/odd position * www.atnyla.com */ #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", !(j%2)); } printf("\n"); } return 0; }
Output:
Enter number of rows: 5 Enter number of columns: 5 10101 10101 10101 10101 10101
Explanation:
The above method is easy to understand and write. However, you can further optimize the previous method by removing the if else condition. The statement j%2 returns 0 if number is even otherwise returns 1. We only need to print complement of value returned by j%2.
Required knowledge
Basic C programming, Loop
Must know - Program to check even number
Logic to print number pattern with 1, 0 at alternate columns
In previous post I explained a similar pattern. Logic to print this is almost similar. If you have noticed the pattern carefully, for every odd columns 0 is printed and for every even columns 1 is printed.
Below is the step by step descriptive logic to print the given pattern.
- 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++).
- To iterate through columns run an inner loop from 1 to cols. The loop structure should look like for(j=1; j<=cols; j++).
- Inside the inner loop print 1 if current column is even otherwise print 0. Means if(j%2==0) then print 1 otherwise print 0.
- Finally move to the next line after printing one column.
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