C Program to Print Odd Numbers from 1 to n

Use a for loop starting from 1 to n incrementing by 2 and print each number with printf, like for(int i=1; i<=n; i+=2) printf("%d\n", i);.

Examples

Input5

Output1 3 5

Input10

Output1 3 5 7 9

Input1

Output1

How to Think About It

To print odd numbers from 1 to n, start counting from 1 and keep adding 2 each time to get the next odd number. Stop when you reach or pass n. This way, you only print odd numbers without checking each number.

Algorithm

1

Get input number n from the user

2

Start a loop with a variable i from 1 to n

3

Increase i by 2 in each step to get only odd numbers

4

Print the current value of i in each loop iteration

5

Stop the loop when i becomes greater than n

Code

c

#include <stdio.h>

int main() {
    int n;
    printf("Enter the value of n: ");
    scanf("%d", &n);
    for (int i = 1; i <= n; i += 2) {
        printf("%d\n", i);
    }
    return 0;
}

Output

Enter the value of n: 5
1
3
5

Dry Run

Let's trace the program with input n=5 through the code

1

Input

User enters n = 5

2

Initialize loop variable

i = 1

3

Check condition

Is i (1) <= n (5)? Yes, continue

4

Print i

Print 1

5

Increment i

i = i + 2 = 3

6

Repeat condition check

Is i (3) <= n (5)? Yes, continue

7

Print i

Print 3

8

Increment i

i = i + 2 = 5

9

Repeat condition check

Is i (5) <= n (5)? Yes, continue

10

Print i

Print 5

11

Increment i

i = i + 2 = 7

12

Repeat condition check

Is i (7) <= n (5)? No, stop loop

i
1
3
5

Why This Works

Step 1: Start from 1

We start the loop at 1 because 1 is the first odd number.

Step 2: Increment by 2

Adding 2 each time ensures we only get odd numbers (1, 3, 5, ...).

Step 3: Stop at n

The loop stops when the number exceeds n, so we print only odd numbers up to n.

Alternative Approaches

Check each number if odd

c

#include <stdio.h>

int main() {
    int n;
    printf("Enter the value of n: ");
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        if (i % 2 != 0) {
            printf("%d\n", i);
        }
    }
    return 0;
}

This method checks every number from 1 to n, which is simpler but less efficient than incrementing by 2.

Use while loop

c

#include <stdio.h>

int main() {
    int n, i = 1;
    printf("Enter the value of n: ");
    scanf("%d", &n);
    while (i <= n) {
        printf("%d\n", i);
        i += 2;
    }
    return 0;
}

Using a while loop works the same way but some may find it easier to read.

Complexity: O(n/2) time, O(1) space

Time Complexity

The loop runs roughly n/2 times because it increments by 2, so time complexity is O(n/2), which simplifies to O(n).

Space Complexity

The program uses a fixed amount of memory for variables, so space complexity is O(1).

Which Approach is Fastest?

Incrementing by 2 is faster than checking each number with a condition because it skips even numbers entirely.

Approach	Time	Space	Best For
Increment by 2	O(n)	O(1)	Efficient and simple for printing odd numbers
Check each number	O(n)	O(1)	Simple logic but less efficient
While loop increment by 2	O(n)	O(1)	Alternative loop style, same efficiency