C Program to Print Even Numbers from 1 to n

Use a for loop from 1 to n and print numbers where i % 2 == 0, like: for(int i=1; i<=n; i++) { if(i % 2 == 0) printf("%d ", i); }.

Examples

Input5

Output2 4

Input10

Output2 4 6 8 10

Input1

Output

How to Think About It

To print even numbers from 1 to n, start counting from 1 up to n. For each number, check if it divides evenly by 2 (no remainder). If yes, print it. This way, only even numbers appear.

Algorithm

1

Get input number n from the user

2

Start a loop from 1 to n

3

For each number, check if it is divisible by 2

4

If divisible, print the number

5

End the loop after reaching 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++) {
        if (i % 2 == 0) {
            printf("%d ", i);
        }
    }
    return 0;
}

Output

Enter the value of n: 10
2 4 6 8 10 

Dry Run

Let's trace input n=5 through the code

1

Input n

User enters n = 5

2

Loop from 1 to n

i starts at 1

3

Check if i is even

i=1, 1 % 2 = 1 (not even), skip print

4

Next i

i=2, 2 % 2 = 0 (even), print 2

5

Continue loop

i=3 (odd, skip), i=4 (even, print 4), i=5 (odd, skip)

6

End loop

Loop ends after i=5

i	i % 2 == 0?	Printed
1	No
2	Yes	2
3	No
4	Yes	4
5	No

Why This Works

Step 1: Loop through numbers

The for loop goes from 1 to n, checking each number.

Step 2: Check even condition

Using i % 2 == 0 tests if the number divides evenly by 2, meaning it is even.

Step 3: Print even numbers

Only numbers passing the even test are printed, so output shows even numbers from 1 to n.

Alternative Approaches

Start loop from 2 and increment by 2

c

#include <stdio.h>

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

This method skips checking odd numbers by starting at 2 and jumping by 2, making it more efficient.

Use while loop with increment

c

#include <stdio.h>

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

Using a while loop with increments of 2 also prints even numbers efficiently without checking each number.

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

Time Complexity

The loop runs from 1 to n, so it executes n times, making the time complexity O(n).

Space Complexity

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

Which Approach is Fastest?

Starting the loop at 2 and incrementing by 2 reduces the number of iterations by half, making it faster than checking each number.

Approach	Time	Space	Best For
Check each number with if	O(n)	O(1)	Simple and clear for beginners
Loop from 2, increment by 2	O(n/2)	O(1)	More efficient, fewer iterations
While loop increment by 2	O(n/2)	O(1)	Alternative loop style, efficient