C++ Program to Print Even Numbers

A C++ program to print even numbers uses a for loop starting from 2 and increments by 2, like for(int i = 2; i <= n; i += 2) std::cout << i << ' '; to print even numbers up to n.

Examples

Input10

Output2 4 6 8 10

Input5

Output2 4

Input1

Output

How to Think About It

To print even numbers up to a given number, start from 2 (the first even number) and keep adding 2 each time until you reach or pass the given number. This way, you only print even numbers without checking each number.

Algorithm

1

Get the maximum number n from the user.

2

Start a loop from 2 to n, increasing by 2 each time.

3

Print the current number in the loop.

4

Stop when the current number exceeds n.

Code

cpp

#include <iostream>

int main() {
    int n;
    std::cout << "Enter the maximum number: ";
    std::cin >> n;
    for (int i = 2; i <= n; i += 2) {
        std::cout << i << " ";
    }
    std::cout << std::endl;
    return 0;
}

Output

Enter the maximum number: 10
2 4 6 8 10 

Dry Run

Let's trace input 10 through the code

1

Input

User enters n = 10

2

Loop start

i = 2, check if i <= 10 (true), print 2

3

Loop continues

i = 4, check if i <= 10 (true), print 4

4

Loop continues

i = 6, check if i <= 10 (true), print 6

5

Loop continues

i = 8, check if i <= 10 (true), print 8

6

Loop continues

i = 10, check if i <= 10 (true), print 10

7

Loop ends

i = 12, check if i <= 10 (false), stop loop

i	Condition i <= n	Action
2	true	print 2
4	true	print 4
6	true	print 6
8	true	print 8
10	true	print 10
12	false	stop

Why This Works

Step 1: Start from 2

We start from 2 because it is the smallest even number and the first number we want to print.

Step 2: Increment by 2

Increasing the loop variable by 2 each time ensures we only get even numbers without extra checks.

Step 3: Loop condition

The loop runs while the current number is less than or equal to the input number, so we don't print numbers beyond the limit.

Alternative Approaches

Check each number if even

cpp

#include <iostream>

int main() {
    int n;
    std::cout << "Enter the maximum number: ";
    std::cin >> n;
    for (int i = 1; i <= n; i++) {
        if (i % 2 == 0) {
            std::cout << i << " ";
        }
    }
    std::cout << std::endl;
    return 0;
}

This method checks every number if it is even, which is less efficient but easy to understand.

Using while loop

cpp

#include <iostream>

int main() {
    int n, i = 2;
    std::cout << "Enter the maximum number: ";
    std::cin >> n;
    while (i <= n) {
        std::cout << i << " ";
        i += 2;
    }
    std::cout << std::endl;
    return 0;
}

Using a while loop instead of for loop achieves the same result with slightly different syntax.

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 and does not use extra space proportional to input size, so space complexity is O(1).

Which Approach is Fastest?

Incrementing by 2 is faster than checking each number for evenness because it skips odd numbers entirely.

Approach	Time	Space	Best For
Increment by 2 in for loop	O(n)	O(1)	Efficient and simple
Check each number if even	O(n)	O(1)	Easy to understand but less efficient
While loop increment by 2	O(n)	O(1)	Alternative syntax, same efficiency