C Program to Check a Leap Year

Problem

Write a program to check whether a given year is a leap year or not according to the Gregorian Calendar System.

Example 1:

Input:
Enter the year: 2024

Output:
It's a leap year.

Explanation:
Year 2024 is a leap year because it is perfectly divisible by 4.

Example 2:

Input:
Enter the year: 2021

Output:
It's not a leap year.

Explanation:
Year 2021 was not a leap year because it is not divisible by 4.

Example 3:

Input:
Enter the year: 2100

Output:
It's not a leap year.

Explanation:
Year 2100 is not a leap year because it is divisible by 100 but not divisible by 400.

Expectations:

You need to write the complete program using the following template as the starter:

#include <stdio.h>

int main() {
    int year;
    
    // The prompt.
    printf("Enter the year: ");
    scanf("%d", &year);
    
    // Write your code here.
    
    return 0;
}

Copy and paste the above code in your favorite IDE and start writing your code after the comment // Write your code here.

Sample Test Cases:

To verify the program you have written, go ahead and test your program using the following test cases:

Don’t forget to try the code on your own before diving into the solution.

Solution

Following are the multiple approaches to the same problem:

The Easiest Approach:

We can use Nested if-else to check multiple conditions as shown in the following code:

#include <stdio.h>

int main() {
    int year;
    
    // The prompt.
    printf("Enter the year: ");
    scanf("%d", &year);
    
    // Write your code here.
    if (year % 4 == 0) {
        if (year % 100 != 0) {
            printf("It's a leap year.\n");
        } else {
            if (year % 400 == 0) {
                printf("It's a leap year.\n");
            } else {
                printf("It's not a leap year.\n");    
            }
        }
    } else {
        printf("It's not a leap year.\n");
    }
    return 0;
}

Output:

Enter the year: 2000
It's a leap year.

You can check the above program for multiple test cases and the code works just fine.

Better Approach:

The previous program can easily be rewritten using the logical conditions as follows:

#include <stdio.h>

int main() {
    int year;
    
    // The prompt.
    printf("Enter a year: ");
    scanf("%d", &year);
    
    // Write your code here.
    if ((year % 4 == 0 && year % 100 != 0) || (year % 400 == 0)) {
        printf("It's a leap year");
    } else {
        printf("It's not a leap year");
    }

    return 0;
}

Output:

Enter a year: 2022
It's not a leap year