Leap year or not

Posted on June 14, 2020

C program to determine whether the year entered by user is a leap year or not.

An year is said to be leap year if
year is a multiple of 4 (except for years evenly divisible by 100, which are not leap years unless evenly divisible by 400).

For example,
year = 1600,
1600 is divisible by both 100 and 400. So, 1600 is a leap year.

year = 1700,
1700 is divisible by 100, but not divisible by 400. So, 1700 is not a leap year.

The above problem can be programmed in two ways:
Using if-else construct, which is explained in method 1 tab.
Using nested if construct, which is explained in method 2 tab.

Program 

#include<stdio.h>
#include<conio.h>
void main()
{
   int year;

    printf("\nEnter the year\t: ");
    scanf("%d",&year);

    if((year%4==0) && (year%100!=0))
    {
	printf("\nLeap year");
    }
    else if((year%4==0) && (year%100==0) && (year%400==0))
    {
        printf("\nLeap year");
    }
    else
    {
 	printf("\nNot a leap year");
    }

    getch();
}

Output 

********** Run1 ********** 
Enter the year  : 1600

Leap year


********** Run2 ********** 
Enter the year  : 1700

Not a leap year

Explanation

A year can be checked whether it is leap year or not by dividing the year by 4, 100 and 400.
If a number is divisible by 4 but not by 100 then, it is a leap year.
Also, if a number is divisible by 4, 100 and 400 then it is a leap year.
Otherwise, the year is not a leap year.

The above conditions are checked using if-else construct

Code to check first condition "If a number is divisible by 4 but not by 100" is
((year%4==0) && (year%100!=0))

For second condition "if a number is divisible by 4, 100 and 400" the code is,
((year%4==0) && (year%100==0) && (year%400==0))

Program 

#include<stdio.h>
#include<conio.h>
void main()
{
    int year;

    printf("\nEnter the year\t: ");
    scanf("%d",&year);

    if(year%4 == 0)
    {
        if(year%100 == 0)
        {
            if(year%400 == 0)
            {
                printf("\nLeap year");
            }
            else
            {
                printf("\nNot a leap year");
            }
        }
        else
        {
            printf("\nLeap year");
        }
    }
    else
    {
        printf("\nNot a leap year");
    }

    getch();
}

Output 

********** Run1 ********** 
Enter the year  : 1600

Leap year


********** Run2 ********** 
Enter the year  : 1700

Not a leap year

Explanation

A year can be checked whether a year is leap year or not by dividing the year by 4, 100 and 400.
If a number is divisible by 4 but not by 100 then, it is a leap year.
Also, if a number is divisible by 4, 100 and 400 then it is a leap year.
Otherwise, the year is not a leap year.

The above conditions are checked using Nested if construct

Here, the year entered by the user is firstly divided by 4. If it is divisible by 4 then it is divided by 100 and then 400.
If year is divisible by all 3 numbers then that year is a leap year.
If the year is divisible by 4 and 100 but not by 400 then it is not a leap year.
If the year is divisible by 4 but not by 100, then it is a leap year. (Remember that if the year is divisible by 4 and not by hundred then the program does not check the last condition, i.e., whether the year is divisible by 400 or not).
If the year is not divisible by 4 then no other conditions are checked and the year is not a leap year.