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Finding a year is leap or not is a bit tricky. We generally assume that if a year number is evenly divisible by 4 is leap year. But it is not the only case. A year is a leap year if −

It is evenly divisible by 100

If it is divisible by 100, then it should also be divisible by 400

Except this, all other years evenly divisible by 4 are leap years.

Let's see how to we can create a program to find if a year is leap or not.

Algorithm of this program is −

START Step 1 → Take integer variable`year`

Step 2 → Assign value to the variable Step 3 → Check if`year`

is divisible by 4 but not 100, DISPLAY "leap year" Step 4 → Check if`year`

is divisible by 400, DISPLAY "leap year" Step 5 → Otherwise, DISPLAY "not leap year" STOP

We can draw a flow diagram for this program as given below −

The pseudocode of this algorithm sould be like this −

procedure leap_year() IF year%4 = 0 AND year%100 != 0 OR year%400 = 0 PRINT year is leap ELSE PRINT year is not leap END IF end procedure

Implementation of this algorithm is given below −

#include <stdio.h> int main() { int year; year = 2016; if (((year % 4 == 0) && (year % 100!= 0)) || (year%400 == 0)) printf("%d is a leap year", year); else printf("%d is not a leap year", year); return 0; }

Output of the program should be −

2016 is a leap year

simple_programs_in_c.htm

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