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C Program To Find Leap Years.

In the Gregorian calendar (the current standard calendar in the world), normally the years that are integer multiples of 4 are called leap years. In each leap year, the month of February has 29 days instead of 28. Adding an extra day to the calendar every four years compensates for the fact that a period of 365 days is shorter than a solar year by almost 6 hours. The Gregorian calendar was first used in 1582.

Some exceptions to this basic rule are required since the duration of a solar year is slightly less than 365.25 days. Over a period of 4 centuries, the accumulated error of adding a leap day every 4 years amounts to about 3 extra days. The Gregorian calendar therefore omits 3 leap days every 400 years, which is the length of its leap cycle. This is done by omitting February 29 in the 3 century years (integer multiples of 100) that are not also integer multiples of 400. For example, 1700, 1800, 1900, 2100 and 2200 are common years, but 1600, 2000 and 2400 are leap years. By this rule, the average number of days per year is 365 + 1/4 − 1/100 + 1/400 = 365.2425.

There is no year 0. The first year in the Gregorian calendar is year 1. The year before 1AD in the western calendar is 1BC, not "year 0".

Method 1:
#include <stdio.h>
int main() {

int year;
printf(" Enter a year: ");
scanf("%d", &year);

while ( year <= 0) {
printf("\n\nWrong Input! The Gregorian Calendar Starts with Year 1.\n\n");
return 0;
}

while ( year > 0) {

if ( year%400 == 0)
printf("\n\n %d is a leap year.\n\n", year);
else if ( year%100 == 0)
printf("%\n\n %d is not a leap year.\n\n", year);
else if ( year%4 == 0 )
printf("\n\n %d is a leap year.\n\n", year);

else
printf("\n\n %d is not a leap year.\n\n", year);
return 0;
}}

Method 2:
#include<stdio.h>
int main() {

int year;
printf(" Enter any year: ");
scanf("%d",&year);

while ( year <= 0) {
printf("\n\nWrong Input! The Gregorian Calendar Starts with Year 1.\n\n");
return 0;
}

while ( year > 0) {
if(((year%4==0)&&(year%100!=0))||(year%400==0))
printf("\n\n %d is a leap year.\n\n",year);
else
printf("\n\n %d is not a leap year.\n\n",year);
return 0;
}}

Method 3: Find the leap years between a given year range
#include<stdio.h>
int main(){

int year;
int min_year,max_year;
printf("\n\n Enter the lowest year: ");
scanf("%d",&min_year);
printf("\n\n Enter the highest year: ");
scanf("%d",&max_year);

if (( min_year < 1)||(max_year <1)) {
printf("\n\nWrong Input! The Gregorian Calendar Starts with Year 1.\n\n");
return 0;
}

if ( min_year > max_year) {
printf("\n\n Wrong Input!\n\n");
return 0;
}
printf("\n\n Leap years in the given range are: ");

for(year = min_year;year <= max_year; year++){
if(((year%4==0)&&(year%100!=0))||(year%400==0))
printf("%d ",year);
}
printf("\n\n");
return 0;
}
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