Simple program to check whether a given year is leap year or not?

Answer 1

#include
#include
void main()
{
int n;
clrscr();
printf("\nEnter the year:");
scanf("%d",&n);
if((n%400==0)((n%4==0)&&(n%100!=0)))
printf("Leap Year");
else
printf("Not a Leap Year");
getch();
}

output:

Enter the year:2003
Not a Leap Year
Enter the year:1312
Leap Year
Enter the year:1300
Not a Leap Year
Enter the year:1600
Leap Year

Answer 2

#include #include
int main()
{
int yr,n;
clrscr();
printf("\nPROGRAM TO CHECK IF THE GIVEN YEAR IS LEAP\n");
printf("Enter the year(example:1989):");
scanf("%d",&yr);
if(yr!=0)
{
n=yr%100;/*Extract tens and units place value*/
if((n%4)==0)
printf("\n%d IS A LEAP YEAR\n",yr);
else
printf("\n%d IS NOT A LEAP YEAR",yr);
}
else
{
printf("\nINVALID INPUT\n");
exit(1);
}
return 0;
getch();
}

Answer 3

# store year
yy=0
isleap="false"

echo -n "Enter year (yyyy) : "
read yy

# find out if it is a leap year or not

if [ $((yy % 4)) -ne 0 ] ; then
: # not a leap year : means do nothing and use old value of isleap
elif [ $((yy % 400)) -eq 0 ] ; then
# yes, it's a leap year
isleap="true"
elif [ $((yy % 100)) -eq 0 ] ; then
: # not a leap year do nothing and use old value of isleap
else
# it is a leap year
isleap="true"
fi
if [ "$isleap" == "true" ];
then
echo "$yy is leap year"
else
echo "$yy is NOT leap year"
fi

Check bellow link for other methods

http://bashscript.blogspot.com/2009/11/shell-script-to-check-whether-given.html

Answer 4

10 CLS: PRINT "Type the year you're interested in:"; : Input yr

20 Leap$ = "No"

30 If Int(yr/4) = yr/4 then Leap$ = "Yes"

40 If Int(yr/100) = yr/100 then Leap$ = "No"

50 If Int(yr/400) = yr/400 then Leap$ = "Yes"

60 PRINT: PRINT "Is the year"; yr; "a leap year ?"; " "; Leap$

Answer 5

First goto notepad and type in commands (each on their own line) as you would in command prompt. Then save the file as a batch file (.bat). Your file is now executable.