27: W.A.P.
to decide whether year is leap year or not.
Program:
#include <stdio.h>
#include <conio.h>
void main()
{
clrscr();
int year;
printf("Enter the 'Year' = ");
scanf("%d", &year);
if (year % 4 != 0)
{
printf("%d is not the 'Leap
Year'.", year);
}
else
{
if (year % 100 == 0)
{
if (year % 400 == 0)
{
printf("%d is
the 'Leap Year'.", year);
}
else
{
printf("%d is
not the 'Leap Year'.", year);
}
}
else
{
printf("%d is the
'Leap Year'.", year);
}
}
getch();
}
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Output:
Enter the 'Year' =
2023
2023 is not the 'Leap
Year'.
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Algorithm:
1.
START
2.
Declare
year
3.
Enter the
value of year.
4.
if (year
% 4 != 0) then goto step 5 else goto step 6.
5.
Display
Not a Leap Year.
6.
if (year
% 100 == 0) then goto step 7 else goto step 10.
7.
if (year
% 400 == 0) then goto step 8 else goto step 9.
8.
Display
It is a Leap Year.
9.
Display
Not a Leap Year.
10. Display It is a Leap Year.
11. STOP
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Flowchart:
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