int i,num;
printf("\n\tEnter the number :");
scanf("%d",&num);
for(i=2;i<=num/2;i++) // only really need to go to sqrt(num) but num/2 works
if(num%i==0{
printf("\n\tNot a Prime Number");
getch();
return;
}
// no need to count them one is enough to be not prime
printf("\n\tIt is a prime number");
getch();
return;
}
void leap_year()
{
int year;
printf("\n\tEnter a year to check if it is a leap year :");
scanf("%d", &year);
if(year%400==0 || (year%100!=0 && year%4==0))
printf("%d is a leap year",year);
else
printf("%d is not a leap year",year);
getch();
return;
}
Objective need: To determine whether the given year is leap or not.
How to ran the program: It depends on which C/C++ Compiler you use.
Limitations: No limitations.
I wrote the following Program:
#include
#include
#include
#include
void even_odd();
void prime_number();
void leap_year();
void main()
{
clrscr();
int choice;
while(1)
{
gotoxy(5,1); printf("-------------------");
gotoxy(5,2); printf("Menu Driven Program");
gotoxy(5,3); printf("-------------------");
gotoxy(5,5); printf("1.Even or Odd");
gotoxy(5,6); printf("2.Prime or Not");
gotoxy(5,7); printf("3.Leap year or Not");
gotoxy(5,8); printf("4.Exit");
printf("\n\n\tEnter Your Choice :");
scanf("%d",&choice);
switch( choice )
{
case 1 :
even_odd();
delay(200);
clrscr();
break;
case 2 :
prime_number();
delay(200);
clrscr();
break;
case 3 :
leap_year();
delay(200);
clrscr();
break;
case 4 :
exit(1);
}
}
}
void even_odd()
{
int number;
printf("\n\tEnter number :");
scanf("%d",&number);
if( number %2 == 0)
printf("\n\t%d is an even number.",number);
else
printf("\n\t%d is an odd number.",number);
getch();
}
void prime_number()
{
int i,num;
printf("\n\tEnter the number :");
scanf("%d",&num);
for(i=2;i<=(num-1);)
{
if(!(num%i==0))
i++;
else
{
printf("\n\tNot a Prime Number");
break;
}
}
if(i==num)
printf("\n\tIt is a prime number");
getch();
}
void leap_year()
{
int year;
printf("\n\tEnter a year to check if it is a leap year :");
scanf("%d", &year);
if ( year%100 == 0)
{
