C program to find Julian Date (day of the year)

This program does not check for validity of date ie. user has to enter a vaild date in the form dd mm yy
eg.31 12 04 ->julian date: 366 since its a leap year

#include<stdio.h>
void main()
{
int dd,mm,yy,julian;
clrscr();
printf("\nenter a valid date (dd mm yy)");
scanf("%d%d%d",&dd,&mm,&yy);
julian=dd;
mm=mm-1;
switch(mm)
{
case 11 : julian=julian+30;
case 10 : julian=julian+31;
case 9 : julian=julian+30;
case 8 : julian=julian+31;
case 7 : julian=julian+31;
case 6: julian=julian+30;
case 5: julian=julian+31;
case 4 : julian=julian+30;
case 3 : julian=julian+31;
case 2 : if(yy%4==0)julian=julian+29;
             else julian=julian+28;
case 1 : julian=julian+31;
}
printf("\nJulian date is %d",julian);
getch();
}