/*
Program to find the day on a certain date entered by the user.

#######################################################################
# (c) 2007 by Krishna Khanna    
#  krishnakhanna@gmail.com
# http://kkonline.org
#######################################################################
*/
# include <iostream.h>
# include <iomanip.h>
# include <conio.h>
# include <ctype.h>

void main()
{
 clrscr();
 int year,month,date,m,r,leap,yeard;
 char cont;

 textcolor(5);
 cout<<"\t\t";
 cprintf("     THE DAY CALCULATOR");
 textcolor(4);
 cout<<endl<<"  ";
 cprintf("\n\n\n\n\n  USING THIS UTILITY YOU CAN KNOW THE DAY ON ANY DATE THAT YOU ENTER");
 cout<<endl<<"  ";
 cprintf("\n  SIMPLY FOLLOW THE INSTRUCTIONS ON YOUR SCREEN.");
 textcolor(138);
 cprintf("\n\n\n\n\n\n\n\n\n\n\n");
 cout<<endl<<"\t\t\t";
 cprintf("PRESS ANY KEY TO CONTINUE ");
 textcolor(4);
 getch();
 clrscr();

 do
 {
 clrscr();
 textcolor(14);
 cprintf("ENTER THE YEAR (e.g. 1978) : ");
 cin>>year;
 cprintf("ENTER THE NUMBER OF THE MONTH ( e.g. 7 FOR JULY ) : ");
 cin>>month;
 cprintf("ENTER THE DATE (1-31, 1-30, 1-29/28 depending on the month) : ");
 cin>>date;
 m=7;// 1 jan 2000 was a saturday
 if(year<2000)
 {
  yeard=2000-year;
  r=yeard/4;  // calculation of number of leap years in between
  r=r+yeard;
  r=r%7;
  m-=r;
 }
 if(year>2000)
 {
  yeard=year-2000 ;
  if (yeard%4==0)
    r=(yeard/4)+1;
  else
    r=(yeard/4)+1 ; // calculation of number of leap years in between & 2000 was a leap year
  r=r+yeard ;
  r=r%7 ;

  m+=r ;
  m=m%7 ;

 }



 r=year%100;
 r%=4; // if r==0 then it is a leap year;

 if((r==0)&&(year>2000))m--;

 switch(month)
 {
  case 12: m+=(30%7);// nov has 30 days.
  case 11: m+=(31%7);// oct has 31 days.
  case 10: m+=(30%7);// sep has 30 days.
  case 9 : m+=(31%7);// aug has 31 days.
  case 8 : m+=(31%7);// july has 31 days.
  case 7 : m+=(30%7);// june has 30 days.
  case 6 : m+=(31%7);// may has 31 days.
  case 5 : m+=(30%7);// apr has 30 days.
  case 4 : m+=(31%7);// march has 31 days.
  case 3: {
	    if(r==0)
	     m+=(29%7); // in a leap year feb has one extra day.
	    else
	     m+=(28%7);
	   }
  case 2: m+=(31%7); // jan has 31 days.


 }


 m=m+(date%7);
 m=m%7  ;
 m--;
 if(m<=0)
 m=7+m;

 // displaying result
 cout<<"\n\n\n\n\n";
 textcolor(8);
 cout<<date<<'/'<<month<<'/'<<year<<" WAS OR WILL BE A ";
 switch(m)
 {
  case 1: cprintf(" SUNDAY "); break;
  case 2: cprintf(" MONDAY "); break;
  case 3: cprintf(" TUESDAY "); break;
  case 4: cprintf(" WEDNESDAY "); break;
  case 5: cprintf(" THURSDAY "); break;
  case 6: cprintf(" FRIDAY "); break;
  case 7: cprintf(" SATURDAY "); break;
  default: cprintf("\n AN EXCEPTIONAL ERROR HAS OCURRED CANNOT FIND DATE");
 }
 cout<<"\n\n";
 textcolor(4);
 cprintf(" DO YOU WANT TO FIND THE DAY OF ANOTHER DATE(Y/N) : ");
 textcolor(4);
 cin>>cont;
 cont=tolower(cont);
 }
 while(cont=='y');
 getch();

 }