//=========================================================================
#define PROGRAMMER "BAHN, William"
#define PROG_CODE  "bahw"
#define COURSE     "ECE-1021"
#define YEAR       (2003)
#define TERM       "Fall"
#define SECTION    (0)
#define ASSIGNMENT "RWA 4.6"
#define REVISION   (0)
#define TITLE      "Real World Application from Section 4.6"
#define SUBTITLE   "Printing a Calandar"
#define EMAIL      "wbahn@eas.uccs.edu"
#define FILENAME   "rwa0406.c"
//=========================================================================

// PROBLEM:
//
// MODIFIED FROM TEXT
//
// Print a calandar for the year entered by the user.
//
// PSEUDOCODE
// 1) REM: Print Header
// 2) GET: Year from User
// 3) REM: Determine if year is a leap year
// 4) REM: Determine the day-of-week that year starts on
// 5) REM: Print Calandar
// 6) EXIT program

//== DETERMINING IF YEAR IS LEAP YEAR======================================
// For the Gregorian Calander:
//   A year is a leap year if ALL of the following conditions apply:
//    The year is divisible by 4
//    The year is NOT divisible by 100 UNLESS the year is divisible by 400
//
// PSEUDOCODE
// 1) IF: (YEAR%400) = 0
// 1.1) YEAR is leap year
// 2) ELSE:
// 2.1) IF: (YEAR%100) = 0
// 2.1.1) YEAR is NOT leap year
// 2.2) ELSE:
// 2.2.1) IF: (YEAR%4) = 0
// 2.2.1.1) YEAR is leap year
// 2.2.2) ELSE:
// 2.2.2.1) YEAR is NOT leap year
//
// Notice that each if() expression is different, therefore a switch()
// statement can't be used (without first setting a condition code)
//
// But also notice that each expression returns a numeric value:
//  
//== DETERMINING THE DAY-OF-WEEK FOR 01 JAN OF YEAR =======================
// For a non-leap year, 01 Jan of the next year is set back one day
//   i.e., If 01 Jan of (year) is TUE then 01 JAN of (year+1) is WED
// For a leap year, 01 Jan of the next year is set back two days
//   i.e., If 01 Jan of (year) is TUE then 01 JAN of (year+1) is THU
// THEREFORE
// If we can determine the day-of-week for 01 Jan 0000 then we can compute
// how many days of cumulative setback there are.
//
// Given YEAR, there are:
//    YEAR days    (setback due to the 365 days each year has)
//   +YEAR/4 days  (additional setback due to the leap years)
//   -YEAR/100     (fewer setback due to century rule)
//   +YEAR/400     (additional setback due to four-century rule)
//
// All of the above are integer divisions with truncation.
//
// HAND EXAMPLE
//
// The year 0012
//
// The 12 years prior to 0007 contained two leap years (0000, 0004, 0008)
// The setback total should be 12+3-0+0 which agrees with formula.
//
// The year 0100
//
// The 100 years prior to 0100 contained 25 leap years.
// The setback total should be 100+25-0+0 which does NOT agree with formula.
//
// The problem is that the formula includes the effect of the current year
// on the setback and we only want the affect of all prior years. At times,
// this error is compensated for by the year 0000 which was a leap year.
//
// New formula:
// Given YEAR, there are:
//    (YEAR-1) days    (setback due to the 365 days each year has)
//   +(YEAR-1)/4 days  (additional setback due to the leap years)
//   -(YEAR-1)/100     (fewer setback due to century rule)
//   +(YEAR-1)/400     (additional setback due to four-century rule)
//
// The year 0012 yields 11+2-0+0 (which is off by the two days of 0000)
// The year 0100 yields 99+24-0+0 (which is off by the two days of 0000)
// The year 0001 yields 0+0-0+0 (which is off by the two days of 0000)
//
// We can therefore either add the two days as a constant or just use
// the formula as is and treat is as the setback from 03 JAN 0000.
//
// To find the day-of-week for 01 JAN 0000 we simply work backwards from
// a known date. 01 JAN 2003 was on a WED.
//
// The year 2003 yields 2002+500-20+5=2487 (+2 = 2489)
// Taking this setback modulo 7 yields a net setback of 4 days.
// Therefore, 01 JAN 0000 would have been on a SAT.
//
// If we code the days of the week starting with SUN = 0 and SAT = 6,
// then we have the code for 01 JAN of YEAR as follows:
//
// Jan1st = (6+2+(year-1)+(year-1)/4-(year-1)/100+(year-1)/400)%7
// Jan1st = (1+(year-1)+(year-1)/4-(year-1)/100+(year-1)/400)%7
//
// Note, since the modulo operator is not well defined for negative
// operands, the results for year 0000 may not be correct.

//== INCLUDE FILES ========================================================
#include <stdio.h> // printf(), scanf(), EOF

//== FUNCTION PROTOTYPES ==================================================
void PrintHeader(void);

//== MAIN FUNCTION ========================================================
int main(void)
{
	int year, leap;
   int month, day, days;
	int dow;

	PrintHeader();
	printf("\n"); // Print a blank line

	// ********** GET THE YEAR FROM USER ************************

	printf("Enter the year you want the calendar for: ");
	scanf("%i", &year);

	if(!(year%400))
		leap = 1;
	else
		if(!(year%100))
			leap = 0;
		else
			if(!(year%4))
				leap = 1;
			else
				leap = 0;

	dow = (1+(year-1)+(year-1)/4-(year-1)/100+(year-1)/400)%7;

	for(month = 1; month <=12; month++)
	{
   	printf("\n");
		switch(month)
		{
			case  1: printf("January   %4i:  ", year); days = 31; break;
			case  2: printf("February  %4i:  ", year); days = 28+leap; break;
			case  3: printf("March     %4i:  ", year); days = 31; break;
			case  4: printf("April     %4i:  ", year); days = 30; break;
			case  5: printf("May       %4i:  ", year); days = 31; break;
			case  6: printf("June      %4i:  ", year); days = 30; break;
			case  7: printf("July      %4i:  ", year); days = 31; break;
			case  8: printf("August    %4i:  ", year); days = 31; break;
			case  9: printf("September %4i:  ", year); days = 30; break;
			case 10: printf("October   %4i:  ", year); days = 31; break;
			case 11: printf("November  %4i:  ", year); days = 30; break;
			case 12: printf("December  %4i:  ", year); days = 31; break;
		}

		printf("SUN MON TUE WED THU FRI SAT");
		if(dow) // Months not starting on Sunday need padding
		{
			printf("\n");
			printf("                ");
			for(day = 1; day <= dow; day++)
				printf("    ");
		}

		for(day = 1; day<=days; day++)
		{
			if(0 == dow)
			{
				printf("\n");
				printf("                ");
			}
			printf("  %2i", day);
			dow++;
			dow%=7;
		}
		printf("\n");


	}
	printf("\n");
	printf("PROGRAM TERMINATED");
	return(0);
}

//== SUPPORT FUNCTIONS ====================================================
void PrintHeader(void)
{
	printf("========================================"
			 "=======================================\n");
	printf("Course....... %s-%i (%s %i)\n", COURSE, SECTION, TERM, YEAR);
	printf("Programmer... %s (%s)\n", PROGRAMMER, PROG_CODE);
	printf("Assignment... %s (Rev %i) (Source Code in %s)\n",
		ASSIGNMENT, REVISION, FILENAME);
	printf("Description.. %s\n", TITLE);
	printf("              %s\n", SUBTITLE);
	printf("========================================"
			 "=======================================\n");

	return;
}