#!/usr/bin/perl -w
require 5.003;
use  POSIX "floor";

#-------------------------- What This Does ----------------------------
# We specify month, day, and year.  This gives the name of the day of the week.
# Note:   Sun = 0,  Jan =  month 1, Jan 1 = day 1

# The idea is to compute the number of days since some reference day (I use
# Jan. 1, 2000, a Saturday) and compute this Mod 7 since 7 days per week.
# Unusable for dates before Sept. 14, 1752 (Gregorian Calendar reformation)

# This example is intended to be clear.  One can make this much shorter,
# but it is then more difficult to understand.

# Asks you for input.
#---------------------------- Main Program ----------------------------

# The following will be useful.
@DoW = ('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
@MoY=('Jan','Feb','Mar','Apr','May','Jun','Jul','Aug','Sep','Oct','Nov','Dec');
# For a leap year February needs to be corrected (done in Step 1).
@MonthLength = (31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);

print "\n Give me a date and I'll tell you the day of the week.
\n\tMonth?  [1,...12]   ";
$Month = <STDIN>;
chomp($Month);
print "\t\t (you picked  $MoY[$Month-1])\n";

print "\n\tDay?  ";
$Day = <STDIN>;
chomp($Day);

print "\n\tYear? [as: 2012]   ";
$Year = <STDIN>;
chomp($Year);

# Step 1: If Year is a leap year, set  $MonthLength[1]=29;
if  ( (($Year % 4) == 0) and
    ( (($Year % 100) != 0) or ( ($Year % 400) == 0) ) ) { $MonthLength[1]=29; }

# Compute the day number of the given date.  For Feb. 2,
# Day_Num = 31 + 2 = 33 while in a non-leap year, the
# Day_Num of March 1 is 31 + 28 + 1 = 60.


# Step 2:  Compute day_number of the specified Month, Day, Year
$Day_Num = 0;		# Initialize
for  ($i=0; $i < ($Month-1); $i++) {	# Days in preceding months
    $Day_Num = $Day_Num +  $MonthLength[$i];
}
$Day_Num = $Day_Num + $Day;            # Add correction for current month.


# Step 3: Number of _complete_ years between $Year and Jan 1, 2000
$YY = $Year - 2000;


# Step 4: Number of leap years between Jan. 1, 2000 and the $Year -1
$Leap_Years_Since_2000 = floor(($YY-1)/4) -floor(($YY-1)/100) + floor(($YY-1)/400);

$Days_since_Jan1_2000 = $Day_Num  + $YY*365 + $Leap_Years_Since_2000;


# Step 5: divide by 7
$Day_of_Week =  ($Days_since_Jan1_2000 + 6) % 7;   # Jan 1, 2000 was a Sat = 6

$Answer = $DoW[$Day_of_Week];
print "\n$Month/$Day/$Year is a $Answer\n";
print "\t$MoY[$Month-1] $Day, $Year is a $Answer\n";

# =================================================================
# Since everything is mod 7, the following is equivalent but shorter.
# It uses $Day_Num from above but replaces steps 3, 4, and 5.

print "\n\n\t(This uses the shorter version)\n"; 
$Y1 = $Year - 1;           # previous year
$Day_of_Week = ($Day_Num + $Y1 + floor($Y1/4) - floor($Y1/100) + floor($Y1/400) ) % 7;

$Answer = $DoW[$Day_of_Week];
print "\t$MoY[$Month-1] $Day, $Year is a $Answer\n\n";