Solves Problem 19.

Some code coming from the internet has been put in utils.
Solution was not used, but seems more elegant.

Signed-off-by: Julien Lengrand-Lambert <julien@lengrand.fr>

e_19.py

@@ -1,4 +1,4 @@
-#!/usr/bin/env python 
+#!/usr/bin/env python
 """
  ##---
  # jlengrand
@@ -18,11 +18,145 @@
  How many Sundays fell on the first of the month during the twentieth century (1 Jan 1901 to 31 Dec 2000)?
  ##---
 """
-def first_sundays():
+class Date:
     """
-    """
-    
-    return 1
+    Date object.
+    Should be initiated with a given year, month and day.
+    Designed to output the value of the day for a given date.
 
-if __name__ == '__main__':
-    print "Answer : %d" % (first_sundays())
+    Should not be used for dates before 1 Jan 1900
+    Month : 1 to 12, should be an int
+    Day : 1 to 31, should be an int
+    """
+    def __init__(self, year, month, day):
+        """
+        Only values needed are day, month and year
+        """
+        self.year = year
+        self.month = month
+        self.day = day
+
+    def days_in_month(self):
+        """
+        Calculates the number of days in the current month
+        """
+        if self.month in [4, 6, 9, 11]:
+            return 30
+        elif self.month is 2:
+            if self.is_leap_year():
+                return 29
+            else:
+               return 28
+        else:
+            return 31
+
+    def _is_end_year(self):
+        """
+        Checks whether this is the end of the current year.
+        Simply checks if we are the 31 December
+        """
+        if (self.month == 12) and (self.day == 31):
+            return True
+        else:
+            return False
+
+    def is_leap_year(self):
+        """
+        Returns true or false depending if this year is a leap year or not.
+        """
+        if self.year % 100 == 0: # we are i a century
+            if self.year % 400 == 0: #special century
+                return True
+            else:
+                return False
+        elif self.year % 4 == 0: # evenly divisible by 4
+            return True
+        else:
+            return False
+
+    def add_day(self):
+        """
+        Adds one day to the current Date.
+        """
+        if ((self.day + 1) > self.days_in_month()):
+            self.day = 1
+            if (self.month + 1) > 12:
+                self.month = 1
+                self.year += 1
+            else:
+                self.month += 1
+        else:
+            self.day += 1
+
+    def __repr__(self):
+        """
+        Nice display for Date Object
+        """
+        return "%d/%d/%d" % (self.day, self.month, self.year)
+
+    def __eq__(self, other_date):
+        """
+        Allows to check if two dates are the same day
+        """
+        if (self.day == other_date.day) and (self.month == other_date.month)  and (self.year == other_date.year):
+                return True
+        else:
+            return False
+
+def ptr_from_start(start_date):
+    """
+    Returns the pointer of day for start_date
+    """
+    #global ref_date
+    #global ref_day
+    #global days
+    ref_date = Date(1900, 1, 1)
+    ref_day = "monday"
+    days = ["monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"]
+
+    day_ptr = days.index(ref_day) # We start from the reference
+    temp_date = ref_date # We suppose the reference is anterior to start_date
+
+    while not(temp_date == start_date):
+            temp_date.add_day()
+            curr_day = days[day_ptr]
+            day_ptr = (day_ptr + 1 ) % len(days)
+
+    return day_ptr
+
+def days_in_period(string_day, nbr_day, start_date, end_date):
+    """
+    Returns the number of string_day that are the nbr_day of the month between
+    start_date and end_date
+
+    WARNING : No checks are made in any way ! Reference is ugly
+    """
+    #global ref_date
+    #global ref_day
+    #global days
+    ref_date = Date(1900, 1, 1)
+    ref_day = "monday"
+    days = ["monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"]
+
+    day_ptr = ptr_from_start(start_date)
+
+    curr_date = start_date
+    end_ptr = 0
+
+    while not (curr_date == end_date):
+        curr_day = days[day_ptr]
+
+        if (curr_day == string_day) and (curr_date.day == nbr_day):
+            end_ptr += 1
+
+        day_ptr = (day_ptr + 1 ) % len(days)        
+        curr_date.add_day()
+
+    return end_ptr
+
+if __name__ == '__main__' :
+    days = ["monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday"]
+    start_date = Date(1901, 1, 1)
+    end_date = Date(2000, 12, 31)
+    print "Answer is : %d" %(days_in_period("sunday", 1, start_date, end_date))
+    raw_input() # USed to keep Windows terminal open