Question

Complete task using python code

Problem 2: Leap Years (2 Points) Your task is to write a program that takes in both a month and year (defined in the definitions file) and outputs the number of days for that month. It should take into account whether the year is a leap year or not. The resulting number of days in a given month/year combo is to be assigned to the variable days. All month names should be treated as capitalized names (e.g., September' is good, but neither september or SEPTEMBER' is allowed). This exercise should be completed using only the Python keywords discussed in class. Note that most years that are divisible by 4 without a remainder are leap years. There are two caveats. A year divisible by 100 without a remainder generally is not a leap year. However, if that same year is also divisible by 00 without a remainder, then it is a leap year. For example, 2004, 2008, 2012, and 2016 are leap years. 1700, 1800, 1900, and 2100 are not leap years. 1600, 2000, 2400, and 2800 are leap years

From definitions file:

month = ‘February’

year = 1200

Answer 1

month-input ("Month") year int (input ("Year") # creating two dictionaries, one for leap yea" and one for non leap years having months and corresponding no. of days dictleap { 'January' :31, . February' :29, "March' :31, 'Apr11' :30, "May. :31, 'June' : 30, ' July' :31, .August' :31, 'September. :30, .october, :31, "November. :30, "December. :31) dict not_leap 'January':31, February:28, March:31'April':30, May31 June':30, Jly:31,August':3, September':30 Otober31, November:30, December' :31 - leapo # condition8 to determine if a year 18 leap or not, leap = 1 mean8 year 18 leap If year 0: if year 100-0: If year 400-0: leap1 else: leapd else: leap-1 else: 1eap0 # displaying days of the month accdg to whether year is leap or not leap-1: pzint (dict_leap[month)) else: print (dict not_leap [month])

Sample Run with Provided Input:

Python Code with Comments:

month = input("Month")
year = int(input("Year"))

# creating two dictionaries, one for leap years and one for non leap years having months and corresponding no. of days
dict_leap = {'January':31, 'February':29, 'March':31, 'April':30, 'May':31, 'June':30, 'July':31, 'August':31, 'September':30, 'October':31, 'November':30, 'December':31}
dict_not_leap = {'January':31, 'February':28, 'March':31, 'April':30, 'May':31, 'June':30, 'July':31, 'August':31, 'September':30, 'October':31, 'November':30, 'December':31}

leap = 0

# conditions to determine if a year is leap or not, leap = 1 means year is leap
if year % 4 == 0:
if year % 100 == 0:
if year % 400 == 0:
leap = 1
else: leap = 0
else: leap = 1
else: leap = 0

# displaying days of the month accdg to whether year is leap or not
if leap == 1:
print(dict_leap[month])
else: print(dict_not_leap[month])