What Day Calculator, get 8 digit numbers of yyyymmdd then prints what day is yyyymmdd.

what_day.py

'''What Day Calculator based on 8 digits (yyyymmdd)

Input of 8 digits(first four for year, next two for month, last two for day)
Output = the exact day (like "Monday" or "Saturday")

Remember, 0001 Jan 1st was Monday!
'''

#Get an input: first 4 digits for year, next 2 digits for month, last 2 digits for day.
#The input will be saved as a string
date = input("Please enter a date an year in 8 digits ex) 19920303:")
print("Your input:", date)

#Split the string of input as a integer.
year = int(date[:4])
month = int(date[4:6])
day = int(date[6:])

#ord_year means ordinary year, weareas leap_year means leap year.
ord_year = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
leap_year = [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]

what_day=['Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday', 'Sunday']

#var num_day = number of days between input and "00010101"

#Firstly, calcultate num_day on year to year basis. 
#Then, calculate surplus days by month and day and add it to the original num_day

leap = 0
#AD1 means Anno Domini 1. The First Year
AD1 = 1

#If input date = 00010101, just print "Monday"
if date == '00010101':
    print("Monday")
#From the input, let's calculate what day was it
else:
    for i in range(AD1, year+1):
        if i % 400 == 0 or i % 4 == 0 and i % 100 != 0:
            leap +=1
    num_day = (year -AD1 - leap)*365 + leap*366
    if year % 400 == 0 or year % 4 == 0 and year % 100 == 0:
        # leap_year
        num_day += sum(leap_year[:(month - 1)]) + (day-1)
    else:
        # ord_year
        num_day += sum(ord_year[:(month - 1)]) + (day-1)
    print(what_day[(num_day%7)])


print("Good Bye~")