pyJamesBond-Explained

# Sean Connery, who played James Bond, 007, died on 31-10-20
# The sum of the digits of that date is 007

# How many dates are there in 2020 that total 007 in this way?

# Mike Kerry - 15-April-2021

mm, dd, ctr = 1, 1, 0
while mm < 13:
    tot = 2         # Start a total containing 2 (as we are only dealing with year "02"
    # combine month and day into one integer e.g. month 12 day 15 becomes 1215
    x = mm * 100 + dd
    # Write a loop which continues until x becomes zero, processing 1 digit at a time
    while x:
        tot += x % 10       # get remainder of x divided by 10 and add it to tot.
                            # This will be the junior digit (5, then 1, then 2, then 1)

        x //= 10            # Floor-Divide x by 10. This drops the junior digit
                            # so, in our example, 1215 becomes 121

    if tot == 7:            # If the total of all digits is 7 ...
        ctr += 1            # ... add 1 to our counter of matching dates
        # and print the counter and the date. I use zfill to get leading zeroes
        print(ctr, str(dd).zfill(2) + "-" + str(mm).zfill(2) + "-20")


    # Line 44 below is an algorithm I developed a few years ago for getting
    # the number of days per month.    I think it's quite fun
    # I had noticed that odd numbered months from Jan to Jly have 31 days
    # But even numbered months from Aug to Dec have 31 days

    # Explanation: Start with (mm > 7)  This is an expression that is either True or False
    # So if mm is August thru December, the expression evaluates to 1 (True)
    # and for months Jan thru July, it evaluates to zero (False)

    # Then the 1 or zero is added to mm so Jan to Jly stay as 1 to 7
    # and Aug to Dec become 9 to 13. Now it's all the odd numbers that have 31 days.

    # Then we use a bitwise AND (&) to determine if the number from above is odd or even
    # This gives 1 for odd numbers, zero for even
    # So Jan, Mar, May, Jly, Aug, Oct, Dec give a 1, the other months give zero
    # The resultant 1 or zero is added to 30, giving us the correct number of days per month
    # (Apart from Feb which is then handled discretely)
    n = 30 + (mm + (mm > 7) & 1)

    if mm == 2:     # Special case for February
        n = 29      # We are only dealing with 2020, which is a leap year

    dd += 1         # Bump day by 1
    if dd > n:      # If beyond month end ...
        mm += 1     # ... go to next month
        dd = 1      # ... and day 1

# Output:-
# 1 04-01-20
# 2 13-01-20
# 3 22-01-20
# 4 31-01-20
# 5 03-02-20
# 6 12-02-20
# 7 21-02-20
# 8 02-03-20
# 9 11-03-20
# 10 20-03-20
# 11 01-04-20
# 12 10-04-20
# 13 04-10-20
# 14 13-10-20
# 15 22-10-20
# 16 31-10-20
# 17 03-11-20
# 18 12-11-20
# 19 21-11-20
# 20 30-11-20
# 21 02-12-20
# 22 11-12-20
# 23 20-12-20