Program that allows to accept only 10 integers but loops if an odd number was entered

[perfringo]

It always make sense to have a plan. In order to make plan we must make inventory i.e. what are the terms and conditions.

- accept only 10 integers on a list
- not accept 3 consecutive odd integers

Now we should try to decompose problem to subproblems and subsolutions.

How translate 'only 10 integers on a list' into Python? Every Python list knows it's length i.e. number of elements. So we can do whatever we need until length of the list is less than ten (note <10. Why so? Because on every iteration we add one value to list and we don't want to add additional item to list which has already 10 items)

nums = []
while len(nums) < 10:
    ...

In order to add items to nums and keep tab on consecutive odd numbers we can write:

consecutive = 0

if num % 2:                   # if odd
    consecutive += 1          # add one to counter
    nums.append(num)
else:                         # if not odd i.e. even
    nums.append(num)
    consecutive = 0           # set counter to zero

We count consecutives, but not taking any action based on that. So we should add condition before we append odd number to list (note that condition is 3 <= consecutive: and not ==. Why? Because counter increases before we check it's value, so with consecutively entering odd numbers counter increases and can be more than 3. Counter is reset only if user enters even number)

if 3 <= consecutive:
    print("You can't enter 3 consecutive odd numbers.")
else:
    nums.append(num)

And putting it all together:

nums = []
consecutive = 0

while len(nums) < 10:
    num = int(input('Enter integer: '))
    if num % 2:
        consecutive += 1
        if 3 <= consecutive:
            print("You can't enter 3 consecutive odd integers.")
        else:
            nums.append(num)
    else:
        nums.append(num)
        consecutive = 0

Note that program crashes if user enters any value which can't be converted into integer.