problem 17

"""
If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.

If all the numbers from 1 to 1000 (one thousand) inclusive were written out in words, how many letters would be used?

NOTE: Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage."""

num_9 = ["", "one" , "two" , "three" , "four" , "five" , "six" , "seven" , "eight" , "nine" , "ten"  , "eleven" , "twelve" , "thirteen" , "fourteen" , "fifteen" , "sixteen" , "seventeen" , "eighteen" , "ninteen"]
num = ["","" , "twenty" , "thirty" , "forty" , "fifty" , "sixty" , "seventy" , "eighty" , "ninety" ]



def NumberLength(i):
  sumof = 0
  if i >= 1 and i < 20 :
    sumof += len(num_9[i])

  # for numbers less than 100
  if i >= 20 and i < 100:
    first = i % 10
    second = ((i % 100) - first) // 10
    if first == 0 :
      sumof += len(num[second])
    if first !=0:
     sumof += len(num_9[first])  + len(num[second])

  # for numbers from  99 to  1000
  if i >= 100 and i < 1000 :
    first = i % 10
    second = ((i % 100) - first) // 10
    third = ((i % 1000) - (i % 100)) // 100
    special = i % 100

    if first == 0 and second ==0:
      sumof += 7 + len(num_9[third])

    if second == 0 and first != 0:
      sumof +=len(num_9[third]) + 10 + len(num_9[first])

    if second == 1:
     print(special)
     sumof += len(num_9[third]) + 10  + len(num_9[special])

    if first == 0 and second > 1:
      sumof +=   len(num_9[third]) + 10   + len(num[second])

    if first != 0 and second > 1:
      sumof +=   len(num_9[third]) + 10 + len(num[second]) + len(num_9 [first])
  return sumof


#print(NumberLength(999))
sumof = 0
for i in  range(1,1000):
 sumof += NumberLength(i)
 print(sumof +21)