Project Euler 17: (Iron)Python
- by Ben Griswold
In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 17.
As always, any feedback is welcome.
# Euler 17
# http://projecteuler.net/index.php?section=problems&id=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.
import time
start = time.time()
def to_word(n):
h = { 1 : "one", 2 : "two", 3 : "three",
4 : "four", 5 : "five", 6 : "six",
7 : "seven", 8 : "eight", 9 : "nine",
10 : "ten", 11 : "eleven", 12 : "twelve",
13 : "thirteen", 14 : "fourteen", 15 : "fifteen",
16 : "sixteen", 17 : "seventeen", 18 : "eighteen",
19 : "nineteen", 20 : "twenty", 30 : "thirty",
40 : "forty", 50 : "fifty", 60 : "sixty",
70 : "seventy", 80 : "eighty", 90 : "ninety",
100 : "hundred", 1000 : "thousand" }
word = ""
# Reverse the numbers so position (ones, tens,
# hundreds,...) can be easily determined
a = [int(x) for x in str(n)[::-1]]
# Thousands position
if (len(a) == 4 and a[3] != 0):
# This can only be one thousand based
# on the problem/method constraints
word = h[a[3]] + " thousand "
# Hundreds position
if (len(a) >= 3 and a[2] != 0):
word += h[a[2]] + " hundred"
# Add "and" string if the tens or ones
# position is occupied with a non-zero value.
# Note: routine is broken up this way for [my] clarity.
if (len(a) >= 2 and a[1] != 0):
# catch 10 - 99
word += " and"
elif len(a) >= 1 and a[0] != 0:
# catch 1 - 9
word += " and"
# Tens and ones position
tens_position_value = 99
if (len(a) >= 2 and a[1] != 0):
# Calculate the tens position value per the
# first and second element in array
# e.g. (8 * 10) + 1 = 81
tens_position_value = int(a[1]) * 10 + a[0]
if tens_position_value <= 20:
# If the tens position value is 20 or less
# there's an entry in the hash. Use it and there's
# no need to consider the ones position
word += " " + h[tens_position_value]
else:
# Determine the tens position word by
# dividing by 10 first. E.g. 8 * 10 = h[80]
# We will pick up the ones position word later in
# the next part of the routine
word += " " + h[(a[1] * 10)]
if (len(a) >= 1 and a[0] != 0 and tens_position_value > 20):
# Deal with ones position where tens position is
# greater than 20 or we have a single digit number
word += " " + h[a[0]]
# Trim the empty spaces off both ends of the string
return word.replace(" ","")
def to_word_length(n):
return len(to_word(n))
print sum([to_word_length(i) for i in xrange(1,1001)])
print "Elapsed Time:", (time.time() - start) * 1000, "millisecs"
a=raw_input('Press return to continue')
