只是稍微关闭项目欧拉＃17

Question

我发布了这个问题几个小时前，但我想我删除它！真的很抱歉...我正在研究Project Euler Problem 17.

虽然还有其他更明显的解决方案，但作为一个学习练习，我接近了想要使用递归解决问题的问题。我也希望某些代码片段稍后可以在其他情况下使用。问题描述本身位于代码顶部的文档字符串中，对于那些不熟悉的代码。

这里是有问题的代码：

"""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 collections 
import re 


SMALLS =  [(0,  "zero"), 
       (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")] 

MULTS_OF_TEN = [(20, "twenty"), 
       (30, "thirty"), 
       (40, "forty"), 
       (50, "fifty"), 
       (60, "sixty"), 
       (70, "seventy"), 
       (80, "eighty"), 
       (90, "ninety")] 

HUNDRED =  [(100, "hundred")] 

BIGS =   [(1000, "thousand"), 
       (10**6, "million"), 
       (10**9, "billion")] 

# other bigs: trillion, quadrillion, quintillion, sextillion, septillion, octillion, nonillion, decillion, 
#    undecillion, duodecillion, tredecillion, quattuordecillion, quindecillion, sexdecillion, 
#    septendecillion, octodecillion, novemdecillion, vigintillion 

SMALLS = collections.OrderedDict(reversed(SMALLS)) 
MULTS_OF_TEN = collections.OrderedDict(reversed(MULTS_OF_TEN)) 
HUNDRED = collections.OrderedDict(reversed(HUNDRED)) 
BIGS = collections.OrderedDict(reversed(BIGS)) 


def int_to_words(num, follows_hundreds=False): 
    """Retuns the text-equivelent of num, using recursion for distinct 
    pieces. 
    """ 
    def do_chunk(n, 
       num_text_map, 
       include_quotient=True, 
       is_hundreds=False): 

     for x in num_text_map: 
      quotient = n // x 
      remainder = n % x 

      if n == x: return num_text_map[x] 

      if quotient and remainder: 
       quotient_text = (int_to_words(quotient) + " " + num_text_map[x]) \ 
           if include_quotient else \ 
           (num_text_map[x]) 
       remainder_text = int_to_words(remainder, follows_hundreds=True) \ 
           if is_hundreds else \ 
           int_to_words(remainder) 
       return quotient_text + " " + remainder_text 

      elif quotient: 
       quotient_text = (int_to_words(quotient) + " " + num_text_map[x]) \ 
           if include_quotient else \ 
           (num_text_map[x]) 
       return quotient_text 

     return False 

    result = do_chunk(num, BIGS) 
    if result: return result 

    result = do_chunk(num, HUNDRED, is_hundreds=True) 
    if result: return result 

    result = do_chunk(num, MULTS_OF_TEN, include_quotient=False) 
    if result and follows_hundreds: return "and " + result 
    if result: return result 

    result = do_chunk(num, SMALLS) 
    if result and follows_hundreds: return "and " + result 
    if result: return result 

def count_letters(string): 
    looking_for = "[a-z]" 
    instances = re.findall(looking_for, string) 
    return len(instances) 

def tally_letters(start, end): 
    total_letters = 0 
    for i in range(start, end + 1): 
     total_letters += count_letters(int_to_words(i)) 
    return total_letters 

这里是程序的输出，对预期的解决方案相比。

>>> answer = tally_letters(1, 1000) 
>>> assert answer == 21124 
Traceback (most recent call last): 
    File "<pyshell#3>", line 1, in <module> 
    assert answer == 21124 
AssertionError 
>>> answer 
1: 21118 

它让我感到困惑，我得到了6分的差异。在此先感谢您的帮助。

Answer 1

我想如果有人给我九个气球，然后再给我一个，我会说我有十个气球。在另一方面，如果我有99气球，然后我得到一个更多，我会说：“我有百个气球”，而不是“我有百个气球”：

>>> int_to_words(10) 
'ten' 
>>> int_to_words(100) 
'hundred' 
>>> int_to_words(1000) 
'thousand' 

len("one")*2 == 6。