python - 欧拉计划 17

Question

我一直在尝试解决欧拉 17 问题，但遇到了一些麻烦。该问题的定义是:

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.

我是用Python写的，代码看了三四遍还是看不出问题出在哪里。它很长(我刚开始学习 python，以前从未编码过)，但我基本上只是定义了不同的函数，这些函数采用不同的数字位数并计算每个字母的数量。我最终得到 21254，实际答案似乎是 21124，所以我差了 130。如有任何帮助，我们将不胜感激。

# create dict mapping numbers to their
# lengths in English

maps = {}
maps[0] = 0
maps[1] = 3
maps[2] = 3
maps[3] = 5
maps[4] = 4
maps[5] = 4
maps[6] = 3
maps[7] = 5
maps[8] = 5
maps[9] = 4
maps[10] = 3
maps[11] = 6
maps['and'] = 3
maps['teen'] = 4
maps[20] = 6
maps[30] = 6
maps[40] = 5
maps[50] = 5
maps[60] = 6
maps[70] = 7
maps[80] = 6
maps[90] = 6
maps[100] = 7
maps[1000] = 8

# create a list of numbers 1-1000
def int_to_list(number):
    s = str(number)
    c = []
    for digit in s:
        a = int(digit)
        c.append(a)
    return c  # turn a number into a list of its digits
def list_to_int(numList):
    s = map(str, numList)
    s = ''.join(s)
    s = int(s)
    return s


L = []
for i in range(1,1001,1):
    L.append(i)

def one_digit(n):
    q = maps[n]
    return q
def eleven(n):
    q = maps[11]
    return q
def teen(n):
    digits = int_to_list(n) 
    q = maps[digits[1]] + maps['teen']
    return q
def two_digit(n):
    digits = int_to_list(n)
    first = digits[0]
    first = first*10
    second = digits[1]
    q = maps[first] + one_digit(second)
    return q
def three_digit(n):
    digits = int_to_list(n)
    first = digits[0]
    second = digits[1]
    third = digits[2]

    # first digit length
    f = maps[first]+maps[100]

    if second == 1 and third == 1:
        s = maps['and'] + maps[11]
    elif second == 1 and third != 1:
        s = digits[1:]
        s = list_to_int(s)
        s = maps['and'] + teen(s)
    elif second == 0 and third == 0:
        s = maps[0]
    elif second == 0 and third != 0:
        s = maps['and'] + maps[third]
    else:
        s = digits[1:]
        s = list_to_int(s)
        s = maps['and'] + two_digit(s)

    q = f + s
    return q
def thousand(n):
    q = maps[1000]
    return q

# generate a list of all the lengths of numbers

lengths = []


for i in L:
    if i < 11:
        n = one_digit(i)
        lengths.append(n)
    elif i == 11:
        n = eleven(i)
        lengths.append(n)
    elif i > 11 and i < 20:
        n = teen(i)
        lengths.append(n)
    elif i > 20 and i < 100:
        n = two_digit(i)
        lengths.append(n)
    elif i >= 100 and i < 1000:
        n = three_digit(i)
        lengths.append(n)
    elif i == 1000:
        n = thousand(i)
        lengths.append(n)
    else:
        pass

# since "eighteen" has eight letters (not 9), subtract 10
sum = sum(lengths) - 10
print "Your number is: ", sum

Answer 1

解释差异

您的代码充满错误:

1. 这是错误的:

   maps[60] = 6
   

   对错误的贡献:+100(因为它影响 60 到 69、160 到 169、...、960 到 969)。

2. 几个青少年误会了:

   >>> teen(12)
   7
   >>> teen(13)
   9
   >>> teen(15)
   8
   >>> teen(18)
   9
   

   对错误的贡献:+40(因为它影响了 12, 13, ..., 112, 113, ..., 918)

3. 以及 x10 形式的任意数字:

   >>> three_digit(110)
   17
   

   对错误的贡献:9(因为 110, 210, ... 910)

4. 数字 20 不计算在内(您考虑 i < 20 和 i > 20 但不考虑 i == 20)。

   对错误的贡献:-6

5. 数字 1000 的英文写法是“一千”，但是:

   >>> thousand(1000)
   8
   

   对错误的贡献:-3

6. 您在最后减去 10 以尝试补偿这些错误之一。

   对错误的贡献:-10

总误差:100 + 40 + 9 − 6 − 3 − 10 = 130。

如何避免这些错误

通过尝试直接使用字母计数，您很难检查自己的工作。 “一百一十”又有多少个字母？是17还是16？如果您采用这样的策略，那么测试您的工作会容易得多:

unit_names = """zero one two three four five six seven eight nine ten
                eleven twelve thirteen fourteen fifteen sixteen seventeen
                eighteen nineteen""".split()
tens_names = """zero ten twenty thirty forty fifty sixty seventy eighty
                ninety""".split()

def english(n):
    "Return the English name for n, from 0 to 999999."
    if n >= 1000:
        thous = english(n // 1000) + " thousand"
        n = n % 1000
        if n == 0:
            return thous
        elif n < 100:
            return thous + " and " + english(n)
        else:
            return thous + ", " + english(n)
    elif n >= 100:
        huns = unit_names[n // 100] + " hundred"
        n = n % 100
        if n == 0:
            return huns
        else:
            return huns + " and " + english(n)
    elif n >= 20:
        tens = tens_names[n // 10]
        n = n % 10
        if n == 0:
            return tens
        else:
            return tens + "-" + english(n)
    else:
        return unit_names[n]

def letter_count(s):
    "Return the number of letters in the string s."
    import re
    return len(re.findall(r'[a-zA-Z]', s))

def euler17():
    return sum(letter_count(english(i)) for i in range(1, 1001))

使用这种方法可以更轻松地检查结果:

>>> english(967)
'nine hundred and sixty-seven'