项目欧拉17

3
我一直在尝试解决欧拉17题,但一直遇到困难。问题的定义如下:
如果将1至5写成单词:one,two,three,four,five,则总共使用了3 + 3 + 5 + 4 + 4 = 19个字母。
如果将从1到1000(包括一千)的所有数字都用单词写出来,将使用多少个字母?
注意:不要计算空格或连字符。例如,342(three hundred and forty-two)包含23个字母,115(one hundred and fifteen)包含20个字母。写数字时使用“and”符号符合英国用法。
我用Python编写了它,即使我已经通过代码三四次了,仍然看不出问题所在。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函数,每个函数都涉及不同数量的数字,并计算每个数字中的字母数。最后我得到的答案是21254,而实际答案似乎是21124,所以我偏差正好是130。任何帮助都将不胜感激。
# 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
1个回答

9

解释这个矛盾

你的代码充满了错误:

  1. This is wrong:

    maps[60] = 6
    

    Contribution to error: +100 (because it affects 60 to 69, 160 to 169, ..., 960 to 969).

  2. Several teenagers are mistaken:

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

    Contribution to error: +40 (because it affects 12, 13, ..., 112, 113, ..., 918)

  3. And any number of the form x10:

    >>> three_digit(110)
    17
    

    Contribution to error: 9 (because 110, 210, ... 910)

  4. The number 20 is not counted (you consider i < 20 and i > 20 but not i == 20).

    Contribution to error: −6

  5. The number 1000 is written "one thousand" in English but:

    >>> thousand(1000)
    8
    

    Contribution to error: −3

  6. You subtract 10 at the end in an attempt to compensate for one of these errors.

    Contribution to error: −10

总误差:100 + 40 + 9 - 6 - 3 - 10 = 130。

如何避免这些错误

通过直接使用字母计数来工作,你让自己很难检查自己的工作。 "one hundred and ten" 里有多少个字母?是17个还是16个?如果你采取了类似这样的策略测试你的工作,那么会更容易:

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))

采用这种方法更容易检查您的结果:

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

1
虽然那可能只是一条评论而不是一个答案...但如果有足够多的这些错误,它们加起来可能会达到130个... - Joran Beasley
我感觉自己像个白痴。非常感谢,那解决了大部分问题。 - user1707302

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