python – 项目Euler 17

我一直在努力解决欧拉17问题并且遇到了一些麻烦.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编写的,即使经过三到四次代码,我仍然看不出问题所在.它很长(我刚刚开始学习python,我以前从未编写过),但我基本上只定义了不同的函数,这些函数占用了不同的位数,并计算了每个函数的字母数.我最终得到了21254,看起来实际答案是21124,所以我完全取消了130.任何帮助都将受到赞赏.

# 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
解释差异

您的代码充满了错误:

>这是错的:

maps[60] = 6

对错误的贡献:100(因为它影响60到69,160到169,……,960到969).
>几个青少年错了:

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

对错误的贡献:40(因为它影响12,13,…,112,113,…,918)
>和任何数量的形式x10:

>>> three_digit(110)
17

对错误的贡献:9(因为110,210,… 910)
>数字20不计算(您认为i <20且i> 20但不是i == 20). 对错误的贡献:-6
>数字1000用英文写成“一千”但是:

>>> thousand(1000)
8

对错误的贡献:-3
>您最后减去10,以尝试补偿其中一个错误.

对错误的贡献:-10

总误差:100 40 9 – 6 – 3 – 10 = 130.

你怎么能避免这些错误

通过尝试直接使用字母计数,您使您很难检查自己的工作.再一次在“一百零一十”中有多少封信?它是17,还是16?如果您采用了这样的策略,那么测试您的工作会更容易:

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))

使用这种方法,可以更轻松地检查结果:

>>> english(967)
'nine hundred and sixty-seven'
https://stackoverflow.com/questions/12647254/project-euler-17

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