Codewars-python每日练习（1）

1 Sum of the first nth term of Series

Task:

Your task is to write a function which returns the sum of following series upto nth term(parameter).

Series: 1 + 1/4 + 1/7 + 1/10 + 1/13 + 1/16 +…

Rules:

You need to round the answer to 2 decimal places and return it as String.
If the given value is 0 then it should return 0.00
You will only be given Natural Numbers as arguments.

Examples:

SeriesSum(1) => 1 = “1.00”
SeriesSum(2) => 1 + 1/4 = “1.25”
SeriesSum(5) => 1 + 1/4 + 1/7 + 1/10 + 1/13 = “1.57”

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我的解决方法：

def series_sum(n):
    if n == 0:
        return '%.2f' %n
    else:
        sum = 0
        for i in range(0, n):
            sum += 1/(1 + i*3)
        return '%.2f' %sum

本题要注意限定浮点数的位数，返回值时应加上'%.2f'

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其他优秀解决方法：

def series_sum(n):
    return '{:.2f}'.format(sum(1.0/(3 * i + 1) for i in range(n)))

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2 Printer Errors

Task

In a factory a printer prints labels for boxes. For one kind of boxes the printer has to use colors which, for the sake of simplicity, are named with letters from a to m.

The colors used by the printer are recorded in a control string. For example a “good” control string would be aaabbbbhaijjjm meaning that the printer used three times color a, four times color b, one time color h then one time color a…

Sometimes there are problems: lack of colors, technical malfunction and a “bad” control string is produced e.g. aaaxbbbbyyhwawiwjjjwwm with letters not from a to m.

You have to write a function printer_error which given a string will output the error rate of the printer as a string representing a rational whose numerator is the number of errors and the denominator the length of the control string. Don’t reduce this fraction to a simpler expression.

The string has a length greater or equal to one and contains only letters from ato z.

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我的解决方法:

def printer_error(s):
    count = 0
    for i in range(0, len(s)):
        if ord("m")< ord(s[i]) <= ord("z"):
            count += 1
    return str(count) + '/' + str(len(s))

将字符串转化为asc码使用ord()函数

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其他优秀解决方法：

from re import sub
def printer_error(s):
    return "{}/{}".format(len(sub("[a-m]",'',s)),len(s))

re.sub()有替换功能

def printer_error(s):
    return "{}/{}".format(len([x for x in s if x not in "abcdefghijklm"]), len(s))

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