Question

Say we have a function that translates the morse symbols:

. -> -.
- -> ...-

如果我们两次行使这一职能,我们就取得了以下成果:

<代码> ->->。 ->...-.



Given an input string and a number of repetitions,  want to know the length of the final string. (Problem 1 from the Flemish Programming Contest VPW, taken from these slides which provide a solution in Haskell).

参考书

4
. 4
.- 2
-- 2 
--... 50


我们期待解决办法。

44
16
20
34028664377246354505728


由于我不知道Haskell,这是我在Sharma的令人厌恶的解决办法,我收到了:

def encode(msg, repetition, morse={ . :  -. ,  - :  ...- }):
    if isinstance(repetition, str):
        repetition = eval(repetition)
    while repetition > 0:
        newmsg =   .join(morse[c] for c in msg)
        return encode(newmsg, repetition-1)
    return len(msg)


def problem1(fn):
    with open(fn) as f:
        f.next()
        for line in f:
            print encode(*line.split())


用于头三个投入的工作,但最后一项投入的记忆有误。

你们如何以更有效的方式来重新制定这一标准?

<><>Edit>>

根据所发表的评论意见:

def encode(p, s, repetition):
    while repetition > 0:
        p,s = p + 3*s, p + s
        return encode(p, s, repetition-1)
    return p + s


def problem1(fn):
    with open(fn) as f:
        f.next()
        for line in f:
            msg, repetition = line.split()
            print encode(msg.count( . ), msg.count( - ), int(repetition))


关于风格和进一步改进的评论仍然值得欢迎。

Answer 1

认为实际上,你只得花很长的时间才能得出结果。还认为,......的顺序和――在座标不影响最后长度(例如“......- 3”和“- 3”产生相同的最后长度。

因此,我将放弃整条雕刻,而是把数量和数量——作为分类。

Answer 2

在您的首字母缩略语中,点数和 da。然后适用:

repetitions = 4
dots = 1
dashes = 0
for i in range(repetitions):
    dots, dashes = dots + 3 * dashes, dashes + dots

想一下为什么这样做。

Answer 3

Per @Hammar (I had the same idea, but he explained it better than I could have ;-):

from sympy import Matrix

t = Matrix([[1,3],[1,1]])

def encode(dots, dashes, reps):
    res = matrix([dashes, dots]) * t**reps
    return res[0,0] + res[0,1]

Answer 4

页: 1

def encode(dots, dashes, repetitions):
    while repetitions > 0:
        dots, dashes = dots + 3 * dashes, dots + dashes
        repetitions -= 1

    return dots + dashes

def problem1(fn):
    with open(fn) as f:
        count = int(next(f))
        for i in xrange(count):
            line = next(f)
            msg, repetition = line.strip().split()
            print encode(msg.count( . ), msg.count( - ), int(repetition))

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