lufeihaidao

# 这是我的 python 解法，相当的 C 呢
astring='''g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp.
 bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle
 qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj.'''
import re
bstring=''
for a in astring:
    if re.match('\w',a):
        a=chr((ord(a)-95)%26+97)
    bstring=bstring+a

lufeihaidao

# 推荐的做法
import string
text='''g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp.
 bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle.
 sqgle qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj.'''
table = string.maketrans(string.ascii_lowercase,
                         string.ascii_lowercase[2:]+string.ascii_lowercase[:2])
text.translate(table)

lufeihaidao

ruby -wle 'n = 12345; print n = $1 while 
          `GET "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=#{n}"`
          =~ /(\d+)$/'

lufeihaidao

require "zip/zip"
zipfilename = "/home/haidao/Downloads/channel.zip"
Zip::ZipFile.open(zipfilename) do |zipfile|
  s = zipfile.get_input_stream("readme.txt").read
  num = s.match(/\d{2,6}/)
  output = Array.new
  while num do
    entry = zipfile.find_entry("#{num}.txt")
    s = entry.get_input_stream.read
    num = s.match(/\d{2,6}/)

lufeihaidao

f = "channel.zip"
z = zipfile.ZipFile(f)
name = '90052.txt'
while 1:
    print z.getinfo(name).comment,
    name = z.read(name).split()[-1] + '.txt'

lufeihaidao

# 推荐的做法
import urllib2, sys, pickle
pickle_data = urllib2.urlopen('http://www.pythonchallenge.com/pc/def/banner.p').read()
data = pickle.loads(pickle_data)
for line in data:
    for char, count in line: sys.stdout.write(char * count)
    sys.stdout.write("\n")

lufeihaidao

require "open-uri"
n=12345
loop do
    flag = ''
    open("http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=#{n}") do |f|
        s=f.read
        flag = s if s=~/htm/
        n=(s=~/div/ ? n/2 : s.split[-1].to_i)
    end
    if !flag.empty?

lufeihaidao

import re
from urllib import urlopen
astring='http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing='
temp='12345'
f = urlopen(astring+temp)
a=f.read()
restring=re.compile('and the next nothing is (\d{3,5})')
for i in range(400):
    temp=re.findall(restring,a)[0]
    print(temp)

lufeihaidao

# 我的代码比较笨，但是方法是一样的，就不贴上来了。
import re
s = '''
copy from the source page
'''
print ''.join(re.findall('[^A-Z][A-Z]{3}([a-z])[A-Z]{3}[^A-Z]', s))

lufeihaidao

gets.tr "a-z", "c-zab"
# 就这么简单！然后输入字符串就好了！真让人喜欢！