yy_gy
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The given hints are useful!
>>> first=[146, 399, 163, 403, 170, 393, 169, 391, 166, 386, 170, 381, 170, 371, 170, 355, 169, 346, 167, 335, 170, 329, 170, 320, 170, 310, 171, 301, 173, 290, 178, 289, 182, 287, 188, 286, 190, 286, 192, 291, 194, 296, 195, 305, 194, 307, 191, 312, 190, 316, 19 ...
If you are familiarly with 'bz2' module, it would be really easy!
>>> import bz2
>>> u=bz2.decompress('BZh91AY&SYA\xaf\x82\r\x00\x00\x01\x01\x80\x02\xc0\x02\x00 \x00!\x9ah3M\x07<]\xc9\x14\xe1BA\x06\xbe\x084')
>>> p=bz2.decompress('BZh91AY&SY\x94$|\x0e\x00\x00\x0 ...
In my opinion, this challenge needs more patience and observation!
>>> import Image
>>> image=Image.open('/homeDownloads/oxygen.png')
>>> data=image.convert('L').getdata()
>>> str=''
>>> s_data=[]
>>> for i in range(19918,39836): # 2/3 of the im ...
download the zip file from ..../channel.zip
extract the channel.zip and read the read.txt
my code:
>>> import re
>>> value=90052
>>> comments=[]
>>> z=zipfile.ZipFile('/home/gaoy/Downloads/channel.zip')
>>> while True:
f=open('/home/gaoy/temp/%s'%value+'. ...
import pickle
f=open('banner.p')
contents=pickle.load(f)
for obj in contents:
... string=''
... for o in obj:
... string=string+o[0]*o[1]
... print string
Answer: channel.html
Reference:http://uil330.blog.163.com/blog/static/1342864882010020103344981/
It is a bit tricky...
>>> pattern='and the next nothing is \d+'
>>> value=12345
>>> while True:
url=Request('http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=%s'%value)
response=urllib2.urlopen(url)
page=response.read()
match=re.search(pattern,page)
if not match:
break;
else:
print p ...