随笔分类 -  Python challenge

Python Challenge is a game in which each level can be solved by a bit of (Python) programming http://www.pythonchallenge.com/
python challenge 17
摘要：View Code 1 #-*- coding:utf8-*- 2 import time 3 start = time.time() 4 5 from urllib2 import Request,urlopen 6 from urllib import quote_plus 7 info='the flowers are on their way' 8 url='http://www.pythonchallenge.com/pc/stuff/violin.php' 9 req = Request(url, headers={'Cookie': 阅读全文
posted @ 2012-02-10 18:35 lcyang 阅读(327) 评论(0) 推荐(0)
python challenge 16
摘要：View Code 1 #-*- coding:utf8-*- 2 import time 3 start = time.time() 4 ''' 5 import xmlrpclib 6 7 server = xmlrpclib.ServerProxy("http://localhost:80") 8 9 words = server.sayHello()10 11 print "result:" + words12 '''13 import xmlrpclib14 server = xmlrpclib. 阅读全文
posted @ 2012-02-10 18:34 lcyang 阅读(203) 评论(0) 推荐(0)
python challenge 13
摘要：View Code 1 #-*- coding:utf8-*- 2 import time 3 start = time.time() 4 ''' 5 import xmlrpclib 6 7 server = xmlrpclib.ServerProxy("http://localhost:80") 8 9 words = server.sayHello()10 11 print "result:" + words12 '''13 import xmlrpclib14 server = xmlrpclib. 阅读全文
posted @ 2012-02-10 18:32 lcyang 阅读(150) 评论(0) 推荐(0)
python challenge 15
摘要：View Code 1 #-*- coding:utf8-*- 2 import time 3 start = time.time() 4 ''' 5 import xmlrpclib 6 7 server = xmlrpclib.ServerProxy("http://localhost:80") 8 9 words = server.sayHello()10 11 print "result:" + words12 '''13 import xmlrpclib14 server = xmlrpclib. 阅读全文
posted @ 2012-02-10 18:32 lcyang 阅读(181) 评论(0) 推荐(1)
python challenge 12
摘要：View Code 1 #-*- coding:utf8-*- 2 import time 3 start = time.time() 4 5 # Get the data from: http://www.pythonchallenge.com/pc/return/evil2.gfx 6 7 h = open("evil2.gfx", "rb") 8 data = h.read() 9 h.close()10 11 new_data = [[], [], [], [], []]12 n = 013 14 for byte in range(len(da 阅读全文
posted @ 2012-02-10 18:31 lcyang 阅读(223) 评论(0) 推荐(0)
python challenge 11
摘要：View Code 1 #-*- coding:utf8-*- 2 import time 3 start = time.time() 4 5 import Image 6 7 # download the image from: http://www.pythonchallenge.com/pc/return/cave.jpg 8 9 image = Image.open("cave.jpg")10 nsize = tuple([x / 2 for x in image.size])11 odd = even = Image.new(image.mode, nsize)1 阅读全文
posted @ 2012-02-10 18:30 lcyang 阅读(249) 评论(0) 推荐(0)
python challenge 9
摘要：1 #-*- coding:utf8-*- 2 import time 3 import re 4 import Image 5 import ImageDraw 6 start = time.time() 7 8 fo = open("10","r+") 9 nums = fo.read()10 num = ''11 for i in nums:12 if i != '\n':13 num += i14 nums = []15 for n in re.findall(r'\d+',num):16 nums 阅读全文
posted @ 2012-02-10 18:28 lcyang 阅读(186) 评论(0) 推荐(0)
python challenge 10
摘要：1 #-*- coding:utf8-*- 2 import time 3 import re 4 start = time.time() 5 ''' 6 a = [1, 11, 21, 1211, 111221, 7 len(a[30]) = ? 8 ''' 9 10 def find_L(st):11 nums = re.findall(r'1+|2+|3+|',st)12 strs = ''13 for i in nums:14 if len(i):15 strs += str(len(i)) + i[0]1 阅读全文
posted @ 2012-02-10 18:28 lcyang 阅读(241) 评论(0) 推荐(0)
python challenge 8
摘要：#-*- coding:utf8-*-import timestart = time.time()#un: 'BZh91AY&SYA\xaf\x82\r\x00\x00\x01\x01\x80\x02\xc0\x02\x00 \x00!\x9ah3M\x07<]\xc9\x14\xe1BA\x06\xbe\x084'#pw: 'BZh91AY&SY\x94$|\x0e\x00\x00\x00\x81\x00\x03$ \x00!\x9ah3M\x13<]\xc9\x14\xe1BBP\x91\xf08''''b 阅读全文
posted @ 2012-02-10 18:27 lcyang 阅读(278) 评论(0) 推荐(0)
python challenge 7
摘要：1 #-*- coding:utf8-*- 2 import time 3 import Image 4 start = time.time() 5 #先将图片 oxygen.png 下载 6 import re, Image 7 8 i = Image.open("oxygen.png") # http://www.pythonchallenge.com/pc/def/oxygen.png 9 row = [i.getpixel((x, 45)) for x in range(0, i.size[0], 7)]10 print row11 ords = [r for r, 阅读全文
posted @ 2012-02-10 18:26 lcyang 阅读(325) 评论(0) 推荐(0)
python challenge 6
摘要：1 #-*- coding:utf8-*- 2 import time 3 import zipfile 4 import re 5 start = time.time() 6 ''' 7 学习 zipfile http://blog.csdn.net/jgood/article/details/4351911 8 ''' 9 '''10 welcome to my zipped list.11 12 hint1: start from 9005213 hint2: answer is inside the zip14 & 阅读全文
posted @ 2012-02-10 18:24 lcyang 阅读(244) 评论(0) 推荐(0)
python challenge 5
摘要：1 ''' 2 将banner.p下载 放在与程序同一文件夹下 3 ''' 4 fo = open("banner.p","r+") 5 #strs = pickle.dumps(fo.read()) 6 st = pickle.load(fo) 7 #print strs 8 for line in st: 9 print "".join([k*v for k,v in line])10 # channel 阅读全文
posted @ 2012-02-10 18:23 lcyang 阅读(227) 评论(0) 推荐(0)
python challenge 4
摘要：import restart = time.time()pattern = "and the next nothing is (\d+)"re_url = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing="nothing = "6711"url = urlopen(re_url + nothing)strs = url.read()obj = re.search(pattern,strs)while obj: nothing = obj.group(1) ur 阅读全文
posted @ 2012-02-10 18:21 lcyang 阅读(391) 评论(0) 推荐(0)
python challenge 3
摘要：1 import re2 fo = open("4","r+")3 strs = fo.read()4 #print strs5 s = ''6 for res in re.findall(r'[a-z][A-Z]{3}([a-z])[A-Z]{3}[a-z]',strs):7 s += res8 print s 9 #xXXXxXXXx 阅读全文
posted @ 2012-02-10 18:20 lcyang 阅读(163) 评论(0) 推荐(0)
python challenge 2
摘要：fo = open("3","r+")print fo.name,fo.modestrs = fo.read()res = ''for x in strs: obj = re.match(r'[a-z]',x) if obj: res += obj.group()print res 阅读全文
posted @ 2012-02-10 18:18 lcyang 阅读(124) 评论(0) 推荐(0)
python challenge 1
摘要：strs = '''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.'''def change(strs): res = [] for x in strs: if x != ' 阅读全文
posted @ 2012-02-10 18:17 lcyang 阅读(373) 评论(0) 推荐(0)
python challenge 0
摘要：print 2**38仅此而已 十分简单！ 阅读全文
posted @ 2012-02-10 18:15 lcyang 阅读(116) 评论(0) 推荐(0)