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Python Challenge

Level 0

In [21]: 1<<38
Out[21]: 274877906944

Level 1

# src:http://www.pythonchallenge.com/pc/def/map.html
 
from string import maketrans
ori = 'abcdefghijklmnopqrstuvwxyz'
des = 'cdefghijklmnopqrstuvwxyzab'
 
tbl = maketrans(ori, des)
 
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."
        
print text.translate(tbl)
 
# So, the answer is:
print 'map'.translate(tbl)
 
# Output:   ocr

Level 2

#src http://www.pythonchallenge.com/pc/def/ocr.html
 
import requests
import re
 
rep = requests.get("http://www.pythonchallenge.com/pc/def/ocr.html")
html = rep.text
 
html = ''.join(html.split()) # strip all newline, tab, space
text = re.findall(r'<!--(.*?)-->', html)[1] # get the second comment
wc = {}
 
for i in text:
    wc[i] = 1 if i not in wc else wc[i]+1
 
l = [i for i in wc if wc[i]==1]
 
# retrive answer in text in order
print ''.join(i for i in text if i in l)
# Output:   equality

Level 3

#src: http://www.pythonchallenge.com/pc/def/equality.html
import requests
import re
 
rep = requests.get("http://www.pythonchallenge.com/pc/def/equality.html")
html = rep.text
 
html = ''.join(html.split()) 
text = re.findall(r'<!--(.*?)-->', html)[0] 
 
res = re.findall(r'[^A-Z][A-Z]{3}([a-z])[A-Z]{3}[^A-Z]', text)
print ''.join(res) 
# Output:   linkedlist

Level 4

# src:  http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=12345
 
import urllib2
import re
link = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=12345"
# link = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=8022"
for i in range(400):
    text = urllib2.urlopen(link).read()
    num = re.findall(r"and the next nothing is (\d+)", text)[0]
 
    link = re.sub('\d+',num, link)
    print num + "\t" + link
    
# Output:   peak.html

Level 5

# src: http://www.pythonchallenge.com/pc/def/peak.html
 
import cPickle as pickle
import urllib2
import re
 
link = "http://www.pythonchallenge.com/pc/def/banner.p"
 
text = urllib2.urlopen(link).read()
data = pickle.loads(text)
 
for line in data:
    print ''.join(i[0]*i[1] for i in line)
 
# Output:  channel