0.求2^38
計算之後用結果替代url中的0
#solution 1
pow(2,38)
#solution 2
2**38
1.字母移位密碼
涉及函數:
ord('a')
chr(97)
trans=string.maketrans('abc','def')
str.translate(trans)
獲取24個小寫/大寫字母:string.ascii_lowercase,string.ascii_uppercase
獲取字母:string.ascii_letters
#solution 1
str="""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. """
result = ''
for letter in str:
if ord(letter)>=ord('a') and ord(letter)<=ord('z'):
code=ord(letter)+2
if code==ord('z')+2: code=ord('b')
if code==ord('y')+2: code=ord('a')
letter=chr(code)
result=result+letter
print result
#solution 2
import string
trans = string.maketrans(string.ascii_lowercase, string.ascii_lowercase[2:] + string.ascii_lowercase[0:2])
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(trans)
print "map".translate(trans)
2.查看網頁源碼,尋找一段亂碼中的字母
import string
text=""" """
text_list = list(text)
text_real=[]
for c in text_list:
if c in string.ascii_letters:
text_real.append(c)
print ''.join(text_real)
3.正則式
import re,urllib
f=urllib.urlopen('http://www.pythonchallenge.com/pc/def/equality.html')
str= f.read()
ans=re.findall('[^A-Z][A-Z]{3}([a-z])[A-Z]{3}[^A-Z]', str)
print ' '.join(ans)
import urllib2
def getNextId(uid):
url="http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing="+uid
page=urllib2.urlopen(url).read()
print page
nextid=page.split(" ")[-1]
return nextid
uid="12345"
while uid.isdigit():
uid=getNextId(uid)
圖片下有這麼一句話“ pronounce it ”,再看源代碼:
“peak hell sounds familiar ”?pickle模塊。按照給定的文件banner.p,使用pickle解序列,將得到一個包含了不定數個元組的列表的列表。經過分析,每個列表的所有元組的第二項相加均爲95,而元組的第一項要麼是‘ ’,要麼是‘#’;那麼,很顯然按一行95個字符的排列輸出將得到有效信息,代碼如下:
import pickle
import urllib
data = urllib.urlopen("http://www.pythonchallenge.com/pc/def/banner.p").read()
list = pickle.loads(data)
string = ''
for metalist in list:
for meta in metalist:
string += meta[0]*meta[1]
string += '/n'
print string
運行程序,將得到channel字樣。下一關網址將是channel.html ,通關!
6.取出zip文件中的comment信息查看源文件其他的信息都是無關的,只有一個zip,將網址html改爲zip,得到了一個zip文件:channel.zip
解壓後看到裏面有個readme.txt文件,裏面內容是:
welcome to my zipped list.
hint1: start from 90052
hint2: answer is inside the zip
根據提示打開90052.txt文件,發現跟第四關是差不多的內容,先遍歷它,python腳本如下:
import re
p = re.compile(r"Next nothing is (\d+)")
prefix = "channel/"
seed = '90052'
while True:
fname = prefix + seed + ".txt"
text = open(fname, 'r').read()
print text
lists = p.findall(text)
if lists:
seed = lists[0]
print " Next is", seed
else:
break
根據該提示和hint2,要取出zip中文件的comment信息,修改上面的腳本:
<span style="font-size:14px;">import re, zipfile
p = re.compile(r"Next nothing is (\d+)")
prefix = "channel/"
seed = '90052'
comments = []
z = zipfile.ZipFile("channel.zip", "r")
while True:
fname = prefix + seed + ".txt"
comments.append(z.getinfo(seed + ".txt").comment)
text = open(fname, 'r').read()
print text
lists = p.findall(text)
if lists:
seed = lists[0]
print " Next is", seed
else:
break
print "".join(comments)</span>
得到結果:
修改url,http://www.pythonchallenge.com/pc/def/hockey.html,回車:
it's in the air. look at the letters.
氧氣在空氣中,oxygen,進入第七關
7.對圖片上的灰度條按ASCII編碼
http://www.pythonchallenge.com/pc/def/oxygen.html
Hint1:圖片中有一道灰度條。
沒有其他提示,必然要從圖片下手了。用到PIL。
該圖像模式是RGBA。每個像素是個四元組,前3個分量分別代表紅(R)、綠(G)、藍(B),對於灰度圖像,R=B=G。
這樣,我們可以找到這段灰度條。將其數值按照ASCII碼映射爲字符(chr 函數)。
import Image
im = Image.open('oxygen.png')
print ('image information:',im.format,im.size,im.mode)
y_begin = 0
while True:
p = im.getpixel((0,y_begin))
if p[0] == p[1] == p[2]:
break
y_begin += 1
x_end = 0
while True:
p = im.getpixel((x_end,y_begin))
if not p[0] == p[1] == p[2]:
break
x_end += 1
print('Y first corrdinate:',y_begin,'X last coordinate:x_end:',x_end)
msg = []
for i in range(0,x_end,7):
p = im.getpixel((i,y_begin))
msg.append(chr(p[0]))
print(''.join(msg))
msg = [105, 110, 116, 101, 103, 114, 105, 116, 121]
print(''.join([chr(x) for x in msg]))
結果:integrity
8.用戶名密碼解密
http://www.pythonchallenge.com/pc/def/integrity.html
網頁源碼裏面有個鏈接,點進入,彈出一個登陸框需要用戶名和密碼,而註釋裏面有一段加密的文字,un,pw。這兩個應該就是賬號和密碼,現在要做的是解密,根據圖像是蜜蜂bees,根據發音有可能是bz2加密的,python解密代碼如下:
import bz2
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'
print bz2.decompress(un)
print bz2.decompress(pw)
輸出結果:
huge
file
輸進去,進入下一關
9.圖片相關操作
http://www.pythonchallenge.com/pc/return/good.html(賬號:huge 密碼:file)
根據提示是要將裏面的點轉換爲線,畫出圖像,使用python腳本如下:
import Image, ImageDraw
img = Image.new('RGB', (640,480))
draw = ImageDraw.Draw(img)
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,
190,321,192,331,193,338,196,341,197,346,199,352,198,360,197,366,197,373,196,380,197,383,196,387,192,
389,191,392,190,396,189,400,194,401,201,402,208,403,213,402,216,401,219,397,219,393,216,390,215,385,
215,379,213,373,213,365,212,360,210,353,210,347,212,338,213,329,214,319,215,311,215,306,216,296,218,
290,221,283,225,282,233,284,238,287,243,290,250,291,255,294,261,293,265,291,271,291,273,289,278,287,
279,285,281,280,284,278,284,276,287,277,289,283,291,286,294,291,296,295,299,300,301,304,304,320,305,
327,306,332,307,341,306,349,303,354,301,364,301,371,297,375,292,384,291,386,302,393,324,391,333,387,
328,375,329,367,329,353,330,341,331,328,336,319,338,310,341,304,341,285,341,278,343,269,344,262,346,
259,346,251,349,259,349,264,349,273,349,280,349,288,349,295,349,298,354,293,356,286,354,279,352,268,
352,257,351,249,350,234,351,211,352,197,354,185,353,171,351,154,348,147,342,137,339,132,330,122,327,
120,314,116,304,117,293,118,284,118,281,122,275,128,265,129,257,131,244,133,239,134,228,136,221,137,
214,138,209,135,201,132,192,130,184,131,175,129,170,131,159,134,157,134,160,130,170,125,176,114,176,
102,173,103,172,108,171,111,163,115,156,116,149,117,142,116,136,115,129,115,124,115,120,115,115,117,
113,120,109,122,102,122,100,121,95,121,89,115,87,110,82,109,84,118,89,123,93,129,100,130,108,132,110,
133,110,136,107,138,105,140,95,138,86,141,79,149,77,155,81,162,90,165,97,167,99,171,109,171,107,161,
111,156,113,170,115,185,118,208,117,223,121,239,128,251,133,259,136,266,139,276,143,290,148,310,151,
332,155,348,156,353,153,366,149,379,147,394,146,399]
second = [156,141,165,135,169,131,176,130,187,134,191,140,191,146,186,150,179,155,175,157,168,157,163,157,159,
157,158,164,159,175,159,181,157,191,154,197,153,205,153,210,152,212,147,215,146,218,143,220,132,220,
125,217,119,209,116,196,115,185,114,172,114,167,112,161,109,165,107,170,99,171,97,167,89,164,81,162,
77,155,81,148,87,140,96,138,105,141,110,136,111,126,113,129,118,117,128,114,137,115,146,114,155,115,
158,121,157,128,156,134,157,136,156,136]
draw.line(first)
draw.line(second)
img.show()
畫出的圖像爲牛,輸入bull,進入下一關。10.look and say 正則表達式
http://www.pythonchallenge.com/pc/return/bull.html
有個sequence.txt文件,點進去:
a = [1, 11, 21, 1211, 111221,
查找規律是:
後面的一個數是前面的數的描述
1
11 代表前面是1個1
21 代表前面是2個1
1211 代表前面是1個2,1個1
111221 代表前面是1個1,1個2,2個1
312211
13112221
然後求第30個元素的長度,python實現如下:
import re
def describe(s):
lists = [str(len(m.group(0))) + m.group(1) for m in re.finditer(r"(\d)\1*", s)]
# print lists
return "".join(lists)
s = "1"
for dummy in range(30):
s = describe(s)
print s
print len(s)
結果:5808
re.finditer(r"(\d)\1*", s)
匹配s中的數字,相鄰數字若相同,則劃分爲一組
返回值m.group(0):相同數字組
返回值m.group(!):該組的首字符