pythonchallenge(0-9)

0.

2**38

1.

方法一：

input

=

"g fmncwmsbgblr rpylqjyrcgr zw fylb. rfyrq ufyr amknsrcpq ypcdmp.\

bmgle gr glzw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.\

kyicrpylq() gq pcamkkclbcb. lmu ynnjw mlrfcspj."

def

mymap(

input

):

output

=

''

for

x

in

input

:

if

'a'

<

=

x

and

'x'

>

=

x:

output

+

=

chr

(

ord

(x)

+

2

)

elif

x

=

=

'y'

:

output

+

=

'a'

elif

x

=

=

'z'

:

output

+

=

'b'

else

:

output

+

=

x

print

output

mymap(

input

)

mymap(

'map'

)

方法二：

import

string

s

=

"g fmncwmsbgblr rpylqjyrcgr zw fylb. rfyrq ufyr amknsrcpq ypcdmp.\

bmgle gr glzw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.\

kyicrpylq() gq pcamkkclbcb. lmu ynnjw mlrfcspj."

leet

=

string.maketrans(

'abcdefghijklmnopqrstuvwxyz'

,

'cdefghijklmnopqrstuvwxyzab'

)

print

string.translate(s,leet)

2. http://www.pythonchallenge.com/pc/def/ocr.htm

import

sys,urllib

import

re

url

=

"http://www.pythonchallenge.com/pc/def/ocr.html"

wp

=

urllib.urlopen(url)

content

=

wp.read()

p

=

re.

compile

(r

'<!--([\s\S]*?)-->'

)

out

=

p.findall(content)[

1

]

res

=

[c

for

c

in

out

if

c.isalpha()]

print

''.join(res)

3.http://www.pythonchallenge.com/pc/def/equality.html

恰好3个大写字母做保镖。

import

sys,urllib

import

re

url

=

"http://www.pythonchallenge.com/pc/def/equality.html"

wp

=

urllib.urlopen(url)

content

=

wp.read()

p

=

re.

compile

(r

'[^A-Z][A-Z]{3}([a-z])[A-Z]{3}[^A-Z]'

)

out

=

p.findall(content)

print

''.join(out)

4. http://www.pythonchallenge.com/pc/def/linkedlist.php

import

sys,urllib

import

re

url

=

"http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=12345"

output

=

open

(

'result.txt'

,

'w'

)

for

i

in

range

(

400

):

print

i

output.write(url

+

'\n'

)

content

=

urllib.urlopen(url).read()

num

=

content.split()[

-

1

]

output.write(content

+

'\n'

)

token

=

url.split(

'='

)

token[

-

1

]

=

'='

+

num

url

=

''.join(token)

output.close()

保存到文本中，再找规律。

5. http://www.pythonchallenge.com/pc/def/peak.html

import

pickle

f

=

open

(

'banner.p'

)

data

=

pickle.load(f)

print

'\n'

.join(''.join(p[

0

]

*

p[

1

]

for

p

in

row)

for

row

in

data)

f.close()

我还以为要"pronounce it"音频处理呢。。结果"peakhell"暗示"pickle"，用pickle load，然后' '与'#'恢复出字符组成的字channel.

6. http://www.pythonchallenge.com/pc/def/channel.html

import

zipfile

currentfile

=

"90052.txt"

comment

=

[]

with zipfile.ZipFile(

'channel.zip'

,mode

=

'r'

) aszf:

while

(

1

):

comment.append(zf.getinfo(currentfile).comment)

data

=

zf.read(currentfile)

nextnum

=

data.split()[

-

1

]

if

str

.isdigit(nextnum): 

currentfile

=

nextnum

+

'.txt'

else

:

print

data

break

print

''.join(comment)

这道题我也是败给它了。先是领会出把url中的html换成zip,这样就可以下载zip文件了（会自动弹出对话框下载），然后找到readme.txt根据提示，类似第5题一样往下搜索，最后结果是Collect the comments，原来每个zip文件都一个信息，信息里包含了comment。

7. http://www.pythonchallenge.com/pc/def/oxygen.html

from

PIL

import

Image

img

=

Image.

open

(

'oxygen.png'

)

pix

=

img.load()

width

=

img.size[

0

]

center

=

img.size[

1

]

/

2

asi

=

''

for

x

in

range

(

0

,width,

7

):

asi

=

asi

+

chr

(pix[x,center][

0

])

print

asi

l

=

[

105

,

110

,

116

,

101

,

103

,

114

,

105

,

116

,

121

]

print

''.join([

chr

(i)

for

i

in

l])

8. http://www.pythonchallenge.com/pc/def/integrity.html

import

bz2

compressed_un

=

"BZh91AY&SYA\xaf\x82\r\x00\x00\x01\x01\x80\x02\xc0\x02\x00 \x00!\x9ah3M\x07<]\xc9\x14\xe1BA\x06\xbe\x084"

compressed_pw

=

"BZh91AY&SY\x94$|\x0e\x00\x00\x00\x81\x00\x03$ \x00!\x9ah3M\x13<]\xc9\x14\xe1BBP\x91\xf08"

decompressed_un

=

bz2.decompress(compressed_un)

decompressed_pw

=

bz2.decompress(compressed_pw)

print

decompressed_un,decompressed_pw

解个压缩就过了。。比想象中容易一点，不过要看出un和pw是bz2压缩后的字符。(1.图片有一直蜜蜂，发音bee有提示 2.对bz2压缩后的形式熟悉)