Gather ye rosebuds while ye may

Python Challenge 1-10

    硬悟学蛇

这几天学校极客大挑战,三道编程题
然后发现这种才有意思嘛。。。
又找了一个:python challenge
感觉这种挑战比看书好多了


第〇关简单计算

求2^38: 2**38 即可


第一关字符串操作

将字符串中得字母移2位
不仅仅是改图上的三个字母,是26个全部改
用的ord()&chr()把26个字符都加二,然后发现yz加二就出去了,然后又加了几个if来限制
黑哥说 maketrans就行了,看上去很方便,还没有研究。。。


第二关字符串操作

找原文件里出现的英文字符

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a = 原文
for i in a:
if ord(i)>95 and ord(i)<122:
print(i,end='')


第三关正则表达式

hint 是这个
One small letter, surrounded by EXACTLY three big bodyguards on each of its sides.
想了半天没有理解他什么意思,特别是那个its因为没有'apostrophe撇号相当恼火
其实是说文字里的每一个字母都被有且仅有的三个大写字母围起来了
这个就是正则的查找:

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r"[a-z][A-Z]{3}[a-z]{1}[A-Z]{3}[a-z]"


第四关 urllib

从网页中获取数字添加在php末尾然后继续访问
就是urllib的操作:

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import urllib.request
nothing = "数字"
for _ in range(500):
url = 'http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing='+nothing
global nothing
nothing = urllib.request.urlopen(url).read().decode()[23:].strip()
print(nothing)


第五关pickle 、字典

叫你 pronunce 一幅山峰的图片

peak hall->pickle
这个pickle文件已经在原文件中给你了banner.pickle
使用pickle:

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import pickle
with open("banner.pickle","rb") as f:
dic = pickle.load(f)

这样pickle里面的文件就储存在 dic 这个字典里面了
字典内容如下:

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[(' ', 95)]
[(' ', 14), ('#', 5), (' ', 70), ('#', 5), (' ', 1)]
[(' ', 15), ('#', 4), (' ', 71), ('#', 4), (' ', 1)]
[(' ', 15), ('#', 4), (' ', 71), ('#', 4), (' ', 1)]
[(' ', 15), ('#', 4), (' ', 71), ('#', 4), (' ', 1)]
[(' ', 15), ('#', 4), (' ', 71), ('#', 4), (' ', 1)]
[(' ', 15), ('#', 4), (' ', 71), ('#', 4), (' ', 1)]
[(' ', 15), ('#', 4), (' ', 71), ('#', 4), (' ', 1)]
[(' ', 15), ('#', 4), (' ', 71), ('#', 4), (' ', 1)]
[(' ', 6), ('#', 3), (' ', 6), ('#', 4), (' ', 3), ('#', 3), (' ', 9), ('#', 3), (' ', 7), ('#', 5), (' ', 3), ('#', 3), (' ', 4), ('#', 5), (' ', 3), ('#', 3), (' ', 10), ('#', 3), (' ', 7), ('#', 4), (' ', 1)]
[(' ', 3), ('#', 3), (' ', 3), ('#', 2), (' ', 4), ('#', 4), (' ', 1), ('#', 7), (' ', 5), ('#', 2), (' ', 2), ('#', 3), (' ', 6), ('#', 4), (' ', 1), ('#', 7), (' ', 3), ('#', 4), (' ', 1), ('#', 7), (' ', 5), ('#', 3), (' ', 2), ('#', 3), (' ', 5), ('#', 4), (' ', 1)]
[(' ', 2), ('#', 3), (' ', 5), ('#', 3), (' ', 2), ('#', 5), (' ', 4), ('#', 4), (' ', 3), ('#', 3), (' ', 3), ('#', 4), (' ', 4), ('#', 5), (' ', 4), ('#', 4), (' ', 2), ('#', 5), (' ', 4), ('#', 4), (' ', 3), ('#', 3), (' ', 5), ('#', 3), (' ', 3), ('#', 4), (' ', 1)]
[(' ', 1), ('#', 3), (' ', 11), ('#', 4), (' ', 5), ('#', 4), (' ', 3), ('#', 3), (' ', 4), ('#', 3), (' ', 4), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 3), (' ', 6), ('#', 4), (' ', 2), ('#', 4), (' ', 1)]
[(' ', 1), ('#', 3), (' ', 11), ('#', 4), (' ', 5), ('#', 4), (' ', 10), ('#', 3), (' ', 4), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 3), (' ', 7), ('#', 3), (' ', 2), ('#', 4), (' ', 1)]
[('#', 4), (' ', 11), ('#', 4), (' ', 5), ('#', 4), (' ', 5), ('#', 2), (' ', 3), ('#', 3), (' ', 4), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 1), ('#', 4), (' ', 7), ('#', 3), (' ', 2), ('#', 4), (' ', 1)]
[('#', 4), (' ', 11), ('#', 4), (' ', 5), ('#', 4), (' ', 3), ('#', 10), (' ', 4), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 1), ('#', 14), (' ', 2), ('#', 4), (' ', 1)]
[('#', 4), (' ', 11), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 3), (' ', 4), ('#', 4), (' ', 4), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 1), ('#', 4), (' ', 12), ('#', 4), (' ', 1)]
[('#', 4), (' ', 11), ('#', 4), (' ', 5), ('#', 4), (' ', 1), ('#', 4), (' ', 5), ('#', 3), (' ', 4), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 1), ('#', 4), (' ', 12), ('#', 4), (' ', 1)]
[(' ', 1), ('#', 3), (' ', 11), ('#', 4), (' ', 5), ('#', 4), (' ', 1), ('#', 4), (' ', 5), ('#', 3), (' ', 4), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 3), (' ', 12), ('#', 4), (' ', 1)]
[(' ', 2), ('#', 3), (' ', 6), ('#', 2), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 3), (' ', 4), ('#', 4), (' ', 4), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 3), ('#', 3), (' ', 6), ('#', 2), (' ', 3), ('#', 4), (' ', 1)]
[(' ', 3), ('#', 3), (' ', 4), ('#', 2), (' ', 3), ('#', 4), (' ', 5), ('#', 4), (' ', 3), ('#', 11), (' ', 3), ('#', 4), (' ', 5), ('#', 4), (' ', 2), ('#', 4), (' ', 5), ('#', 4), (' ', 4), ('#', 3), (' ', 4), ('#', 2), (' ', 4), ('#', 4), (' ', 1)]
[(' ', 6), ('#', 3), (' ', 5), ('#', 6), (' ', 4), ('#', 5), (' ', 4), ('#', 2), (' ', 4), ('#', 4), (' ', 1), ('#', 6), (' ', 4), ('#', 11), (' ', 4), ('#', 5), (' ', 6), ('#', 3), (' ', 6), ('#', 6)]
[(' ', 95)]

想了半天发现每一个都是字符加数字的组合
再想了半天发现每一行的数字加起来都一样
猜到了就是这样的。。。


第六关zipfile

进去那个paypal跟题没有关系
源代码里面有个提示zip 就直接下载channel.zip
解压出来跟第四关的规则差不多
关键就是读文件:

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nothing = '90052'
for i in range(5000):
text = nothing+'.txt'
nothing = open("%s"%text,"rb").read().decode()[15:].strip()
print(nothing)

最后一个是46145.txt:
Collect the comments.
然后又想了半天,原来 zipfile里面有一个zipinfo就叫做comments
怎么用呢?

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import datetime
import zipfile
def print_info(archive_name):
zf = zipfile.ZipFile(archive_name)
for info in zf.infolist():
print info.filename
print '\tComment:\t', info.comment
print '\tModified:\t', datetime.datetime(*info.date_time)
print '\tSystem:\t\t', info.create_system, '(0 = Windows, 3 = Unix)'
print '\tZIP version:\t', info.create_version
print '\tCompressed:\t', info.compress_size, 'bytes'
print '\tUncompressed:\t', info.file_size, 'bytes'
print

if __name__ == '__main__':
print_info('channel.zip')

显示如下:

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README.txt
Comment:
Modified: 2007-12-16 10:08:52
System: 3 (0 = Windows, 3 = Unix)
ZIP version: 23
Compressed: 63 bytes
Uncompressed: 75 bytes

然后结合最开始的顺序读出来:

这一关略麻烦呢。。。


第七关图片操作

从这一关开始后面都是用python2写的

应该是灰度处理中间的一行灰色
然后读取出这个灰度值组成的数据
不出意料应该是包含了信息的。
使用Image来处理图片: 加入第三方库文件
然后对这张图片进行切割,只留下灰色的部分:

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import Image
im = Image.open('oxygen.png').crop((0,45,608,47))
print im.size

用预览打开,发现每一个宽7 pixel,就开始读数据了:

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for i in range(0,87):
L.append(im.getpixel(((1+7*i),1))[0])
print L


这个是aksii哟~

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for i in L:
k+=chr(i)
print k

果然是这样的:


第八关bz2

这一关没办法不是自己做的
就是一个解压:

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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'
un=bz2.decompress(un)
pw=bz2.decompress(pw)
print un,pw


第九关un:huge pw:file

这一关没有使用 python
源文件跟上一关的 coords一样的,就是连线
于是我把上一关的html的值改成这一关的了:

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<html>
<head>
<title>working hard?</title>
<link rel="stylesheet" type="text/css" href="../style.css">
</head>
<body>
<br><br>
<center>
<img src="integrity.jpg" width="640" height="480" border="0" usemap="#notinsect"/>
<map name="notinsect">
<area shape="poly"
coords="146,399,163,403,170,393,169,391,166,386,170,381,170,371,170,355,169,346,167,335,170,329,170,320,170,310,171,301,173,290,178,289,182,287,188,286,190,286,192,291,194,296,195,305,194,307,191,312,190,316,190,321,192,331,193,338,196,341,197,346,199,352,198,360,197,366,197,373,196,380,197,383,196,387,192,389,191,392,190,396,189,400,194,401,201,402,208,403,213,402,216,401,219,397,219,393,216,390,215,385,215,379,213,373,213,365,212,360,210,353,210,347,212,338,213,329,214,319,215,311,215,306,216,296,218,290,221,283,225,282,233,284,238,287,243,290,250,291,255,294,261,293,265,291,271,291,273,289,278,287,279,285,281,280,284,278,284,276,287,277,289,283,291,286,294,291,296,295,299,300,301,304,304,320,305,327,306,332,307,341,306,349,303,354,301,364,301,371,297,375,292,384,291,386,302,393,324,391,333,387,328,375,329,367,329,353,330,341,331,328,336,319,338,310,341,304,341,285,341,278,343,269,344,262,346,259,346,251,349,259,349,264,349,273,349,280,349,288,349,295,349,298,354,293,356,286,354,279,352,268,352,257,351,249,350,234,351,211,352,197,354,185,353,171,351,154,348,147,342,137,339,132,330,122,327,120,314,116,304,117,293,118,284,118,281,122,275,128,265,129,257,131,244,133,239,134,228,136,221,137,214,138,209,135,201,132,192,130,184,131,175,129,170,131,159,134,157,134,160,130,170,125,176,114,176,102,173,103,172,108,171,111,163,115,156,116,149,117,142,116,136,115,129,115,124,115,120,115,115,117,113,120,109,122,102,122,100,121,95,121,89,115,87,110,82,109,84,118,89,123,93,129,100,130,108,132,110,133,110,136,107,138,105,140,95,138,86,141,79,149,77,155,81,162,90,165,97,167,99,171,109,171,107,161,332,155,348,156,353,153,366,149,379,147,394,146,399,"
href="" />
<area shape="poly"
coords="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"
href="" />
</map>
<br><br>
<font color="#303030" size="+2">Where is the missing link?</font>
</body>
</html>

然后应该是个牛,要不就是羊。。。
我输入一个http://www.pythonchallenge.com/pc/return/cow.html他还告诉我hmm. it's a male.


第十关

1 11 21 1211 111221 …
len(a[30])
1:一个一
11:两个一
21:一个二一个一
1211:一个一一个二两个一
这个顺序来的
直接写程序就行了:

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def next(arr):
last = arr[0]
count = 1
s =''
for i in arr[1:]:
if last == i:
count += 1
else:
s += str(count)+last
last = i
count = 1
s += str(count)+last
print len(s)
return s

arr = "1"
for i in range(30):
arr = next(arr)

我代码写的不好,但是正确了的。。。
把上面的print s改成print len(s)就可以得出答案了


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