pythonchallenge

提示，查看前一关解决方案将url中的pc->pcc

要解决这里面的问题需要有一下蛋疼的精神
    1. 随时查看源代码
    2. 熟悉英语，例如里面给个图是个拉链要坑爹的联想到zip
    3. 熟悉python模块，例如有peak提示让你发音,
       熟悉python的人才知道pickle

0

http://www.pythonchallenge.com/pc/def/0.html
这个题没什么说的 套入计算即可
2 ** 38 = 274877906944

Hint: try to change the URL address
-->
http://www.pythonchallenge.com/pc/def/274877906944.html

1

everybody thinks twice before solving this. thick twice，如果直接按照这三个字母对应的东西解密的话接出来的依然是 乱码,回头一想找到规律，每个小写字母移动了两位，这就是think twice

k -> m o -> q e -> g 给出的提示又都是小写 可以看出规律， s = '''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.''' import string from = string.ascii_lowercase to = from[2:] + from[:2] table = string.maketrans(from, to) print string.translate(s, table) --> Out[21]: "i hope you didnt translate it by hand. thats what computers are for. doing it in by hand is inefficient and that's why this text is so long. using string.maketrans() is recommended. now apply on the url."

In [22]: string.translate('map', table) Out[22]: 'ocr'

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

2

recognize the characters. maybe they are in the book, 
but MAYBE they are in the page source.
page source不是书上的玩意，是源代码
源代码里面有很大一段的注释

hint:find rare characters in the mess below:

from collections import Counter

In [55]: Counter(s)
Out[55]: Counter({')': 6186, '@': 6157, '(': 6154, ']': 6152, '#':
6115, '_': 6112, '[': 6108, '}': 6105, '%': 6104, '!': 6079, '+':
6066, '$': 6046, '{': 6046, '&': 6043, '*': 6034, '^': 6030, '\n':
1219, 'a': 1, 'e': 1, 'i': 1, 'l': 1, 'q': 1, 'u': 1, 't': 1, 'y':
1})

rare --> aeilquty --> equality
http://www.pythonchallenge.com/pc/def/equality.html

3

标题是re，明显是正则,hit又提示一个小写字母在三个大写中间
继续看源码，里面又有一堆注释
import re
p = re.compile('[^A-Z][A-Z]{3}[a-z][A-Z]{3}[^A-Z]')
m = p.search(s)
res = p.findall(s)
text = ''.join(t[4] for t in res)
-->
linkedlist

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

4

linkedlist 居然是一个也面一个页面的找，输入俩页面之后会发现提示你
是不是手累了，确实需要写个脚本的，不过里面往下走还有坑。

import requests
import re
p = re.compile('\d+')
url = 'http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing='
next_num = '12345'
with open('pych', 'aw') as f:
    while 1:
        r = requests.get(url + next_num)
        f.write(r.text)
        next_num = p.search(r.text).group()
        print r.text
        print url + next_num

坑
        16044
        Yes. Divide by two and keep going.
        82682
        There maybe misleading numbers in the text. One example is 82683.
        Look only for the next nothing and the next nothing is 63579
        82683
        You've been misleaded to here. Go to previous one and check.
        66831
        peak.html

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

5

hit:pronounce it --> pickle
源码里面有个链接，banner.p打开一看是被pickle过的东西，
解压出来的东西比接特别,十一个二维数组，看一眼每一行
数字加和为95，很自然的想到把这玩儿打印出来看看。

import requests
import pickle
url = 'http://www.pythonchallenge.com/pc/def/banner.p'
r = requests.get(url)
text = pickle.loads(r.text)
with open('f', 'wa') as f:
    for line in text:
        for symbol, num in line:
            f.write(symbol * int(num))
        f.write('\n')

打开文件一看是channel

              #####                                                                      ##### 
               ####                                                                       #### 
               ####                                                                       #### 
               ####                                                                       #### 
               ####                                                                       #### 
               ####                                                                       #### 
               ####                                                                       #### 
               ####                                                                       #### 
      ###      ####   ###         ###       #####   ###    #####   ###          ###       #### 
   ###   ##    #### #######     ##  ###      #### #######   #### #######     ###  ###     #### 
  ###     ###  #####    ####   ###   ####    #####    ####  #####    ####   ###     ###   #### 
 ###           ####     ####   ###    ###    ####     ####  ####     ####  ###      ####  #### 
 ###           ####     ####          ###    ####     ####  ####     ####  ###       ###  #### 
####           ####     ####     ##   ###    ####     ####  ####     #### ####       ###  #### 
####           ####     ####   ##########    ####     ####  ####     #### ##############  #### 
####           ####     ####  ###    ####    ####     ####  ####     #### ####            #### 
####           ####     #### ####     ###    ####     ####  ####     #### ####            #### 
 ###           ####     #### ####     ###    ####     ####  ####     ####  ###            #### 
  ###      ##  ####     ####  ###    ####    ####     ####  ####     ####   ###      ##   #### 
   ###    ##   ####     ####   ###########   ####     ####  ####     ####    ###    ##    #### 
      ###     ######    #####    ##    #### ######    ###########    #####      ###      ######


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

6

打开一看是拉链 --> zip
然后看标题是channel, python面貌似没有channel + zip有关的东西啊
源代码里有一段捐赠的注释，想了好久实在没有办法
http://www.pythonchallenge.com/pc/def/channel.zip
google下发现居然把url后缀改为.zip
下载下来解压出来就是,linkedlist里面的那些东西
有个readme
    welcome to my zipped list.

    hint1: start from 90052
    hint2: answer is inside the zip
应该用python的zipfile
看了下里面有个ZipFile, ZipInfo
试了下用ZipFile channel.zip后有个 NameToInfo
显示了一个dict = {
    '99254.txt': <zipfile.ZipInfo at 0x374b050>,
    '99460.txt': <zipfile.ZipInfo at 0x374b118>,
    ...
}
去了90052.txt查看了下，属性里面有个comment，貌似这里会有信息
是个'*'

import zipfile
import re

p = re.compile('\d+')
next = '90052'
suffix = '.txt'
z = zipfile.ZipFile('channel.zip')
infos = z.NameToInfo
result = []
try:
    while 1:
        next_file = next + suffix
        comment = infos[next_file].comment
        result.append(comment)
        print next_file, comment
        with z.open(next_file) as zf:
            next = p.search(zf.read()).group()
except:
    with open('out', 'wa') as out:
        out.write(''.join(result))

打开文件查看是
hockey
http://www.pythonchallenge.com/pc/def/hockey.html

it's in the air. look at the letters.

在天上，重新组合这个单词么？
妹的，英语不好行了吧，用python的全排列看看
from itertools import permutations
with open('word', 'wa') as f:
f.write('\n'.join(''.join(word) for word in permutations('HCOKEY')))

我甚至找了这俩网站来查字母
http://www.scrabblefinder.com/word/
http://www.allscrabblewords.com/unscramble/

我擦，妹的，最后居然不是,而是用里面的字母oxygey

****************************************************************
****************************************************************
**                                                            **  
**   OO    OO    XX      YYYY    GG    GG  EEEEEE NN      NN  **  
**   OO    OO  XXXXXX   YYYYYY   GG   GG   EEEEEE  NN    NN   **  
**   OO    OO XXX  XXX YYY   YY  GG GG     EE       NN  NN    **  
**   OOOOOOOO XX    XX YY        GGG       EEEEE     NNNN     **  
**   OOOOOOOO XX    XX YY        GGG       EEEEE      NN      **  
**   OO    OO XXX  XXX YYY   YY  GG GG     EE         NN      **  
**   OO    OO  XXXXXX   YYYYYY   GG   GG   EEEEEE     NN      **  
**   OO    OO    XX      YYYY    GG    GG  EEEEEE     NN      **  
**                                                            **  
****************************************************************
 **************************************************************

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

7

这一关是用pil
学了下pil的库,不过不想做这个题怎么办
http://danmcewan.com/python-challenge-7
http://holger.thoelking.name/python-challenge/7

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

8

看了下源代码，里面有个链接，打开需要用户名密码，
源代码下面有用户名密码的东西，不过貌似是加密的，
不过都有BZh91AY&SY这个东西，应该是某种算法的格式
google了下原来是python的bz2模块

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 --> huge file
登录进去之后是
http://www.pythonchallenge.com/pc/return/good.html

9

依然是pil的题目
http://holger.thoelking.name/python-challenge/9

http://www.pythonchallenge.com/pc/return/bull.html

10

读出来的数字

1 11 21

1个1 --> 11
两个1 --> 21
一个2一个1 --> 1211

ch n
[(n,ch),....]

from itertools import groupby

next = '1'
result = ['1']
for i in xrange(30):
    _generator = groupby(next)
    stat = [str(len(list(_iter))) + key for key, _iter in _generator]
    next = ''.join(stat)
    result.append(next)
print len(result[30])

http://www.pythonchallenge.com/pc/return/5808.html

11

import Image

# download the image from: http://www.pythonchallenge.com/pc/return/cave.jpg

image = Image.open("cave.jpg")
nsize = tuple([x / 2 for x in image.size])
odd = even = Image.new(image.mode, nsize)

for x in range(image.size[0]):
    for y in range(image.size[1]):
        if x % 2 == 0 and y % 2 == 0:
            even.putpixel((x / 2, y / 2), image.getpixel((x, y)))
        elif x % 2 == 0 and y % 2 == 1:
            odd.putpixel((x / 2, (y - 1) / 2), image.getpixel((x, y)))
        elif x % 2 == 1 and y % 2 == 0:
            even.putpixel(((x - 1) / 2, y / 2), image.getpixel((x, y)))
        else:
            odd.putpixel(((x - 1) / 2, (y - 1) / 2), image.getpixel((x, y)))

even.save("even.jpg")
odd.save("odd.jpg")

http://www.pythonchallenge.com/pc/return/evil.html

12

查看源代码，发现里面图片为evil1.jpg说明还有2,3,4
第二张里面说not jpg --> gfx,将url后缀改为gfx

下面是网上的解决方案，不知道为毛是5
evil = open('evil2.gfx','r').read()
img1 = open('img1.jpg','w')
img2 = open('img2.jpg','w')
img3 = open('img3.jpg','w')
img4 = open('img4.jpg','w')
img5 = open('img5.jpg','w')

for b in range(0,len(evil),5):
    img1.write(evil[b])
    img2.write(evil[b+1])
    img3.write(evil[b+2])
    img4.write(evil[b+3])
    img5.write(evil[b+4])

img1.close()
img2.close()
img3.close()
img4.close()
img5.close()

图片需要用火狐打开
dis pro port ional ity
disproportionality
ity被一条线画掉了
disproportional

13

#! /usr/bin/env python
'''python challenge level 13
question url: http://www.pythonchallenge.com/pc/return/disproportional.html
answer url: http://www.pythonchallenge.com/pcc/return/italy.html
'''

import xmlrpclib
xmlrpc = xmlrpclib.ServerProxy("http://www.pythonchallenge.com/pc/phonebook.php")
print xmlrpc.system.listMethods()
print xmlrpc.system.methodHelp('phone')
print xmlrpc.phone('Bert')

http://www.pythonchallenge.com/pc/return/italy.html

14

import Image, math

def get_pixel(t):
    t = (100 * 100 - 1) - t
    shell = int((math.sqrt(t) + 1) / 2)
    leg = 0 if (shell == 0) else int((t - (2 * shell - 1) ** 2) / 2 / shell)
    elt = t - (2 * shell - 1) ** 2 - 2 * shell * leg - shell + 1

    if leg == 0:
        x, y = shell, elt
    elif leg == 1:
        x, y = -elt, shell
    elif leg == 2:
        x, y = -shell, -elt
    else:
        x, y = elt, -shell

    return (49 + x, 50 - y)

# Download the image from: http://www.pythonchallenge.com/pc/return/wire.png

input = Image.open("wire.png")
output = Image.new("RGB", (100, 100))

for i, px in enumerate(input.getdata()):
    output.putpixel(get_pixel(i), px)

output.save("new.jpg")

http://www.pythonchallenge.com/pc/return/cat.html
cat