The Python Challenge 0-4

项目地址：http://www.pythonchallenge.com/

Level-0

提示Hint: try to change the URL address.，修改0.html为1.html，提示2**38 is much much larger.，打开python控制台，计算后替换1.html为274877906944.html，进入下一关。

>>> 2**38
274877906944

Level-1

提示everybody thinks twice before solving this.，观察图片发现是一道找规律的数学题，字母间位置相差2，由此猜测位移计算，将字符串中所有字母右移两位。

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. "

def right_2(str):
    if str is 'y':
        return 'a'
    if str is 'z':
        return 'b'

    if str.isalpha() and str not in 'yz':
        return chr(ord(str)+2)
    return str
s = map(right_2, str)
print ''.join(s)

ord() 函数以一个字符（长度为1的字符串）作为参数，返回对应的 ASCII 数值。

chr() 用一个范围在 range（256）内的（就是0～255）整数作参数，返回一个对应的字符。

isalpha() 方法检测字符串是否只由字母组成。

map() 会根据提供的函数对指定序列做映射。第一个参数 function 以参数序列中的每一个元素调用 function 函数，返回包含每次 function 函数返回值的新列表。

程序运行结果为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.，根据提示添加str = "map"，运行得到结果ocr。

解法二：

根据提示的函数，maketrans() 方法用于创建字符映射的转换表，第一个参数是字符串，表示需要转换的字符，第二个参数也是字符串表示转换的目标。两个字符串的长度必须相同，为一一对应的关系。

生成字母表

>>> import string
>>> string.ascii_lowercase
'abcdefghijklmnopqrstuvwxyz'

操作字符串自定义规则

import string
from string import maketrans
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. "

def right_2(str):
    
    s = string.ascii_lowercase
    s1 = s
    s2 = s[2:26] + s[0:2]

    trantab = maketrans(s1,s2)
    if str.isalpha():
        return str.translate(trantab)
    return str
str = map(right_2,str)
print ''.join(str)

Level-2

提示recognize the characters. maybe they are in the book,but MAYBE they are in the page source.，直接查看源代码提示find rare characters in the mess below，复制后保存到文件，程序如下，运行结果为equality。

import string

l = []
with open('/Users/markzhang/Desktop/level3.txt','r') as f:
    str = f.read()
    for s in str:
    
        if s in string.ascii_lowercase:
            l.append(s)
        if s in string.ascii_uppercase:
            l.append(s)
    print ''.join(l)

Level-3

提示One small letter, surrounded by EXACTLY three big bodyguards on each of its sides.，要求查找类似xXXXxXXXx的小写字母，直接查看源代码，复制后保存到文件，程序如下，运行结果为linkedlist。

import re

with open('/Users/markzhang/Desktop/level4.txt','r') as f:
    str = f.read()
    pattern = re.compile(r'[^A-Z][A-Z]{3}([a-z])[A-Z]{3}[^A-Z]')
    result = re.findall(pattern,str)
    print ''.join(result)

Level-4

打开是一张图片，点击图片提示and the next nothing is 44827，替换为nothing=44827后提示and the next nothing is 45439，继续替换时提示Your hands are getting tired and the next nothing is 94485，猜测是请求发包次数的问题，回到显示图片页面，查看源码提示urllib may help. DON'T TRY ALL NOTHINGS, since it will never end. 400 times is more than enough.，尝试获取noting的值，循环发包获取答案。

import requests
import re

p = '12345'

def next_index(p):
    url = 'http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=%s' %p
    res = requests.get(url).content
    pattern = re.compile(r'\d+')
    index = re.search(pattern,res).group()
    return index

for i in range(1,401):
    print 'the %d time' %i
    p = next_index(p)
    print p
    

运行到86次报错，访问第85次的nothing值16044，提示Yes. Divide by two and keep going.，

根据提示除以二后使用8022继续访问，提示and the next nothing is 25357，赋值p后继续运行上面的程序，运行到55次出错，访问此时结果82683，提示You've been misleaded to here. Go to previous one and check.，

根据提示继续访问上一次运行结果82682，提示There maybe misleading numbers in the text. One example is 82683. Look only for the next nothing and the next nothing is 63579，继续访问63579，提示and the next nothing is 37278，赋值p后继续运行上面的程序，运行到109次报错，使用108次的值66831继续访问，得到peak.html。