【2019.06.01】33关 python游戏 pythonchallenge 通关记录

地址http://www.pythonchallenge.com/

1. 2的38次方

print(pow(2, 38))

将原url中的 0替换为 274877906944回车就会进入下一关

2. 字符映射

图片中的笔记本给了三组字母，很容易发现规律：前面的字母往后移动两位就是后面的字母。
使用 maketrans 和 translate

text = '''g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amk
   ...: nsrcpq ypc dmp. bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q u
   ...: fw rfgq rcvr gq qm jmle. sqgle qrpgle.kyicrpylq() gq pcamkkclbc
   ...: b. lmu ynnjw ml rfc spj.'''

import string
i = string.ascii_lowercase
t = str.maketrans(i, i[2:] + i[:2])
print(text.translate(t))

print('map'.translate(t))

结果是：

i hope you didnt translate it by hand. thats what com
…: puters are for. doing it in by hand is inefficient and that’s w
…: hy this text is so long. using string.maketrans() is recommende
…: d. now apply on the url.

将url中的map也通过该映射进行转换为ocr，进入下一关

3. 统计字符串中字符频率

在网页源代码里。那就右键查看源代码往下拉看到绿色乱码区域

意思就是：要在下面这一大串字符里找到出现次数最少的几个字符

import re

import requests
url = 'http://www.pythonchallenge.com/pc/def/ocr.html'
res = requests.get(url).text
text = re.findall('.*?.*',res,re.S)

text_str = ''.join(text)

lst = []
key =[]
for i in text_str:
    lst.append(i)
    if i not in key:
        key.append(i)

for items in key:
    print(items, lst.count(items))

可以看到出现次数最少的就是最后几个字符，合起来是「equality」，替换 url 字符进入下一关

4. One small letter, surrounded by EXACTLY three big bodyguards on each of its sides.

一个小写字母两边有三个大写字母

import re

import requests
url = 'http://www.pythonchallenge.com/pc/def/equality.html'
res = requests.get(url).text
text = re.findall('.*',res,re.S)
print(text)
p = re.compile("[a-z]+[A-Z]{3}([a-z])[A-Z]{3}[a-z]+")
print("".join(p.findall(str(text))))

得到linkedlist
进入http://www.pythonchallenge.com/pc/def/linkedlist.html提示要进linkedlist.php。进入下一关

5.

点开图片到达http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=12345提示你下一个nothing是44827，
这一关作者弄了个小恶作剧，需要手动输入数值到 url 中然后回车，你以为这样就完了么？并没有它有会不断重复弹出新的数值让你输入，貌似无穷尽。一段简单的爬虫加正则就能搞定。思路很简单，把每次网页中的数值提取出来替换成新的 url 再请求网页，循环下去。不过有个问题是当，结果i到85的时候告诉我要除以2。代码中处理一下

import requests
import re
url = 'http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing='
resp = requests.get(url=url+'12345').text
count = 0
nexturl = ''
nextid = ''
while True:
    try:
        nextid = re.search('\d+', resp).group()
        count += 1
        nextid = int(nextid)
    except Exception as msg:
        # print(msg)
        if "Divide by two" in resp:
            nextid = nextid/2
        else:
            print(f'最后一个url为 {nexturl}')
            break
        
        
    nexturl = url + str(nextid)
    print(f'url {count}:{nexturl}')
    
    # 重复请求
    resp = requests.get(nexturl).text
    print(resp)
    i = 0
    while len(resp) < 5:
        i += 1
        print(f'再次访问{i}')
        resp = requests.get(nexturl).text
    

最后得到250 peak.html，进入下一关

6

import requests

resp = requests.get('http://www.pythonchallenge.com/pc/def/banner.p').content
with open('banner.p', 'wb') as f:
    f.write(resp)

import pickle
fname="banner.p"
with open(fname,'rb') as fs:
    data=pickle.load(fs)

for line in data:
       print("".join(x[0]*x[1] for x in line))