【2019.06.01】33关 python游戏 pythonchallenge 通关记录

柔情只为你懂 2022-01-23 00:13 144阅读 0赞

# 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))

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4NTM0MTA3_size_16_color_FFFFFF_t_70]

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