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  • Pythonchallenge一起来闯关

    http://www.pythonchallenge.com/是一个在线的python过关游戏,一共有33关.玩这个游戏对熟悉python用法及相关库的使用都很有好处.

    目前做到了第九关.python版本3.4. 操作系统版本32位win7,64位win7.

    0.http://www.pythonchallenge.com/pc/def/0.html

    计算2的38次方

    1.http://www.pythonchallenge.com/pc/def/274877906944.html

    提示已经很清楚了,简单的字符转换.可以使用str模块中的maketrans()和translate()方法.https://docs.python.org/3/library/stdtypes.html#str.translate

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

    页面源码里有注释:

    <!--
      find rare characters in the mess below:
      -->

    用到了re模块,这里有很好的关于re模块的使用说明,我觉得比官方文档要好读一些http://www.cnblogs.com/huxi/archive/2010/07/04/1771073.html

    import urllib.request as ur
    import re
    
    url = "http://www.pythonchallenge.com/pc/def/ocr.html"
    
    def main():
        global url
    
        response = ur.urlopen(url)
        body = response.read()
    
        text = re.search("<!--
    %(.|s)+",body.decode())
        dic={}
        #print(text.group(0))
        for x in text.group(0):
            if x not in dic:
                dic[x] = 1
            else:
                dic[x] += 1 
    
        print(dic)
        for i in dic:
            if (dic[i]== 1 and 'a' <= i <='z'):
                print(i)
    
    if __name__ == '__main__':
        main()

    3.http://www.pythonchallenge.com/pc/def/equality.html

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

    import urllib.request as ur
    import re
    
    url = "http://www.pythonchallenge.com/pc/def/equality.html"
    
    def main():
        response = ur.urlopen(url)
        body = response.read()
    
        pattern="[^A-Z][A-Z][A-Z][A-Z]([a-z])[A-Z][A-Z][A-Z][^A-Z]"
    
        result = re.findall(pattern,body.decode())
        print(result)
    
    if __name__ == '__main__':
        main()

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

      查看页面源代码,可以看到提示有follow the chain,又有一个href,点击之后页面显示“next nothing is ....”猜想是不断地访问下一个url直到答案出现.

    使用re模块时,注意match()与search()的区别:
    Python offers two different primitive operations based on regular expressions: re.match() checks for a match only at the beginning of the string, while re.search() checks for a match anywhere in the string (this is what Perl does by default).
     
    import urllib.request as ur
    import re
    
    url = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=8022"
    
    def main():
        global url
        global body
        while 1:    
            print(url)
            response = ur.urlopen(url)
            #print(type(response))
    
            body = response.read()
            #print(type(body))
            #print(body)
            #print(body.decode())
    
            pattern="and the next nothing is ([0-9]+)"
            #match从string的开头查找
            #search从string的anywhere开始找
            result = re.search(pattern,body.decode())
            #print(result)
            #m=re.match(r'and the next nothing is ([0-9]+)', body)
            #print(m)
            if result:
                num = result.group(1)
                print(result.group(0))
                print(num)
                url = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=%s" % num
            else:
                #print(type(result),result)
                break
        print(body)
    
    if __name__ == '__main__':
        main()

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

      查看页面源代码有提示.peak hell sounds familiar.去翻翻标准库,看到一个模块叫pickle的.
      pickle模块用来实现python对象的序列化,反序列化.
    import urllib.request as ur
    import pickle
    
    
    url = "http://www.pythonchallenge.com/pc/def/banner.p"
    
    def main():
        global url
    
        httpresponse = ur.urlopen(url)
        dataBytes = httpresponse.read()
    
        obj = pickle.loads(dataBytes)
        #print(type(obj),obj)
        for ele in obj:
            #print(ele)
            s=""
            for i in ele:
                #print(i)
                #print(i[0]*i[1])
                s += i[0]*i[1]
            print(s)
            
    if __name__ == '__main__':
        main()

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

        查看页面源代码,注释里有zip字样,去标准库里查查看有没有zip相关。
        输入urlhttp://www.pythonchallenge.com/pc/def/channel.zip发现有一个channe.zip文件可以下载
        解压后读一下readme.txt。是类似于04的一直读文件下去,直到出现提示,collect comments。去翻标准库,发现一个zipfile.getinfo()的方法,原来打包的每个文件还可以有注释的,
        还真是第一次知道.
       
    import urllib.request as ur
    import zipfile
    import pathlib
    import re
    
    url = "http://www.pythonchallenge.com/pc/def/channel.zip"
    
    
    def main():
        # global url
    
        #下载zip文件并解压
        # httpresponse = ur.urlopen(url)
        # dataBytes = httpresponse.read()
        # with open("channel.zip", "wb") as local_file:
        #     if local_file.write(dataBytes):
        #         z = zipfile.ZipFile("channel.zip")
        #         z.extractall("./channel")
    
        # # p = pathlib.Path("./channel")
        # # for files in  p.iterdir():
        # #     #print(files,type(files))
        # #     #str = open(files).read()
        # #     str = files.open().read()
        # #     print(str)
        # #     break;
    
        #读取解压的文档获取提示
    
        z = zipfile.ZipFile("channel.zip")
        z.extractall("./channel")
    
        orderedFilePath=[]
        filepath = "./channel/90052.txt"
        filename = "90052.txt"
        while 1:
            orderedFilePath.append(filename)
            f = open(filepath)
            content = f.read()
            result = re.match("Next nothing is (d+)",content)
            if result:
                num = result.group(1)
                #print(num)
                filepath = "./channel/%s.txt" % num
                filename = "%s.txt" % num
            else:
                #print(content)
                break
    
        print(orderedFilePath)
        for file in orderedFilePath:
            c = z.getinfo(file).comment.decode()
            print(c,end="")
    
    if __name__ == '__main__':
        main()

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

    这一关要用到图形库PIL.首先PIL官方版不支持py3,其次官方版不支持64位python.一番google之后,下载http://pillow.readthedocs.org/en/latest/installation.htmlpillow代替PIL

    这题没什么思路,google以后知道是要读那些灰色块的像素值,再将其转码为对应的字符.
    import urllib.request as ur
    from PIL import Image
    
    def ReadUrl(url):
        httpresponse = ur.urlopen(url)
        dataBytes = httpresponse.read()
    
        return dataBytes
    
    def main():
        url = "http://www.pythonchallenge.com/pc/def/oxygen.png"
        databytes = ReadUrl(url)
        f = open("oxygen.png","wb")
        f.write(databytes)
        im = Image.open("oxygen.png")
        #print(im.size)
        (width,height) = im.size
    
        for i in range(0,width,1):    
            pos = (i,height/2)
            #pixel(49,49,49,255)前三位代表的是rgb的值.
            pixel = im.getpixel(pos)
            #ascii码转字符
            char = chr(pixel[0])
            print(char,end="")
    
        print(''.join(map(chr, [105, 110, 116, 101, 103, 114, 105, 116, 121])))
    
    if __name__ == '__main__':
        main()

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

    页面源码里有一个链接,href="../return/good.html",点击之后发现要用户名和密码,恰好当前页面源码里有注释

    <!--
      un: 'BZh91AY&SYAxafx82 x00x00x01x01x80x02xc0x02x00 x00!x9ah3Mx07<]xc9x14xe1BAx06xbex084'
      pw: 'BZh91AY&SYx94$|x0ex00x00x00x81x00x03$ x00!x9ah3Mx13<]xc9x14xe1BBPx91xf08'
      -->
    un,pw这不是username,password的缩写吗.最开始认为是和加密解密有关的,看了hashlib模块没有头绪,上网一搜原来是和压缩解压缩有关,BZh91AY&SY这是代表一种bzip2的压缩算法的.应该使用bz2模块..bz后缀的压缩文件还真没见过......
    开始时把un,pw作为str处理发现不对,要作为bytes处理
    import bz2
    
    un=b'BZh91AY&SYAxafx82
    x00x00x01x01x80x02xc0x02x00 x00!x9ah3Mx07<]xc9x14xe1BAx06xbex084'
    pw=b'BZh91AY&SYx94$|x0ex00x00x00x81x00x03$ x00!x9ah3Mx13<]xc9x14xe1BBPx91xf08'
    
    def main():
           #username = bz2.decompress(un.decode("utf-8"))
            username=bz2.decompress(un)
            password=bz2.decompress(pw)
            print(username,password)
    
    if __name__ == '__main__':
        main()

    9.http://www.pythonchallenge.com/pc/return/good.html

    提示1.要将这些点连起来,2.first+second=?
    考虑获取点的坐标,重新画图。
    import urllib.request as ur
    from PIL import Image
    
    def SaveImg(url,imgname):
        httpresponse = ur.urlopen(url)
        dataBytes = httpresponse.read()
        f = open(imgname,"wb")
        f.write(databytes)
        f.close()
    
    first = [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,
    111,156,113,170,115,185,118,208,117,223,121,239,128,251,133,259,136,266,139,276,143,290,148,310,151,
    332,155,348,156,353,153,366,149,379,147,394,146,399]
    
    second = [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]
    
    def main():
        coordinates = first + second
        im = Image.open("good.jpg")
        w,h = im.size
        mode = im.mode
        bands = im.getbands()
        print(type(mode),mode)
        print(type(bands),bands)
    
        #两两为一个像素点位置
        postions = []
        for x in range(0,len(coordinates),2):
            postion = (coordinates[x],coordinates[x+1])
            postions.append(postion)
    
        im2 = Image.new(mode,(w,h))
        for pos in postions:
            im2.putpixel(pos,(255,255,255))
        im2.save('09answer.jpg')
    
    if __name__ == '__main__':
        main()
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  • 原文地址:https://www.cnblogs.com/sdu20112013/p/4364507.html
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