利用第8关的账号和密码进入第10关, 显示是一张黄牛的图片, bull? 图片下面是一个 Python 语句:

len(a[30]) = ?

将鼠标放在牛的身上, 原来是一个链接, 点击打开, 显示一个未完的列表:

a = [1, 11, 21, 1211, 111221,

看来这题的题目就是根据那个未完的列表提示, 推测 a[30] 的长度.

再看看那个未完的列表有什么规律:

a[0] = ’1′
a[1] = ’11′ 每两个字符拆分理解为: 1个’1′ 等于 a[0]
a[2] = ’21′ 每两个拆分理解为: 1个’2′ 等于 a[1]
a[3] = ’1211′ 每两个拆分理解为: 1个’2′ + 2个’1′ 等于 a[2]
a[4] = ’111221′ 第两个字符拆分理解为: 1个’1′ + 1个’2′ + 2个’1′ 等于 a[3]
…

规律出来了, 剩下的可以交给 Python :)

import re

def repl(match_obj):
    return '%s%s' % (len(match_obj.group()), match_obj.groups()[0])

if __name__ == '__main__':
    a = '1'
    reg = re.compile(r'(\d)\1*')
    for i in range(30):
        a = reg.sub(repl, a)
    print('a[30] = %s\n' % a)
    print('len(a[30]) = %d' % len(a))

可知, len(a[30]) 的答案是5808. 下一关的网址就是: http://www.pythonchallenge.com/pc/return/5808.html

Have fun~~

利用第8关得到的账号和密码成功进入第9关, 显示一幅河岸的图片, 图片上有若干黑点, 网页没有其它文字提示. 再看看网页的标题, “connect the dots”???难道是将图片上的黑点连接起来?

先别帮着动手, 看看网页的源代码吧, 毕竟以往的经验说明提示经常出现在网页源代码里:) 果然, 网页源代码的注释里出现两组数字—-first 和 second, 而且提示 “first+second”. 再回看标题”connect the dots”, 难道是将这些数字代表的坐标点连接起来?? 利用 PIL 的 ImageDraw 进行画图:

from PIL import Image, ImageDraw

first_points = (    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_points = (    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
)

if __name__ == '__main__':
    img = Image.new('RGBA', (500,500))
    draw = ImageDraw.Draw(img)
    draw.polygon(first_points)
    draw.polygon(second_points)
    img.save('good_ok.jpg')

打开生成的图片 good_ok.jpg, 明显看到那些点连接起来是一头牛, cow? 于是尝试打开http://www.pythonchallenge.com/pc/return/cow.html,显示:

hmm. it’s a male.

呵呵,原来是头公牛—-bull, 于是成功得到下一关的网址: http://www.pythonchallenge.com/pc/return/bull.html

Have fun~~

zip.exe -FF omni.jar –out omni_fix.jar

运行上面命令, 将会生成一个修复好的omni_fix.jar 文件, 将其重命名为 omni.jar 即可, 这时, 你就可以用你熟悉的解压工具打开它了.

备注:
Info-zip 的下载地址: ftp://ftp.info-zip.org/pub/infozip/win32/zip300xn.zip

Have fun~~

1. 最普遍的算法是利用递归:

def fibonacci(n):
    return n>=2 and fibonacci(n-2)+fibonacci(n-1) or n

看上去非常美妙,但其时间复杂度是非常惊人的: O(2^n) :(

2. 另一个常见的算法, 就是用循环取代递归算法:

def fibonacci(n):
    a, b = 0, 1
    for i in range(n):
        a, b = b, a+b
    return a

其时间复杂度为: O(n), 相比第1种递归算法, 已经是一种”飞跃”了 :> 它也有其变种, 就是利用 Python 的 generator, 其原理是一样, 只是应用的场合不一样罢了

def fibonacci():
    a, b = 0, 1
    while 1:
        yield a
        a, b = b, a+b

f = fibonacci()
n = 100
for i in range(n+1):
    print(next(f))

3. 最后一种算法是利用矩阵运算来加速:

import numpy

fibonacci_matrix = numpy.matrix([[1,1],[1,0]], dtype=numpy.ndarray)
def fibonacci(n):
    return fibonacci_matrix**(n-1)[0, 0]

需要用到 numpy 库, 其时间复杂度为: O(logn).

4. 通过 Fibonacci 序列的通用公式, 可以利用简单的数学方法求其近似值:

from math import sqrt

def fibonacci(n):
    return int(((1+sqrt(5))/2)**n/sqrt(5))

虽然这是4种算法中速度最快的, 但当n越大,其计算精度偏差越大 :>

一打开第8关的网页, 显示的是一张蜜蜂辛勤(Working hard)采花图片, 将鼠标放在蜜蜂上, 发现蜜蜂的区域是一个链接: http://www.pythonchallenge.com/pc/return/good.html
打开那链接,发觉是需账号认证才能打开的. 用户名是和密码是什么呢? 将网页退后到第8关的网页, 查看其源代码, 发现了二行特别的注释:

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'

看来, 将上面的注释解密就行了. 留意那两行注释, 都是以”BZh91AY&SY”开头的,还有图片是蜜蜂?Bee?猜想是不是用 bz2 加密的呢? 立即试试:

import bz2

if __name__ == '__main__':
    un = b'BZh91AY&SYA\xaf\x82\r\x00\x00\x01\x01\x80\x02\xc0\x02\x00 \x00!\x9ah3M\x07<]\xc9\x14\xe1BA\x06\xbe\x084'
    pw = b'BZh91AY&SY\x94$|\x0e\x00\x00\x00\x81\x00\x03$ \x00!\x9ah3M\x13<]\xc9\x14\xe1BBP\x91\xf08'
    print('Username: %s' % bz2.decompress(un).decode('ascii'))
    print('Password: %s' % bz2.decompress(pw).decode('ascii'))

通往下一关: http://www.pythonchallenge.com/pc/return/good.html 的用户名是huge, 密码是file.

Have fun:)

打开网页,显示的是一张河岸的图片, 特别的是, 图片中央有一细条”灰色带”.将图片另存为”oxygen.png”, 用图片编辑器打开, 放大到最大的倍数.可以看到,那色带是一长方形的灰度色带(left: 0px, top: 43px, right: 607px, bottom: 51px), 它又是由大小为7px * 9px的小长方形灰度块(除了第1块是 5px * 9px)组成的. 灰度的值是0~255, 正好跟 Ascii 相对应, 思路出来了, 下面交给 Python 来处理:

from PIL import Image

if __name__ == '__main__':
    img = Image.open('oxygen.png')
    for p in range(4, 607, 7):
        print(chr(int(sum(img.getpixel((p,43))[:3])/3)), end='')
    print()

注意, 处理图片需要用到 PIL 库, Python 3的 Windows 编译版可以到以下网址下载: http://www.lfd.uci.edu/~gohlke/pythonlibs/#pil

运行结果:

smart guy, you made it. the next level is [105, 110, 116, 101, 103, 114, 105, 116, 121]

可知要对 “[105, 110, 116, 101, 103, 114, 105, 116, 121]“进行映射就得到答案了;现将上面的代码改写,一次性的得到答案:)

from PIL import Image
import re

if __name__ == '__main__':
    img = Image.open('oxygen.png')
    map_string = ''.join(map(lambda p: chr(int(sum(img.getpixel((p, 43))[:3])/3)), range(4, 607, 7)))
    #print(map_string)
    print('\nKey Words: ', end='')
    for i in re.finditer(r'\d+', map_string):
        print(chr(int(i.group())), end='')
    else:
        print()

显示结果:

Key Words: integrity

可知下一关的网址是: http://www.pythonchallenge.com/pc/def/integrity.html

Have fun :)

#!bin/pyhton3
# coding=utf8

import os, sys, getopt, functools
from PIL import Image

# 源图片路径
SRC = os.getcwd()
# 目标保存图片的路径
DEST = os.getcwd()
# 缩略图保存的文件夹名称, 最终缩略图的保存路径为 sys.path.join(DEST, THUMB_DIR).
THUMB_DIR = '缩略图'
# 缩放比例
RATE = 0.15
# 缩略图的命名, 在原文件名前加上 THUMB_PREFFIX, 如"thumb_"
THUMB_PREFFIX = ''
# 是否递归搜索子文件夹
IS_RECURSIVE = True

def coroutine(func):
    """ 用于协程的装饰器 """
    @functools.wraps('func')
    def start(*args, **kwargs):
        g = func(*args, **kwargs)
        next(g)
        return g
    return start

@coroutine
def thumb(save_path, rate, thumb_preffix=''):
    img_exts = ('.jpg', '.jepg', '.gif', '.png', '.bmp')
    saved_filenames = set()
    repeatd_times = 0
    while 1:
        img_path = (yield)
        img_ext = os.path.splitext(img_path)[1].lower()
        if not img_ext in img_exts:
            continue
        print(img_path)
        save_filename = '%s%s%s' % (thumb_preffix, os.path.splitext(os.path.split(img_path)[1])[0], img_ext)
        if save_filename in saved_filenames:
            repeatd_times += 1
            save_filename = '%s_%s%s' % (os.path.splitext(save_filename)[0], repeatd_times, img_ext)
        img = Image.open(img_path)
        img.thumbnail((int(img.size[0]*rate), int(img.size[1]*rate)), 1)
        img.save(os.path.join(save_path, save_filename))
        saved_filenames.add(save_filename)
        print('%s %s' % ('Thumb', save_filename))

def process(src, func, is_recursive=True):
    files = []
    dirs = []
    for f in os.listdir(src):
        path = os.path.join(src, f)
        if os.path.isfile(path):
            files.append(path)
        elif is_recursive and os.path.isdir(path) and f.lower() not in ('缩略图', 'thumb', 'thumbs', 'thumbnail', 'thumbnails'):
            dirs.append(path)
    for path in files:
        func.send(path)
    for path in dirs:
        process(path, func, is_recursive)

if __name__ == '__main__':
    src, dest, thumb_dir, rate, thumb_preffix, is_recursive = SRC, DEST, THUMB_DIR, RATE, THUMB_PREFFIX, IS_RECURSIVE
    opts, args = getopt.getopt(sys.argv[1:], 'r:d:p:s')
    print(opts, args)
    for k,v in opts:
        if k == '-r':
            rate = float(v)
        elif k == '-d':
            thumb_dir = v
        elif k == '-p':
            thumb_preffix = v
        elif k == '-s':
            is_recursive = False
    if args:
        args.append(os.getcwd())
        src, dest = args[:2]

    print(src, dest, thumb_dir, rate, thumb_preffix, is_recursive)
    sys.exit()
    save_path = os.path.join(dest, thumb_dir)
    if not os.path.exists(save_path):
        os.makedirs(save_path)

    my_thumb = thumb(save_path, rate, thumb_preffix)
    process(src, my_thumb, is_recursive)
    print('\nDone!')
    input()

脚本依然是基于 PIL 的, 但目前 PIL 的官方还未发布兼容 Python 3 的 Windows 编译版, 幸运的是, 网上可以下载到非官方的编译版本:

http://www.lfd.uci.edu/~gohlke/pythonlibs/#pil

脚本默认递归搜索当前文件夹及子文件夹的图像文件, 然后将生成的缩略图文件放在当前文件夹的”缩略图”文件夹下.当然, 也可修改脚本的参数, 或者在命令行下指定参数:

python3 thumbnial_batch.py -r 0.5 -d ‘thumbs’ -p ‘thumb’ -s 源文件夹 目标文件夹

参数 -r: 指定压缩图片大小的比率;参数 -d: 指定保存缩略图的文件夹名称; 参数 -p: 指定生成的缩略图的文件名添加的前缀; 参数 -s: 只搜索源文件夹,不进行递归搜索子文件夹.
Have fun :>

#!/bin/python
#coding=utf-8

import filecmp, shutil, os, sys

SRC = r'C:/a'
DEST = r'C:/b'

IGNORE = ['Thumbs.db']

def get_cmp_paths(dir_cmp, filenames):
    return ((os.path.join(dir_cmp.left, f), os.path.join(dir_cmp.right, f)) for f in filenames)

def sync(dir_cmp):
    print(dir_cmp.left)
    for f_left, f_right in get_cmp_paths(dir_cmp, dir_cmp.right_only):
        if os.path.isfile(f_right):
            os.remove(f_right)
        else:
            shutil.rmtree(f_right)
        print('删除 %s' % f_right)
    for f_left, f_right in get_cmp_paths(dir_cmp, dir_cmp.left_only+dir_cmp.diff_files):
        if os.path.isfile(f_left):
            shutil.copy2(f_left, f_right)
        else:
            shutil.copytree(f_left, f_right)
        print('复制 %s' % f_left)
    for sub_cmp_dir in dir_cmp.subdirs.values():
        sync(sub_cmp_dir)

def sync_files(src, dest, ignore=IGNORE):
    if os.path.isfile(src) or os.path.isfile(dest):
        print('只能对文件夹进行同步, 请正确输入源文件夹和目标文件夹...')
        return
    dir_cmp = filecmp.dircmp(src, dest, ignore=IGNORE)
    sync(dir_cmp)
    print('同步完成!')

if __name__ == '__main__':
    src, dest = SRC, DEST
    if len(sys.argv) == 3:
        src, dest = sys.argv[1:3]
    sync_files(src, dest)
    input()

脚本设置了默认的 src 和 dest 参数(源文件夹和目标文件夹), 也可以在运行脚本时指定:

python3 syncfiles_py3k.py your_src your_dest

Have fun~~

Have fun~~

回归正题, Firefox Nightbuild 版有它的好处, 问题是很多扩展未能兼容, 但有一部分扩展是兼容的,只是它支持的最大版本号未修改过来,这样,就需要人工手动修改其扩展(xpi文件)的install.rdf. (当然,除了修改最大版本号,还有很多方法可以让其兼容,网上都有介绍)

下面的 Python 脚本可以达到批量修改xpi文件的目的:

#!bin/python3
# coding=utf8

import re, zipfile, os, sys

maxversion = 100
ff_maxversion_reg = re.compile(br'(ec8030f7-c20a-464f-9b0e-13a3a9e97384.*?em:maxVersion.*?)([^>< ="/]+)', re.S+re.I)

if __name__ == '__main__':
    if len(sys.argv) > 1:
        maxversion = int(sys.argv[1])
    print('Target Maxversion: %s\n' % maxversion)
    sys.exit()

    for filename in os.listdir():
        print('%s ' % filename, end='')
        if os.path.isdir(filename) or not filename.lower().endswith('.xpi'):
            print('skip.')
            continue
        zin = zipfile.ZipFile(filename)
        rdf = zin.read('install.rdf')
        version = 0
        for item in ff_maxversion_reg.finditer(rdf):
            match_obj = re.search(br'\d+', item.groups()[1])
            if match_obj and int(match_obj.group()) > version:
                version = int(match_obj.group())
        if version >= maxversion:
            zin.close()
            print('skip.')
            continue

        zout = zipfile.ZipFile('new.xpi','w')
        rdf = ff_maxversion_reg.sub(br'\g<1>'+ str(maxversion).encode('utf8'), rdf)
        zout.writestr('install.rdf', rdf)
        for item in zin.infolist():
            if item.filename.lower() == 'install.rdf':
                continue
            else:
                buffer = zin.read(item.filename)
                zout.writestr(item, buffer)
        zin.close()
        zout.close()
        os.remove(filename)
        os.rename('new.xpi', filename)
        print('done!')

将脚本放于 Firefox 配置文件夹的 extensions 目录下运行即可.也可以带1个参数运行,用于指定目标最大版本号,如:

python3 xpi_rdf_batch.py 200

值得注意的是,目前它仅可以修改 xpi 文件, 对于 unpack 的扩展不能修改.

Have fun :)