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Archive for 2011年9月

The Python Challenge Lv.10

Posted by Ross Wan 于 2011/09/08

Lv.10

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~~

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The Python Challenge Lv.9

Posted by Ross Wan 于 2011/09/07

Lv.9

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~~

Posted in Python | Tagged: , , , | 1 Comment »

Firefox:修复 omni.jar

Posted by Ross Wan 于 2011/09/06

自从 Firefox 4 开始, Firefox 许多核心文件都放在一个名为 omni.jar 的文件里, 但是, 目前很多流行的解压工具都无法正常浏览或者解压它. 究其原因, 它是一个非标准的 zip 格式文件, 有人曾经就此向 Mozill 提交过 bug, 不过直到目前为止, Firefox 9.0a1 测试版, 问题依旧存在.现提供一个解决方法: 利用 Info-zip 来修复 omni.jar 文件的 header:

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~~

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Fibonacci(斐波那契)序列的4种求解算法

Posted by Ross Wan 于 2011/09/02

Fibonacci(斐波那契)序列, 其求解算法, 很多编程书都会以其作为例子.

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 :>

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The Python Challenge Lv.8

Posted by Ross Wan 于 2011/09/01

Lv.8

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:)

Posted in Python | Tagged: , , | 2 Comments »