9 第九关
是一张图,上面有很多的黑点,查看网页源代码发现了上一关的提示:
还发现了一大串的数字
感觉又是一个使用PIL库进行图像处理的题,百度后知道要将这些点连接起来并重新画图。但是不能在原始图上修改,我们应该重新画图并进行修改。
编写代码:
from PIL import Image
def main(three):
img = Image.open('good.jpg')
# 利用img的模式和大小创建新图片
img1 = Image.new(img.mode, img.size)
# 两两为一个像素位置
positions = []
for i in range(0, len(three), 2):
position = (three[i], three[i+1])
positions.append(position)
for j in positions:
img1.putpixel(j, (255, 255, 255))
img1.show()
if __name__ == '__main__':
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]
main(first+second)
得到图像:
图像上是个牛,就是英文的bull
那么第十关的url为:http://www.pythonchallenge.com/pc/return/bull.html
10 第十关
看标题好像是求len(a[30])是多少,查看网页源代码
发现一大串数据,并且发现一个txt文件
点开后发现是一个列表
百度后才知道含义
11— 表示前一个数“1”是 1 个 1;
21— 表示前一个数“11”是 由 2 个 1 组成;
1211— 表示前一个数“21”是 由 1 个 2、1 个 1 组成;
111221— 即 11 12 21 ,表示前一个数“1211”是依次由 1 个 1,1 个 2,
2 个 1组成;
那么后一组的数据就会是312211
于是编写程序求解低30位是什么元素:
def main(): # 申请2个字符串,一个为1,一个为空,a是用来存放数据的 a = '1' sub1 = '' for i in range(30): j = k = 0 while j <len(a): while k < len(a) and a[k] == a[j]: k += 1 sub1 += str(k-j) + a[j] j = k a = sub1 sub1 = '' print(len(a)) if __name__ == '__main__': main()
得到答案:
所以低11关的url是:http://www.pythonchallenge.com/pc/return/5808.html
11 第十一关
查看网页源代码
标题叫做奇偶,应该和图像处理有关,修改制定位置的像素,我们把图片下载到本地,修改奇数位置的像素:
from PIL import Image im = Image.open('cave.jpg') odd = Image.new(im.mode, (im.size[0], im.size[1])) for x in range(1, im.size[0], 2): for y in range(1, im.size[1], 2): odd.putpixel(((x-1),(y-1)),im.getpixel((x, y))) odd.show()
得到结果:
有个单词evil,试了下
则12关的url是:http://www.pythonchallenge.com/pc/return/evil.html
如果要求输入账号密码,还是输入 huge file
12 第十二关
这题的源代码也没有任何提示,估计又是和图像梳理有关
发现图形的url为:http://www.pythonchallenge.com/pc/return/evil1.jpg
试着把evil1改为evil2,出现
图上说not jpg 而是 gfx,于是后缀修改为gfx,自动就把gfx文件下载下来了
将evil2修改为evil3出现
看来关键点就在下载下的evil.gfx文件
evil1图上是把扑克分成了5份,于是我们把evil.gfx文件也分成5份,evil.gfx的第一个字节给第一个文件,第二个字节给第二个文件,第三个字节给第三个文件,第四个字节给第四个文件,第五个字节给第五个文件,第六个字节给第一个文件,第七个字节给第二个文件……
编写代码:
with open('evil2.gfx', 'rb') as f: data = f.read() for i in range(5): with open('evil%d.jpg'%i, 'wb') as f1: f1.write(data[i::5])
得到五张图:
连起来的单词是:disproportional
所以第13关的url是:http://www.pythonchallenge.com/pc/return/disproportional.html
13 第十三关
查看网页源代码
点击php文件,发现提示信息
百度得到提示 是使用xmlrpclib,即xml的远程方法调用(https://www.jb51.net/article/44515.htm)
import xmlrpc.client s = xmlrpc.client.ServerProxy('http://www.pythonchallenge.com/pc/phonebook.php') print(s.system.listMethods())
得到输出:
看到有个phone方法,于是修改代码:
由上题的提示得知,这个恶魔的名字叫Bert,必须在ie浏览器打开
import xmlrpc.client s = xmlrpc.client.ServerProxy('http://www.pythonchallenge.com/pc/phonebook.php') print(s.system.methodHelp('phone')) print(s.phone('Bert'))
得到答案:
第14关的url是:http://www.pythonchallenge.com/pc/return/italy.html
14 第十四关
打开网页源代码
发现提示提多的:
1 walk around
2 remember: 100*100 = (100+99+99+98) + (...
3 有个图片wire.png
from PIL import Image im = Image.open('wire.png') print(im.size)
则它的像素为:(10000, 1)
那么这道题要做的是将10000*1的图像改成100*100的图像,像绕圈圈一样,先向右走100步,然后再向下走99步,再向左走99步,再向上走98步。典型的蛇形矩阵
from PIL import Image im = Image.open('wire.png') im1 = Image.new(im.mode, (100, 100)) # 代表四个方向 right down left up direction = [(1, 0), (0, 1), (-1, 0), (0, -1)] # 初始位置 x , y = -1 ,0 k = 0 # 100*100 = [[100, 99, 99, 98],[98, 97, 97, 96]]...应该有200个元素, steps = 200 while steps/2>0: # 做一圈的迭代 for vector in direction: step = steps//2 for i in range(step): x = x + vector[0] y = y + vector[1] pixel = im.getpixel((k, 0)) im1.putpixel((x, y), pixel) k += 1 steps -= 1 im1.show()
得到一个图像:
是个猫。对应的英文应该是cat.
第14关的url为:http://www.pythonchallenge.com/pc/return/cat.html
输入发现是张图,有提示:
所以第14关的url应该为:http://www.pythonchallenge.com/pc/return/uzi.html
15 第十五关
有一张日历图,仔细看26号这天是被圈出来的
查看网页源代码
收到了2个提示:
1 他不是最年轻的吗,他是第二位的
2 要做的事,为了明天买花
再看日历右下角,二月有29号,所以这一年是闰年。需要我们求出闰年时,1月27号是星期二的日期。
于是编写代码:
import datetime import calendar for year in range(1800, 2000): t = datetime.date(year,1,27) if calendar.isleap(year) and t.weekday() == 1: print(t)
得到结果:
百度出1756-1-27莫扎特出生
所以答案是:mozart
第十六关的url是:http://www.pythonchallenge.com/pc/return/mozart.html
16 第十六关
查看网页源代码:
上面的语句意思是让我们做吧,是句俚语
打印每行像素的值,我们发现每行像素都有连续5个195的值
我们以195为基准位置,反转一下色素,得到新的图片
编写程序:
from PIL import Image im = Image.open('mozart.gif') widt, heig = im.size for y in range(heig): line = [im.getpixel((x,y)) for x in range(widt)] idx = line.index(195) lines = line[idx:] + line[:idx] for x in range(widt): im.putpixel((x, y), lines[x]) im.show()
得到图片:
第17关的url为:http://www.pythonchallenge.com/pc/return/romance.html