这一篇来闯关10-15.感觉这几关比先前的难了不少,有的题目完全没思路.
页面源码中的链接点击后有a = [1, 11, 21, 1211, 111221,
google之后知道这是一个叫做http://en.wikipedia.org/wiki/Look-and-say_sequence的数列.下面是一种比较简洁的解法.
import re
x="1"
for each in range(5):
print(x)
l = re.findall(r"(\d)(\1*)", x)
print(l)
x="".join([str(len(i+j))+i for i,j in re.findall(r"(\d)(\1*)", x)])
#re.findall()返回一个元祖的list,元祖的元素个数为表达式中group的个数
#re中\number表示匹配group(number)匹配到的内容,比如说(\d)(\1)如果(\d)匹配的是2,那么这里的\1则要匹配数字2
#如果(\d)匹配到的是1,那么这里的\1则要匹配数字1
#print(x)
print(len(x))
注意一下正则表达式中\number的意思.我看完文档后写了好几个例子才搞清楚是什么意思.
>>> re.findall(r‘(\d)(\w)\1‘,"1a1234")
[(‘1‘, ‘a‘)]
>>> re.findall(r‘(\d)(\w)\1‘,"2a1234")
[]
>>> re.findall(r‘(\d)(\w+)\1‘,"2a1234")
[(‘2‘, ‘a1‘)]
根据提示 “odd even"猜想取出坐标(x，y)一个为奇数一个为偶数的像素点.按这个思路写完之后发现不对,换个思路,将原图中坐标(x，y)一个为奇数一个为偶数的像素点去除(将该点像素设成黑色(0，0，0)).得到答案。
import urllib.request as ur
from PIL import Image
def main():
#需要用户名密码访问url
url = "http://www.pythonchallenge.com/pc/return/cave.jpg"
username = ‘huge‘
password = ‘file‘
p = ur.HTTPPasswordMgrWithDefaultRealm()
p.add_password(None, url, username, password)
handler = ur.HTTPBasicAuthHandler(p)
opener = ur.build_opener(handler)
ur.install_opener(opener)
dataBytes = ur.urlopen(url).read()
f = open("11_cave.jpg","wb")
f.write(dataBytes)
im = Image.open("11_cave.jpg")
w,h = im.size
mode = im.mode
bands = im.getbands()
print(w,h)
print(type(mode),mode)
print(type(bands),bands)
#imAnswer = Image.new(mode,(w,h),(255,255,255))
for i in range(w):
for j in range(h):
if (i + j)%2 == 1:
pos = (i,j)
im.putpixel(pos,0)
im.save(‘11answer.jpg‘)
if __name__ == ‘__main__‘:
main()
完全没有思路的一道题.搜索之后知道是有5张图片,二进制的数据被写到了一个叫做evil2.gfx的文件中,需要像挨个发牌一样一字节一字节的重新"发回到"5个图片文件....
太考验发散思维了...
import urllib.request as ur
def main():
#需要用户名密码访问url
url = "http://www.pythonchallenge.com/pc/return/evil2.gfx"
username = ‘huge‘
password = ‘file‘
p = ur.HTTPPasswordMgrWithDefaultRealm()
p.add_password(None, url, username, password)
handler = ur.HTTPBasicAuthHandler(p)
opener = ur.build_opener(handler)
ur.install_opener(opener)
dataBytes = ur.urlopen(url).read()
f = open("12_evil2.gfx","wb")
f.write(dataBytes)
print(len(dataBytes))
types = [‘jpg‘,‘png‘,‘gif‘,‘png‘,‘jpg‘]
for i in range(5):
imagedata = dataBytes[i::5]
image = open(‘12answer_%d.%s‘ % (i,types[i]),‘wb‘)
image.write(imagedata)
if __name__ == ‘__main__‘:
main()
执行代码后,有5张图片,前面4张字母连起来就是答案.用二进制的方式打开图片文件.可以看到.jpg文件的首字节0xff,.png文件的首字节为0x89,猜测应该是首字节或者是前几字节
代表图片的格式的字段,没有去深究.
页面源码中有一个链接指向phonebook.php,点击进入以后出现提示:This XML file does not appear to have any style information associated with it. The document tree is shown below.google之后知道是少了某些描述文件,可以简单理解为类似于html中缺少了css,所以浏览器就无法正常展示格式吧.然而还是不知道要如何继续下去,去标准库中查找xml相关的内容,还是没啥思路.继续google页面中的节点名称<methodResponse>,找到了xmlrpc---xml远程调用,到这里问题就简单了.代码如下:
import xmlrpc.client
def main():
x = xmlrpc.client.ServerProxy("http://www.pythonchallenge.com/pc/phonebook.php")
print(x.system.listMethods())
print(x.phone(‘Bert‘))
if __name__ == ‘__main__‘:
main()
XML-RPC is a Remote Procedure Call method that uses XML passed via HTTP as a transport. With it, a client can call methods with parameters on a remote server (the server is named by a URI) and get back structured data.类似于webservice,server端实现了一些功能,并提供一些接口给客户端使用.
这一题没啥思路,google了之后知道是要将下面的那个size为10000*1的条形码像上面的那个便便一样的面包一样不断旋转,最终得到一个100*100的
图.有了思路之后就好办了,我的代码如下：
import urllib.request as ur
from PIL import Image
l = [[i, i-1, i-1, i-2] for i in range(100, 0, -2)]
#print(l)
def main():
im = Image.open("wire.png")
#print(im.size)
(width,height) = im.size
mode = im.mode
#pos = [(x,y) in range(width,height)]
#print(pos)
#print(width,height)
#print(im.getpixel((639,486)))
x,y=0,0
imAnswer = Image.new(mode,(100,100),(255,255,255))
#print(imAnswer.size)
#左上角为坐标(0,0),左下角为坐标(0,99),右上角坐标为(99,0),右下角坐标为(99,99)
#im.getpixel((i,0))
#imAnswer.putpixel((0,0),(255,0,0))
#imAnswer.putpixel((99,0),(0,255,0))
#imAnswer.putpixel((0,99),(0,0,255))
#imAnswer.putpixel((99,99),(255,255,255))
#imAnswer.save(‘14Answer.jpg‘)
lastPos=(-1,0) #上一个坐标位置
maxStep=100 #当前最多走到多少步需要变向
currentSteps=0 #当前方向已经走了多少步
currentDirection=1 #1:向右,2:向下,3:向左,4向上
for i in range(0,10000):
pixel = im.getpixel((i,0))
if currentDirection == 1:
pos = (lastPos[0] + 1,lastPos[1])
if currentDirection == 2:
pos = (lastPos[0],lastPos[1] + 1)
print(lastPos,pos)
if currentDirection == 3:
pos = (lastPos[0] - 1,lastPos[1])
if currentDirection == 4:
pos = (lastPos[0],lastPos[1] - 1)
#print(currentDirection,lastPos,pos)
try:
imAnswer.putpixel(pos,pixel)
except:
pass
#print(pos)
#更新点的位置信息
lastPos = pos
currentSteps += 1
#处理到达边界的情况
if (currentSteps == maxStep):
if currentDirection == 1:
currentDirection = 2
currentSteps = 0
maxStep -= 1
#print("turn to down!",currentDirection)
elif currentDirection == 2:
currentDirection = 3
currentSteps = 0
elif currentDirection == 3:
currentDirection = 4
currentSteps = 0
maxStep -= 1
elif currentDirection == 4:
currentDirection = 1
currentSteps = 0
imAnswer.save(‘14Answer.jpg‘)
if __name__ == ‘__main__‘:
main()
执行完后得到一张猫的图片,输入http://www.pythonchallenge.com/pc/return/cat.html得到提示:and its name is uzi. you‘ll hear from him later.把cat换成uzi即是下一关地址.
这一题比较简单
标题是whom?图片是一个日历.应该是找出某个日期的某个人.图片中右下角的图仔细看可以看到2月29日,说明是闰年.
import calendar
from datetime import date
yearlist = []
for x in range(0,10):
for y in range(0,10):
year = 1000 + x*100 + y*10 + 6
if calendar.isleap(year):
yearlist.append(year)
#print(yearlist)
for year in yearlist:
d = date(year,1,1)
if d.weekday() == 3: #周一到周日:0-6
print(year)
#第二年轻的,是1756年.google 1756年1月26日 第二天是莫扎特诞生