hax01tips
hax01tips
注意:本文之后极有可能设为私有
题目
hax01
Your mission is the following: Simply enter a URL into the box. The domain of the URL must be or end with 'nasa.gov'. The URL will be fetched right away. The content returned should contain the string: "2200178118" in the first 10 Kbytes of data. 404/403/etc error pages are not accepted. Remember, do not do anything illegal. Make sure you type the right URL, do not guess.
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Hint: google is your friend.
http://google.com/search?q=site:nasa.gov
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当时我的思路是找出所有以nasa.gov结尾的域名,然后遍历这些网址。之后我真的写了个python程序,取了google检索出来的前1000个页面,取出域名,保存起来,去除重复的有500多个。接着,读取html页面,判断是否还有字符串。其间,遇到了个网速的问题,超时后经常会跑到电信的114搜索上去。验证了170多个页面后,我发现自己理解错题目了,这里的URL并不是指URL以nasa.gov结尾,而是指URL的域名以nasa.gov结尾。我无语了,这相当于域名下的所有网页都有可能。这个工作量巨大得几乎是不可能的。暂时中止。
以下代码可供参考,修改了n次,可能现在已经没法直接运行。
2.5
1
from urllib import FancyURLopener
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import urllib2
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import sys
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import re
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import locale
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"""
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class MyOpener(FancyURLopener):
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version = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; it; rv:1.8.1.11)Gecko/20071127 Firefox/2.0.0.11'
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res = re.compile(r'(([a-zA-Z]+\.)+nasa.gov)')
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myopener = MyOpener()
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url = 'http://www.google.co.jp/search?&num=100&as_qdr=all&as_occt=any&as_sitesearch=nasa.gov'
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li = []
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for i in range(0, 10):
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url = url + '&start=' + str(i*100)
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page = myopener.open(url)
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str1 = page.read()
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for aItem in res.findall(str1):
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if not aItem[0] in li:
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li.append(aItem[0])
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"""
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with open('nasa.txt') as li:
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#li = open('nasa.txt')
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#print li.count
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m = 0
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for a in li:
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#print 'http://'+a
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m = m + 1
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print m
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url = a
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req = urllib2.Request(url)
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try:
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response = urllib2.urlopen(req)
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the_page = response.read()
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with open(url + '.txt') as nasa:
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write(the_page)
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if the_page.find(r'daohang.118114.cn') <> -1 :
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print '114'
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elif the_page.find('2200178118', 0, 10240) <> -1 :
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print url
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else :
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print '
'
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except urllib2.URLError, e:
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print e.reason
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"""
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#gUrl = 'http://www.google.co.jp/search?hl=ja&source=hp&q=site%3Anasa.gov&lr=&aq=f&oq='
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#google = urllib.urlopen(gUrl)
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#str = google.read()
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for str in open('sitenasa_gov.htm'):
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for aItem in res.findall(str):
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print aItem[0]
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#print str
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str = 'www.xxx.nasa.gov/wwf.nasa.gov'
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"""
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61
#2200178118
62
3.1
1
from urllib.request import FancyURLopener
2
import urllib
3
import sys
4
import re
5
import locale
6
"""
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class MyOpener(FancyURLopener):
8
version = 'Mozilla/5.0 (Windows; U; Windows NT 5.1; it; rv:1.8.1.11)Gecko/20071127 Firefox/2.0.0.11'
9![]()
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res = re.compile(r'(([a-zA-Z]+\.)+nasa.gov)')
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myopener = MyOpener()
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url = 'http://www.google.co.jp/search?&num=100&as_qdr=all&as_occt=any&as_sitesearch=nasa.gov'
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li = []
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for i in range(0, 10):
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url = url + '&start=' + str(i*100)
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page = myopener.open(url)
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str1 = page.read()
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for aItem in res.findall(str1):
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if not aItem[0] in li:
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li.append(aItem[0])
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"""
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fiPath = sys.argv[1]
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with open(fiPath) as li:
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#li = open('nasa.txt')
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#print li.count
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m = 0
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for a in li:
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#print 'http://'+a
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m = m + 1
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#print m
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url = a
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req = urllib.request.Request(url)
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try:
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response = urllib.request.urlopen(req)
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the_page = response.read()
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with open(url[7:-1] + '.txt', 'wb') as nasa:
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nasa.write(the_page)
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nasa.flush()
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if the_page.decode('utf8').find(r'icc.qonc.com') != -1:
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print('114')
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elif the_page.decode('utf8').find('2200178118', 0, 10240) != -1:
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print(url)
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else :
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print('
')
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except urllib.error.URLError as e:
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print(e.code)
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except UnicodeDecodeError as UDE:
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print(UDE)
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"""
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#gUrl = 'http://www.google.co.jp/search?hl=ja&source=hp&q=site%3Anasa.gov&lr=&aq=f&oq='
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#google = urllib.urlopen(gUrl)
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#str = google.read()
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for str in open('sitenasa_gov.htm'):
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for aItem in res.findall(str):
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print aItem[0]
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#print str
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str = 'www.xxx.nasa.gov/wwf.nasa.gov'
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"""
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#2200178118
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而后,过了大概几个月,变换思路,解决,意外的简单……
事实上,只要向服务器提交数据,一般服务器也会将该数据返回到页面上。该题最后的hint不是让我们来搜该域名,而是告诉我们怎样在google.com的页面上显示我们想要的数据。譬如 http://www.google.co.jp/search?q=2200178118 该页面的前10K里应该包含了该字符串。接下来,我们只需要在nasa.gov上找个页面提交数据就行了。
over