[pythonchallenge]-- explanation of 0~4 levels

I found an exercise today python Good project , It's called pythonchallenge, Through the form of game level , We can learn better Python！！

Hzy The blog of

Today, let's talk about the front 5 Turn off , What have you met .

The first 0 Turn off

See the picture , Ask us to find out 2 Of 38 Power

print(2**38)# 274877906944

so easy, We copied the results to .html You can go to the next level .

The first 1 Turn off

It's also a picture , You can see

• K -> M
• O-> Q
• E -> G

Observe the pattern , It seems that they all press 26 Letters , The result of counting two digits forward , Here is another paragraph , Let's translate it with our rules .

## The following text
instr = "abcdefghijklmnopqrstuvwxyz"
outstr = "cdefghijklmnopqrstuvwxyzab"
strans = str.maketrans(instr,outstr)
test = "g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp. bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj."
print(test.translate(strans))
# http://www.pythonchallenge.com/pc/def/map.html
print("map".translate(strans))
# The next link is ：http://www.pythonchallenge.com/pc/def/ocr.html

The first 2 Turn off

The second level is a book , Let's right click , Open the page source code to view , I found a comment below , Ask us to find in a pile of garbled code Key letters .

import requests
import re
content = requests.get('http://www.pythonchallenge.com/pc/def/ocr.html').text
anwser = re.sub('\n|\t|\r', '', content) # Remove some spaces
anwser = re.findall("find.*?<!--(.*?)-->", anwser) # Take out the notes
anwser = list(str(anwser)) # Convert to list
key = (s for s in anwser if s.isalpha()) # Find the letters inside
print("".join(key))
# result equality

The first 3 Turn off

The first 3 Guan Hedi 2 Close very much , We directly open the page source code .

• According to the requirements of the topic , There are three capital letters around a small letter
• We use regular ’[a-z]+[A-Z]{3}([a-z])[A-Z]{3}[a-z]+' matching

import re
import requests
content = requests.get('http://www.pythonchallenge.com/pc/def/equality.html').text
notes = re.search('<!--(.*?)-->', re.sub('\n', '', content))
if notes:
ret = re.findall('[a-z]+[A-Z]{3}([a-z])[A-Z]{3}[a-z]+', notes.group())
print(ret)
# answer ：linkedlist

The first 4 Turn off

This question , There is also a graph , Then we can click in , Found only a string of characters

• and the next nothing is 44827
• Right click on the source , The notes tell us , By always looking for the next number , We are the most 400 The answer can be found once .

Then in the process of writing, you will find the following problems

• Yes. Divide by two and keep going. You need to divide the number by 2, Instead of using regular direct matching

• There maybe misleading numbers in the text. One example is 82683. Look only for the next nothing and the next nothing is 63579

• Regular will match two numbers , You have to decide which number you need .

import requests
import re
url = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=77864"
text = requests.get(url).text
for i in range(400):
number_list = re.findall("\d+", text)
if len(number_list) == 1:
url = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=" + number_list[0]
elif len(number_list) == 0:
number = url.split('nothing=')[1]
url = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=" + str(int(number) / 2)
elif len(number_list) == 2:
url = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=" + number_list[0]
content = requests.get(url).text
if content.find("You've been misleaded to here.") != -1:
print(" The correct number is {}".format(number_list[1]))
url = "http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=" + number_list[1]
r = requests.get(url)
if r.status_code == 200:
text = r.text
print(str(i) + ": ", text)
# http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=66831 Your next answer ：peak.html

Come back tomorrow and take over the topic in ！