Question

我有一个小请求。我在学校有一个分配，我需要通过nc连接到服务器，然后猜测号码以便找到秘密消息。我有这个服务器的源代码，我上周试图破解这个。如果你能指出我的方向是正确的那么好。

#!/usr/bin/python

import socket
import os, sys
import ast
import random

def write(msg):
  sys.stdout.write(msg)
  sys.stdout.flush()

def read(prompt=''):
  if prompt:
    write(prompt)
  return sys.stdin.readline().strip()

def loop():
  RANGE = 1000000
  num = read('Select a number (0, 10000000000000000000000): ')
  try:
    num = ast.literal_eval(num) + 0
    assert 0 < num < 10000000000000000000000
  except:
    write("Oops! Wrong number!\n\n")
    return
  rand = num + int(random.random()*RANGE)
  guess = read('Guess a number in range: [{:n}, {:n}]: '.format(num, num+RANGE))
  try:
    guess = ast.literal_eval(guess)
  except:
    write("Oops! You probably didn't enter a number!\n\n")
    return
  if guess == rand:
    write('Wow! The secret is [...].\n')
    sys.exit(0)
  write('Better luck next time!\n\n')  

def main():
  write('Hello there in "Guess the number!"\nYou have 30 seconds.\n')
  while True:
    loop()

main()

Answer 1

关键是ast.literal_eval。它允许输入浮点数。浮点数的精度为<16位。输入0.999999999999999e22后，您也会在求和后0.999999999999999e22得到最终结果，因此您可以输入两次以进入。

示例会话：

Hello there in "Guess the number!"
You have 30 seconds.
Select a number in range (0, 10000000000000000000000): 0.999999999999999e22
Now guess a number in range: [1e+22, 1e+22]: 0.999999999999999e22
Wow! The secret is [...].

原因是

>>> 0.999999999999999e22 + 1000000 == 0.999999999999999e+22
True

然而令人惊讶的是，0.9e22可能会失败，但是也不会花费很多时间：

>>> 0.9e22 + 999999 == 0.9e22
False

试图破解猜测服务器

1 个答案: