python 使用特定函数的加密程序

Encryption program with python using a specific function

我正在尝试编写一个 python 程序,该程序将使用用户也给出的方程式来加密用户的输入消息。这将是我的第一个程序("hello world" 和 "number guesser" 除外)。我会尝试更具体。我认为要做的第一件事是创建一个列表并写下所有字母及其对应的数字。要求输入字符串,并要求用户写出像 3x+2 这样的等式。我想要做的是,对于字符串输入的每个字母,从列表中找到相应的数字并使用等式产生另一个输出。我想我应该拆分要加密的消息并找到每个数字值,但我如何让 python 将这些数字放入等式中?感谢您的帮助

您可以在 for-loop 中转到 char-by-char 并使用 ord() 获取 ascii-character 的序号。要计算算术表达式,您可以使用 ast(请参阅 Evaluating a mathematical expression in a string)。在此之前,您需要将 "x" 替换为之前收集的序号。

这是一个最简单的例子,希望能做到这一点。

用法:

$ python mycrypt.py "Stack Overflow is Awesome!" "3x+2"
:251:350:293:299:323:98:239:356:305:344:308:326:335:359:98:317:347:98:197:359:305:347:335:329:305:101:

代码:

#!/usr/bin/python
import sys
import ast
import operator as op

# supported operators
operators = {ast.Add: op.add, ast.Sub: op.sub, ast.Mult: op.mul,
             ast.Div: op.truediv, ast.Pow: op.pow, ast.BitXor: op.xor,
             ast.USub: op.neg}

def eval_expr(expr):
    """
    >>> eval_expr('2^6')
    4
    >>> eval_expr('2**6')
    64
    >>> eval_expr('1 + 2*3**(4^5) / (6 + -7)')
    -5.0
    """
    return eval_(ast.parse(expr, mode='eval').body)

def eval_(node):
    if isinstance(node, ast.Num): # <number>
        return node.n
    elif isinstance(node, ast.BinOp): # <left> <operator> <right>
        return operators[type(node.op)](eval_(node.left), eval_(node.right))
    elif isinstance(node, ast.UnaryOp): # <operator> <operand> e.g., -1
        return operators[type(node.op)](eval_(node.operand))
    else:
        raise TypeError(node)

# Your first argument (e.g. "Stack Overflow is Awesome!")
text = sys.argv[1]
# Your second argument (e.g. "3x+2")
calc = sys.argv[2]

# Print beginning
sys.stdout.write(":")   
# For each character in text do your calculation                
for c in text:
    # Generate expression by replacing x  with ascii ordinal number 
    #  prefixed with a multiplicator. 
    # -> character "a" has ordinal number "97"
    # -> expression "3x+2" will be changed to "3*97+2"
    #  likewise the expression "3*x+2" will be changed to "3**97+2" 
    #  which might not be the expected result.
    expr = calc.replace("x", "*" + str(ord(c))) 
    # Evaluate expression
    out = eval_expr(expr)
    # Print cipher with trailing colon and without newline                          
    sys.stdout.write(str(out) + ":")
# Print ending newline
print("")