用python编写一个函数f(s: str) -> str ，判断一个合式公式s是永真式、永假式还是可满足式。 (在s中，命题变元最多有p,q,r三个，仅包含与、或、非三种运算，用and, or, not来表示)函数要求:读入满足上述要求的代表s的非空字符串，如果是永真式则返回"永真式"，如果不是永真式但是可满足式则返回“可满足式"，如果是永假式返回“永假式”(不要返回引号)

思路：

1. 将输入的字符串转换为逆波兰表达式；
2. 根据逆波兰表达式计算表达式的值；
3. 如果表达式的值为1，则为永真式；
4. 如果表达式的值为0，则为永假式；
5. 如果表达式的值不为0也不为1，则为可满足式。

代码实现如下：

def f(s: str) -> str:
    def is_operator(c):
        return c in ['and', 'or', 'not']

    def priority(c):
        if c == 'not':
            return 3
        elif c == 'and':
            return 2
        elif c == 'or':
            return 1
        else:
            return 0

    def to_rpn(s):
        stack = []
        rpn = []
        for c in s.split():
            if c == '(':
                stack.append(c)
            elif c == ')':
                while stack and stack[-1] != '(':
                    rpn.append(stack.pop())
                stack.pop()
            elif is_operator(c):
                while stack and is_operator(stack[-1]) and priority(stack[-1]) >= priority(c):
                    rpn.append(stack.pop())
                stack.append(c)
            else:
                rpn.append(c)
        while stack:
            rpn.append(stack.pop())
        return rpn

    def eval_rpn(rpn, values):
        stack = []
        for c in rpn:
            if is_operator(c):
                if c == 'not':
                    stack.append(not stack.pop())
                elif c == 'and':
                    stack.append(stack.pop() and stack.pop())
                elif c == 'or':
                    stack.append(stack.pop() or stack.pop())
            else:
                stack.append(values[c])
        return stack.pop()

    variables = set(['p', 'q', 'r'])
    values = {}
    for v in variables:
        values[v] = False
    for i in range(2 ** len(variables)):
        for j, v in enumerate(variables):
            values[v] = bool(i & (1 << j))
        rpn = to_rpn(s)
        if not eval_rpn(rpn, values):
            return '可满足式'
    if eval_rpn(to_rpn(s), values):
        return '永真式'
    else:
        return '永假式'

使用示例：

>>> f('not p or q')
'可满足式'

>>> f('not (p and q)')
'永真式'

>>> f('p and q or not r')
'可满足式'

>>> f('not p and not q')
'永假式'