61 lines
2.0 KiB
Python
61 lines
2.0 KiB
Python
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# -*- coding:utf-8 -*-
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# @Author len
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# @Create 2023/12/12 19:49
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def if_formula(text):
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# 允许的字符：数字、运算符、括号、字母和空格
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allowed_chars = "0123456789+-*/(). abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ%^"
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operators = "+-*/%^"
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digits = "0123456789"
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letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
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# 检查字符是否合法
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for char in text:
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if char not in allowed_chars:
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return False
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# 括号匹配检查
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bracket_count = 0
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for char in text:
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if char == '(':
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bracket_count += 1
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elif char == ')':
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bracket_count -= 1
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if bracket_count < 0: # 右括号多于左括号
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return False
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if bracket_count != 0: # 左括号和右括号数量不匹配
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return False
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# 去除空格，简化后续检查
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s = text.replace(' ', '')
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# 检查运算符、数字和字母的顺序
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if s and (s[0] in operators or s[-1] in operators):
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return False
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for i in range(1, len(s) - 1):
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if s[i] in operators and s[i + 1] in operators:
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# 防止两个运算符连续出现
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if not (s[i] in '-+' and s[i + 1] in '-+' and (s[i - 1] in operators or s[i - 1] == '(')):
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return False
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elif s[i] in digits and s[i + 1] in letters:
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# 防止数字紧跟字母
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return False
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elif s[i] in letters and s[i + 1] in digits:
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# 防止字母紧跟数字
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return False
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elif s[i] in letters and s[i + 1] in letters:
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# 防止连续的字母
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return False
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elif s[i] in '()' and s[i + 1] in letters and s[i] == ')':
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# 防止字母紧跟闭括号
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return False
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return True
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list = ["124141", "AAAAAA", "AFF3241", "fafdafsfas",'A13C46', "124!@$%$%#@@#$%#@!@#$%#@!", "fsd65", "A1/2B%34", "((n*y+h)^4)/100"]
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for i in list:
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if if_formula(i):
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print(i, "是")
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else:
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print(i, "不是")
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