import re#乘除运算
def numl_mod(args):resurt  = re.compile('(\d+\.?\d*)([*/])(\-?\d+\.?\d*)')#正则匹配,将匹配出来的结果赋给resurtwhile resurt.search(args):#while循环判断直到算出最终结果new_num = resurt.search(args)num = new_num.group()#取出原值赋给numnum1 = float(new_num.group(1))num2 = new_num.group(2)#取出来的是 * / 号，在if 处进行判断num3 = float(new_num.group(3))if num2 == '*':result = num1 * num3#返回 * 的结果elif num2 == '/':result = num1 / num3#返回 / 的结果args = args.replace(num, str(result), 1)#将计算的结果替换返回return args
# print(numl_mod("10*2/5"))# #加减运算
def poor_add(args):resurt = re.compile('(\-?\d+\.?\d*)([-+])(\-?\d+\.?\d*)')while resurt.search(args):new_num = resurt.search(args)num = new_num.group()num1 = float(new_num.group(1))num2 = new_num.group(2)num3 = float(new_num.group(3))if num2 == '-':result = num1 - num3elif num2 == '+':result = num1 + num3args = args.replace(num, str(result), 1)return args# print(poor_add("3-5+6"))#检查判断去括号# origin = "1-2*((60-30+(-9-2-5-2*3-5/3-40*4/2-3/5+6*3)*(-9-2-2*5/3+7/3*99/4*2998+10*568/14))-(-4*3)/(16-3*25))"
# origin = input("Enter number >>>>>")
#检测式子的合法性
def check(args):flag = True#默认标识位为真if re.findall("[a-zA-Z]",args):#检查式子的合法性，不能含有大小写字母print("error")flag = Falsereturn flag#检测运算
def format(args):#检测式子，将 含有空格/++/+-/--等进行替换，返回一个新的式子new_args1 = args.replace("++", "+")new_args2 = new_args1.replace("+-", "-")new_args3 = new_args2.replace("--", "+")new_args4 = new_args3.replace("-+", "-")new_args5 = new_args4.replace(" ", "")new_args6 = new_args5.replace("*+","*")new_args = new_args6.replace("/+","/")return new_argsprint("--------------")
print("欢迎使用计算机")
print("--------------")
#主代码块
#执行语句块
while True:choice = input("请选择 q/Q【退出】 y/Y 【执行】")if choice == "q" or choice == "Q":print("谢谢您的使用")breakorigin = input("Enter number >>>>>")if check(origin):new_origin = format(origin)while re.search("\(",new_origin):#匹配如果有括号，执行下面代码new_new_origin = re.search("\([^()]+\)",new_origin).group()#匹配出最里面的括号里的式子# print(new_new_origin)s = numl_mod(new_new_origin)# print(s)s1 = poor_add(s)s2 = s1[1:-1]new_origin = new_origin.replace(new_new_origin, s2, 1)# print(new_origin)# print(s2, type(s2))# breakelse:#括号执行完后执行的代码块resurt = numl_mod(new_origin)# print(resurt, "resurt")# print(resurt)new_resurt = poor_add(resurt)# a2 = a2[1:-1]print(new_resurt)#返回的最终结果，也就是我们要得到的最终计算结果