题解 | #简单计算器#
简单计算器
http://www.nowcoder.com/practice/b8f770674ba7468bb0a0efcc2aa3a239
python3 直白解放，其他实在想不出来了。
s = input()
for i in range(len(s)):
if s[i] in ("+","-","*","/"):
re = s.split(s[i])
if s[i] == "+":
pls = float(re[0])+float(re[1])
print("%.4f"%float(re[0])+s[i]+"%.4f"%float(re[1])+"="+"%.4f"%pls)
elif s[i] == "-":
pls = float(re[0]) - float(re[1])
print("%.4f" % float(re[0]) + s[i] + "%.4f" % float(re[1]) + "=" + "%.4f" % pls)
elif s[i] == "*":
pls = float(re[0]) * float(re[1])
print("%.4f" % float(re[0]) + s[i] + "%.4f" % float(re[1]) + "=" + "%.4f" % pls)
elif s[i] == "/":
if re[1] == "0.0":
print("Wrong!Division by zero!")
else:
pls = float(re[0]) / float(re[1])
print("%.4f" % float(re[0]) + s[i] + "%.4f" % float(re[1]) + "=" + "%.4f" % pls)
else:
if s[i] not in ("+","-","*","/","0","1","2","3","4","5","6","7","8","9","."):
print("Invalid operation!")
break
学历配不上这个行业
，专升本性价比比考研高
搞这么长干什么！还有你面试要先针对那个公司的技术栈专门去准备，别一问三不知，他需要你会他们公司对口的技术，写这么多不对口没有用