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#练习01：给定一个不超过5位的正整数,判断其有几位,使用input函数
#code01: 列出每种情况
#!~/.pyenv/shims/python
num = int(input("please input 5bit num: "))
if num >= 0 and num <= 9:
print("The num is 1 bit")
elif num > 9 and num <= 99:
print("The num is 2 bit")
elif num >99 and num <=999:
print("The num is 3 bit")
elif num > 999 and num <= 9999:
print("The num is 4 bit")
elif num > 9999 and num <= 99999:
print("The num is 5 bit")
else:
print("input error, please enter input!")
#code02: 将最坏的情况放在最前面,防止无用的数据进行判断
num = int(input("please input 5bit num: "))
if num < 0:
print("input error, please enter input!")
else:
if num >= 0 and num <= 9:
print("The num is 1 bit")
elif num > 9 and num <= 99:
print("The num is 2 bit")
elif num >99 and num <=999:
print("The num is 3 bit")
elif num > 999 and num <= 9999:
print("The num is 4 bit")
elif num > 9999 and num <= 99999:
print("The num is 5 bit")
#code03: 二分法判断，嵌套用法，解决效率问题
num = int(input("please input 5bit num: "))
if num < 0:
print("input error, please enter input!")
else:
if num > 99999:
print("input error, please enter input!")
else:
if num >= 100:
if num >= 100 and num < 1000:
print("The num is 3 bit")
elif num >= 1000 and num < 10000:
print("The num is 4 bit")
else:
print("The num is 5 bit")
else:
if num >= 10:
print("The num is 2 bit")
else:
print("The num is 1 bit")