By Python6 ways: given an integer of more than 5, it is determined by several

Method One: compare

a=int(input(">>>>"))
if a<10:            
    print(1)
elif a<100:   #第一个条件已经过滤了大于9，所以这里区间是11到100
    print(2)
elif a<1000:
    print(3)
elif a<10000:
    print(4)
else:
    print(5)

Method Two: Use divisible achieve, except after if was not a 0 or 0, this method is introduced to calculate the efficiency will be reduced, it is possible to add, do not subtract, multiply it can not in addition to, not counting can not be calculated

'''
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'''
i = int(intput('>>>')
if i // 10000:
    print(5):
elif i // 1000:
    print(4)
elif i // 100:
    print(3)
elif i // 10:
    print(2)
else:
    print(1)

Analysis: The assumption is that in five cases, regardless of other conditions

In [1]: 6666 // 10000
Out[1]: 0                除以10000为零证明是小于5位数

In [2]: 6666 // 1000
Out[2]: 6                但是如果能被1000整除，它就是一个4位数

In [3]: 6666 // 100
Out[3]: 66

In [4]: 6666 // 10
Out[4]: 666

In [5]: 6666 // 1
Out[5]: 6666

Method three:

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'''
a=int(input(">>>"))
if a<0:
    print("Format is wrong")
elif a<100000:    ##限定5位
    if a<10:
        print(1)
    elif a<100:
        print(2)
    elif a<1000:
        print(3)
    elif a<10000:
        print(4)
    else:  
        print(5)
else:
    print("请输入一个不超过5位的数")

Method four: string processing implemented

#!/usr/bin/python3
nnumber=input(">>>>")
length=len(nnumber)
if length>4:
    print(5)
elif length>3:
    print(4)
elif length>2:
    print(3)
elif length>1:
    print(2)
else:
    print(1)

Method five: binary realization

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'''
#!/usr/bin/python3
number = int(input("number >> "))
if number >= 100:       ##直接从100开始折
    if number >= 10000:
        print("5")
    elif number >= 1000:
        print("4")
    else:
        print("3")
else:
    if number >= 10:
        print("2")
    else:
        print("1")

Method six: math realized, this method is slower than the division, if the cycle 100 million times quite clear

number=int(input("输入一个不超过5位的正整数： ")
if a<=0 or a>=100000:
    print('请输入一个不超过5位的正整数')
else:
    import math
    b=int(math.log10(a)+1)
    print(b)