```
S = input()
count = 0
for i in range(len(S)-1):
if S[i] != S[i+1]:
count += 1
print((count + 1) // 2)
```

This one and

```
S = int(input())
count = 0
for i in range(len(S)-1):
if S[i] != S[i+1]:
count += 1
print((count + 1) // 2)
```

This phrase and why doesn't it work at the bottom, s is an integer value

python

2022-09-20 10:47

The problem is, perhaps, asking about the **each digit** of any natural number.

If you treat that natural number as a string, the code becomes really simple.

But if you really treat that natural number as a natural number and solve the problem, it gets complicated.

That's why your code doesn't work. At least this much should be done so that `S`

can be used as an integer.

```
# Not tested. Just look at how cumbersome and troublesome it is to take each digit of an integer.
S = int(input())
max10sq = 1
while S > max10sq :
max10sq *= 10
max10sq /= 10
count = 0
while S > 0 :
greaterDigit = ceil(S / max10sq)
S %= max10sq
smallerDigit = ceil(S / max10sq)
max10sq /= 10
if greaterDigit != smallerDigit :
count += 1
```

I don't know if that's the answer.

2022-09-20 10:47

input returns the str object. You have to provide an int object, but it doesn't work, so it works differently.

```
a = input()
print(type(a)
print(type(int(a))
```

2022-09-20 10:47

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