Codeforces Round 555 (Div. 3) |
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Finished |
Let's denote a function $$$f(x)$$$ in such a way: we add $$$1$$$ to $$$x$$$, then, while there is at least one trailing zero in the resulting number, we remove that zero. For example,
We say that some number $$$y$$$ is reachable from $$$x$$$ if we can apply function $$$f$$$ to $$$x$$$ some (possibly zero) times so that we get $$$y$$$ as a result. For example, $$$102$$$ is reachable from $$$10098$$$ because $$$f(f(f(10098))) = f(f(10099)) = f(101) = 102$$$; and any number is reachable from itself.
You are given a number $$$n$$$; your task is to count how many different numbers are reachable from $$$n$$$.
The first line contains one integer $$$n$$$ ($$$1 \le n \le 10^9$$$).
Print one integer: the number of different numbers that are reachable from $$$n$$$.
1098
20
10
19
The numbers that are reachable from $$$1098$$$ are:
$$$1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1098, 1099$$$.
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