A13884 | Strange Functions
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题目描述
Let's define a function $f(x)$ ( $x$ is a positive integer) as follows: write all digits of the decimal representation of $x$ backwards, then get rid of the leading zeroes. For example, $f(321) = 123$ , $f(120) = 21$ , $f(1000000) = 1$ , $f(111) = 111$ .
Let's define another function $g(x) = \dfrac{x}{f(f(x))}$ ( $x$ is a positive integer as well).
Your task is the following: for the given positive integer $n$ , calculate the number of different values of $g(x)$ among all numbers $x$ such that $1 \le x \le n$ .
Let's define another function $g(x) = \dfrac{x}{f(f(x))}$ ( $x$ is a positive integer as well).
Your task is the following: for the given positive integer $n$ , calculate the number of different values of $g(x)$ among all numbers $x$ such that $1 \le x \le n$ .
输入格式
The first line contains one integer $t$ ( $1 \le t \le 100$ ) — the number of test cases.
Each test case consists of one line containing one integer $n$ ( $1 \le n < 10^{100}$ ). This integer is given without leading zeroes.
Each test case consists of one line containing one integer $n$ ( $1 \le n < 10^{100}$ ). This integer is given without leading zeroes.
输出格式
For each test case, print one integer — the number of different values of the function $g(x)$ , if $x$ can be any integer from $[1, n]$ .
输入输出样例
输入 #1
5 4 37 998244353 1000000007 12345678901337426966631415
输出 #1
1 2 9 10 26
Explanations for the two first test cases of the example:
1. if $n = 4$ , then for every integer $x$ such that $1 \le x \le n$ , $\dfrac{x}{f(f(x))} = 1$ ;
2. if $n = 37$ , then for some integers $x$ such that $1 \le x \le n$ , $\dfrac{x}{f(f(x))} = 1$ (for example, if $x = 23$ , $f(f(x)) = 23$ , $\dfrac{x}{f(f(x))} = 1$ ); and for other values of $x$ , $\dfrac{x}{f(f(x))} = 10$ (for example, if $x = 30$ , $f(f(x)) = 3$ , $\dfrac{x}{f(f(x))} = 10$ ). So, there are two different values of $g(x)$ .
1. if $n = 4$ , then for every integer $x$ such that $1 \le x \le n$ , $\dfrac{x}{f(f(x))} = 1$ ;
2. if $n = 37$ , then for some integers $x$ such that $1 \le x \le n$ , $\dfrac{x}{f(f(x))} = 1$ (for example, if $x = 23$ , $f(f(x)) = 23$ , $\dfrac{x}{f(f(x))} = 1$ ); and for other values of $x$ , $\dfrac{x}{f(f(x))} = 10$ (for example, if $x = 30$ , $f(f(x)) = 3$ , $\dfrac{x}{f(f(x))} = 10$ ). So, there are two different values of $g(x)$ .
C++ 编辑器
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评测结果:Accepted