CF1455A.Strange Functions

普及/提高-

通过率:0%

AC君温馨提醒

该题目为【codeforces】题库的题目,您提交的代码将被提交至codeforces进行远程评测,并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供,我们无法保证该服务的稳定性,若提交后无反应,请等待一段时间后再进行重试。

题目描述

Let's define a function f(x)f(x) ( xx is a positive integer) as follows: write all digits of the decimal representation of xx backwards, then get rid of the leading zeroes. For example, f(321)=123f(321) = 123 , f(120)=21f(120) = 21 , f(1000000)=1f(1000000) = 1 , f(111)=111f(111) = 111 .

Let's define another function g(x)=xf(f(x))g(x) = \dfrac{x}{f(f(x))} ( xx is a positive integer as well).

Your task is the following: for the given positive integer nn , calculate the number of different values of g(x)g(x) among all numbers xx such that 1xn1 \le x \le n .

输入格式

The first line contains one integer tt ( 1t1001 \le t \le 100 ) — the number of test cases.

Each test case consists of one line containing one integer nn ( 1n<101001 \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)g(x) , if xx can be any integer from [1,n][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

说明/提示

Explanations for the two first test cases of the example:

  1. if n=4n = 4 , then for every integer xx such that 1xn1 \le x \le n , xf(f(x))=1\dfrac{x}{f(f(x))} = 1 ;
  2. if n=37n = 37 , then for some integers xx such that 1xn1 \le x \le n , xf(f(x))=1\dfrac{x}{f(f(x))} = 1 (for example, if x=23x = 23 , f(f(x))=23f(f(x)) = 23 , xf(f(x))=1\dfrac{x}{f(f(x))} = 1 ); and for other values of xx , xf(f(x))=10\dfrac{x}{f(f(x))} = 10 (for example, if x=30x = 30 , f(f(x))=3f(f(x)) = 3 , xf(f(x))=10\dfrac{x}{f(f(x))} = 10 ). So, there are two different values of g(x)g(x) .
首页