Problem:
Score : 200 points

Problem Statement

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();

    token.parse().ok().expect("failed to parse token")
}

fn main() {

    let input = read::<String>();
    let i: i32 = input.parse().unwrap();
    let n: Vec<char>= input.chars().collect();

    let keta = n.len();

    let ans: i32 = match keta {
        1 => i,
        2 => 9,
        3 => 9 + i - 99,
        4 => 909,
        5 => 909 + i - 89999,
        6 => 90909,
        _ => 0,
    };
    println!("{}",ans);

}