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: Accepted)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::io::*;
use std::str::FromStr;

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

use std::cmp::{min};
use std::num;

fn main() {
    
    let n = read::<usize>();
    let mut cnt = 0;
    let mut tmp = 1;
    
    loop {
        if tmp < n {
            cnt += 1;
            tmp += 1;
            if tmp == 10 {
                tmp = 100;
            }
            else if tmp == 1000{
                tmp = 10000;
            }
        }
        else if tmp == n && tmp != 100000{
            cnt += 1;
            break;
        }else{
            break;
        }
    }
    
    println!("{}", cnt);
    
}