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 (1.42.0) (judge verdict: Accepted)

use text_io::*;
use std::collections::*;
use std::process::exit;
use std::cmp::max;

fn main() {
    let mut n:usize = read!();
    let mut sum = 0;
    for i in 1..n+1 {
        if i<10||(100<=i&&i<1000)||(10000<=i&&i<100000) {
            sum+=1;
        }
    }
    println!("{}",sum);
}