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)

#![allow(non_snake_case)]
use proconio::input;
// use std::collections::HashMap;

fn digs(n: i64) -> i64 {
    if 0 <= n && n < 10 {
        return 1;
    }

    return digs(n / 10) + 1;
}

fn main() {
    input! {
        N: i64,
    }

    let mut ans = 0;
    for n in 1..N + 1 {
        let d = digs(n);
        if d % 2 == 1 {
            ans += 1;
        }
    }
    println!("{}", ans);
}