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)


fn main() {
    // N
    let n = read_line::<u32>();
    solve(n);
}

fn solve(n: u32) {
    let ans = (0..n)
        // 桁数が奇数
        .filter(|d| {
            (((d + 1) as f64).log10() as u32) % 2 == 0
        })
        .count();
    println!("{}", ans);
}

#[allow(dead_code)]
fn read_line<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read_line::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}