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)

use std::io;

fn main() {
	let mut buf = String::new();
	io::stdin().read_line(&mut buf).expect("");
	let v = buf.trim().chars().collect::<String>();
	
	let mut ret = 0;
	if v.len() % 2 == 1 {
		let b:i32 = match buf.trim().parse() {
			Ok(n) => n,
			_ => panic!(""),
		};
		ret = ret + b - 10_i32.pow((v.len() - 1) as u32) + 1
	}
	for n in 1..(v.len() as u32) {
		if n % 2 == 1 {
			ret = ret + 10_i32.pow(n) - 10_i32.pow(n-1);
		}
	}
	println!("{}", ret);
}