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: Wrong Answer)

use std::io::*;
use std::str::FromStr;
 
/* 
The read function reads a string according to each one. 
*/
fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char) 
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn keta(x: i64) -> i64{
    let mut y:i64 = x;
    let mut count:i64 = 0;
    while y!=0{
	y /= 10;
	count+=1;
    }
    count
}

//main function
fn main() {
    let n: i64 = read();
    let mut sum: i64 = 0;
    let mut exp: i64 = 10;
    let mut count = 1;
    for _i in 0..keta(n){
	if count % 2 != 0{
	    if n < exp {
		if exp == 10 {
		    sum += n;
		}else{
		    sum += n%(exp/10)+1;
		}
	    }else {
		sum += exp*9/10;
	    }
	    exp *= 100;
	}
	count += 1;
    }
    println!("{}",sum);
}