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;
use std::cmp;
use std::collections::BinaryHeap;
use std::collections::HashMap;
use std::collections::HashSet;

fn read_line() -> String {
    let mut buf = String::new();
    io::stdin().read_line(&mut buf).ok();
    buf
}


fn main() {
    let n: usize = read_line().trim().parse().unwrap();
    let mut ans = 0;

    let mut m = 0;
    let mut tmp = n;
    while tmp > 0 {
        tmp /= 10;
        m += 1;
    }
    let mut tmp = 1;
    for i in 0..m {
        if i%2 > 0 {
            ans += tmp - tmp / 10;
        }
        tmp *= 10;
    }
    if m %2 > 0 {
        ans += n - tmp/10 + 1;
    }

    println!("{}", ans);
}