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 PHP (7.4.4) (judge verdict: Accepted)

<?php

$N = trim(fgets(STDIN));
$count = 0;
$len_check = [];
for($i=1; $i<=$N; $i*=10){
    $digit = count(str_split($i));
        
    if($digit%2 == 1){
        if ($digit == 1 ) {
            for($j=1; $j<10; $j++){
                if($N < $j){
                    break;
                }
                $count += 1;
            }
        }else{
            $tmp = 1;
            for($d=1; $d<=($digit-1); $d++){
                $tmp *= 10;
            }
            for($j=$tmp; $j<$tmp*10; $j++){
                if($N < $j){
                    break;
                }
                $count += 1;
            }
        }
        
    }
}
echo $count;
?>