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 PHP7 (7.0.15) (judge verdict: Wrong Answer)

<?php
fscanf(STDIN, "%d", $n);
$arr = str_split($n);

if(count($arr) < 2){
    echo $n;
}

$sum = [];
for($i = 0; $i < count($arr)-1; $i++){
    if($i % 2 == 0 || $i == 0){
        $sum[] = 9 * (10 ** $i);
    }
}

if(count($arr) % 2 != 0){
    $a = 10 ** (count($arr)-1);
    echo $a;
    $sum[] = $n - $a +1;
}

echo array_sum($sum);
?>