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 (5.6.30) (judge verdict: Accepted)

<?php
  
fscanf(STDIN, "%d", $a);

echo countOdd($a);

function countOdd($a) {
	$len = strlen($a);
	if ($len === 1) {
  		return $a;
	}

  	$odds = 0;
	$b = intval(str_repeat('9', $len - 1));
	$is_odd = $len % 2; 
	if ($is_odd) {
		$odds = $a - $b;
	}
	$odds += countOdd($b);

  	return $odds;
}