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", $n);
$num = 0;
if($n >= 10){
	$num += 9;
}else{
	$num += $n;
}
 
if($n >= 1000){
	$num += 900;
}elseif($n >= 100){
  	$num = $num + $n - 99;
}else{
	;
}
 
if($n === 100000){
	$num += 90000;
}elseif($n >= 10000){
  	$num = $num + $n - 9999;
}else{
	;
}
 
echo $num . "\n";
?>