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: Accepted)

<?php
  
$input = (int)fgets(STDIN);
$count = 0;

for($i=1; $i<=$input; $i++){
  if($i < 10){
    $count += 1;
  }elseif($i >= 100 and $i < 1000){
    $count += 1;
  }elseif($i >= 10000 and $i < 100000){
    $count += 1;
  }
}

echo $count;