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
  $input=explode(' ',trim(fgets(STDIN)));
  $a=intval($input[0]);
//A,B,Cそれぞれを$a,$b,$cに格納
  if(($a >= 1)&&($a <= 100000)){
    if($a == 100000){
      $result = 90909;
      echo $result;
    }else if(($a >=10000)&&($a <= 99999)){
      $b = $a - 9999;
      $c = 900;
      $d = 9;
      $result = $b + $c + $d;
      echo $result;
    }else if(($a >=1000)&&($a <= 9999)){
      $b = 0;
      $c = 900;
      $d = 9;
      $result = $b + $c + $d;
      echo $result;
    }else if(($a >=100)&&($a <= 999)){
      $b = 0;
      $c = $a - 99;
      $d = 9;
      $result = $b + $c + $d;
      echo $result;
    }else if(($a >=10)&&($a <= 99)){
      $b = 0;
      $c = 0;
      $d = 9;
      $result = $b + $c + $d;
      echo $result;
    }else if(($a >=0)&&($a <= 9)){
      $b = 0;
      $c = 0;
      $d = $a;
      $result = $b + $c + $d;
      echo $result;
    }
  }
?>