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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
class uneven{
  public static void Main(){
    String str = Console.ReadLine();
    int count=0;
    if(str.Length%2==0){
      for(int i=1; i<str.Length; i+=2){
        count+=9*(int)Math.Pow(10,i-1);
      }
    }
    else{
      for(int i=1; i<str.Length-1; i+=2){
        count+=9*(int)Math.Pow(10,i-1);
      }
      count+=int.Parse(str)-int.Parse(str[0].ToString())*(int)Math.Pow(10,str.Length-1);
      count++;
    }
    Console.WriteLine(count);
  }
}