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: Accepted)

using System;

public class Hello{
    public static void Main(){
        int n = int.Parse(Console.ReadLine());
        
        int num = (int)Math.Log10(n) + 1;
        int result1 = num % 2 == 1 ? 1 + n - (int)(Math.Pow(10, num - 1)) : 0;
        num--;
        
        while(true){
            if(num % 2 == 0){
                num--;
            }else{
                int res = (int)(Math.Pow(10, num) - Math.Pow(10, num - 1));
                result1 += res;
                num -= 2;
            }
            
            if(num <= 0){
                break;
            }
        }
        
        Console.WriteLine(result1);
    }
}