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 A009{
    public static void Main(){
        int n = int.Parse(Console.ReadLine());
        
        int ans = 0;
        
        if(10<=n) ans += 9;
        else{
            Console.WriteLine(n);
            return;
        }
        
        if(1000<=n) ans += 900;
        else{
            if(99<=n) ans += n-99;
            Console.WriteLine(ans);
            return;
        }
        
        if(100000<=n) ans += 90000;
        else if(9999<=n) ans += n-9999;
        
        Console.WriteLine(ans);
    }
}