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;
using System.Text;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using static System.Console;
using static System.Math;


class Program
{
    static void Main(string[] args)
    {

        int N = int.Parse(Console.ReadLine());
        int ans;
        //出力
        if (N < 10){
            ans = N;
            Console.WriteLine(ans);

        }
        else if (10 <= N && N < 100)
        {
            ans = 9;
            Console.WriteLine(ans);
        }
        else if(100<=N && N<1000){
            ans = N-99+9;
        Console.WriteLine(ans);
        }
        else if (1000 <= N && N < 10000)
        {
            ans = 9+999;
            Console.WriteLine(ans);
        }
        else if (10000 <= N && N < 100000)
        {
            ans = N-99-9999+9+999;
            Console.WriteLine(ans);
        }
        else if (N ==100000)
        {
            ans = 90909;
            Console.WriteLine(ans);
        }
    }
}