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;
using System.Text;
using System.Linq;

class Program
{
    static void Main(string[] args) {
#if DEBUG
        System.Diagnostics.Debug.WriteLine("-----TextFile1");
        System.Diagnostics.Debug.WriteLine(process("../TextFile1.txt"));
        System.Diagnostics.Debug.WriteLine("-----TextFile2");
        System.Diagnostics.Debug.WriteLine(process("../TextFile2.txt"));
        System.Diagnostics.Debug.WriteLine("-----TextFile3");
        System.Diagnostics.Debug.WriteLine(process("../TextFile3.txt"));
#endif

#if !DEBUG
        String ret = process(new System.IO.StreamReader(Console.OpenStandardInput()));
        Console.WriteLine(ret);
#endif
    }

    static string process(string filePath) {
        using (var input = new System.IO.StreamReader(filePath)) {
            if(input.EndOfStream) return "NO PROC";
            return process(input);
        }
    }

    static string process(System.IO.TextReader input) {
        int n     = int.Parse(input.ReadLine());

        int ans = 0;

        for(int i=1; i<=n; i++) {
            int keta = i.ToString().Length;
            if (keta % 2 == 1) ans++;
        }
        return ans.ToString();
    }
}