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.Collections.Generic;
using System.ComponentModel;
using System.Linq;
using System.Threading.Tasks;
using System.IO;

class Solve
{

	const int mod = 1000000007;
	static void Main(string[] args)
	{
        var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
		Console.SetOut(sw);

		int N = int.Parse(Console.ReadLine());
		long ans = 0;
		if (N < 10) { ans = N; }
        else if (N < 100) { ans = 9; }
        else if (N < 1000) { ans = N - 90; }
        else if (N < 10000) { ans = 909; }
        else if (N < 100000) { ans = N - 9090; }
		else { ans = 90909; }
		Console.WriteLine(ans);


		Console.Out.Flush();
	}
}