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.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
	static void Main()
	{
		long n = long.Parse(Console.ReadLine());
    long answer = 0;
    for(long i = 1; i <= n; i++)
    {
      if(whatDigit(i) % 2 == 1) answer++;
    }

		Console.WriteLine(answer);
	}

  static long whatDigit (long n)//nは何桁か
    {
      for(long i = 1; i < 20; i++)
      {
        n /= 10;
        if(n == 0)
        {
          return i;
          break;
        }
      }
      return 0;
    }
  
}