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.Collections.Generic;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApp1
{
    class Program
    {
        static void Main(string[] args)
        {
            long N = long.Parse(Console.ReadLine());
            int cnt = 0;
            for (int i = 1; i <= N; i++)
            {
                if (i.ToString().Length % 2 != 0)
                {
                    cnt++;
                }
                else
                {
                    i *= 10;
                }
            }
            Console.WriteLine(cnt);
        }
    }
}
// 文字列の入力
//string s = Console.ReadLine();

// 整数の入力
//long n = long.Parse(Console.ReadLine());

// 文字列配列の入力
//string[] inputStrArray = Console.ReadLine().Split(' ');

// 整数配列の入力
//long[] inputLongArray = Console.ReadLine().Split(' ').Select(i => long.Parse(i)).ToArray();