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.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Text.RegularExpressions;
using static System.Console;

namespace AtCoder_1
{
    class Program
    {
        static void Main(string[] args)
        {
            var a = CinT();
            int c = 0;
            for(int i = 1;i <= a; i++)
            {
                if((i + "").Length % 2 == 1)
                {
                    c++;
                }
            }
            Cout(c);
            int res = 0;
            //if(a.Length % 2 == 1)
            //{
            //    res = a.Length / 2;
            //    res *= 9;
            //    S2i(a[0] + "")
            //}
            //else
            //{
            //    res = (a.Length / 2) * 9;
            //}

        }



        static readonly int MOD1E9 = 10000007;
        static bool isInt(double d) { return (int)d == d; }
        static string Cin() { return ReadLine(); }
        static string[] CinAr() { return ReadLine().Split(' '); }
        static int CinT() { return S2i(Cin()); }
        static int[] CinTAr() { return CinAr().Select((x, i) => { return S2i(x); }).ToArray(); }
        static long CinL() { return S2L(Cin()); }
        static long[] CinLAr() { return CinAr().Select((x, i) => { return S2L(x); }).ToArray(); }
        static double CinD() { return S2D(Cin()); }
        static double[] CinDAr() { return CinAr().Select((x, i) => { return S2D(x); }).ToArray(); }
        static int S2i(string s) { return int.Parse(s); }
        static long S2L(string s) { return long.Parse(s); }
        static double S2D(string s) { return double.Parse(s); }
        static void Cout(object s) { WriteLine(s); }
        static int[] yakusuu(int n)
        {
            if (n == 1) { return new int[] { 1 }; }
            HashSet<int> res = new HashSet<int>();
            for (int i = 1; i < Math.Ceiling(n / 2.0); i++)
            {
                if (n % i == 0)
                {
                    res.Add(i);
                    res.Add(n / i);
                }
            }
            return res.ToArray();
        }
        /// <summary>
        /// Debug時のみの使用を推奨
        /// </summary>
        /// <typeparam name="Type"></typeparam>
        /// <param name="types"></param>
        static void CoutAr<Type>(IEnumerable<Type> types)
        {
            string text = "";
            foreach (var i in types)
            {
                text += " " + i;
            }
            WriteLine(text.Substring(1));
        }

        public static long nCr(int n, int r)
        {
            // naive: return Factorial(n) / (Factorial(r) * Factorial(n - r));
            return nPr(n, r) / Factorial(r);
        }

        public static long nPr(int n, int r)
        {
            // naive: return Factorial(n) / Factorial(n - r);
            return FactorialDivision(n, n - r);
        }

        private static long FactorialDivision(int topFactorial, int divisorFactorial)
        {
            long result = 1;
            for (int i = topFactorial; i > divisorFactorial; i--)
                result *= i;
            return result;
        }

        private static long Factorial(int i)
        {
            if (i <= 1)
                return 1;
            return i * Factorial(i - 1);
        }
    }
}