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;
using System.Text;
using System.Collections.Generic;

public class Hello{
    public static void Main(){
        int N = int.Parse(Console.ReadLine());
        int digit = (int)Math.Floor(Math.Log(N,10)) + 1;
        int sum = 0;
        if(digit%2==0){
            for(int i=1;i<=digit-1;i+=2){
                sum += 9 * (int)Math.Pow(10,i-1);
            }
        }else{
            for(int i=1;i<=digit-2;i+=2){
                sum += 9 * (int)Math.Pow(10,i-1);
            }
            sum += N - (int)Math.Pow(10,digit-1) + 1;
        }
        Console.WriteLine(sum);
    }
}