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.Collections.Generic;

public class Program{
    public static void Main(){
        int N=int.Parse(Console.ReadLine());
        int ans;
        if(N>0 && N<10){
            ans=N;
        }else if(N>=10 && N<100){
            ans=9;
        }else if(N>=100 && N<1000){
            ans=N-90;
        }else if(N>=1000 && N<10000){
            ans=909;
        }else if(N>=10000 && N<100000){
            ans=N-9090;
        }else{
            ans=90909;
        }
        Console.WriteLine(ans);
    }
}