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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;

class Prog{
    static void Main(){
        int x = int.Parse(Console.ReadLine());
        string y = x.ToString();
        double a = 0;
        if(y.Length % 2 != 0){
            a += x - Math.Pow(10,(y.Length/2)+1) + 1;
        }
        for(int i = 0;i < y.Length-1;i = i + 2){
            a += 9*Math.Pow(10,i);
        }
        Console.WriteLine(a);
    }
}