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;
class B_Uneven{
	static void Main(string[] args){
		double N = double.Parse(Console.ReadLine());
		int Score = 0;
		if(Math.Floor(N / 10) > 0){
			Score += 9;
			if(Math.Floor(N / 1000) > 0){
				Score += 900;
				if(N > 10000)Score += ((int)Math.Floor(N) - 9999);
				if(N >= 100000)Score --;
			}else if(N >= 100)Score += ((int)Math.Floor(N) - 99);
		} else Score += (int)Math.Floor(N);
		Console.WriteLine(Score);
	}
}