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;

class Program {
    static void Main(string[] args) {
        //入力
        int N = int.Parse(Console.ReadLine());
        //Nの桁数
        int digit = (int)Math.Floor(Math.Log10(N)) + 1;

        //回答となる数値
        int ans = 0;

        for (int i = 1; i < digit; i++) {
            if (i % 2 == 1) {
                ans += 9 * (int)Math.Pow(10, i - 1);
            }
        }
        //最後に、奇数桁ならば端数分を足す
        if (digit % 2 == 1) {
            ans += N + 1 - (int)Math.Pow(10, digit - 1);
        }

        //出力
        Console.WriteLine(ans);
    }
}