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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Akash
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BUnevenNumbers solver = new BUnevenNumbers();
        solver.solve(1, in, out);
        out.close();
    }

    static class BUnevenNumbers {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int count = 0;
            for (int i = 1; i <= n; i++) {
                if (digits(i) % 2 == 1)
                    count++;
            }
            out.println(count);
        }

        int digits(int n) {
            if (n <= 0) {
                return 0;
            }
            return 1 + digits(n / 10);
        }

    }
}