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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int digits  = String.valueOf(N).length();
        int count = 0;
        for (int i = 1; i<=digits; i++){
            if (i%2!=0){
                if(i==digits){
                    count = count + N-(int)Math.pow(10,digits-1)+1;
                }else {
                    count = count + (int) Math.pow(10, i - 1) * 9;
                }
            }
        }
        System.out.println(count);
    }
}