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: Wrong Answer)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int n = std.nextInt();
        int count = (int) Math.ceil(Math.log10(n));
        int ans = 0;
        for (int i = 1; i <= count; i++) {
            if (i % 2 == 0) {
                continue;
            }

            if (i == count) {
                int base = (int) Math.pow(10, count - 1);
                int last;
                if (n % base == 0) {
                    last = n - base;
                } else {
                    last = n - base + 1;
                }
                ans += last;
            } else {
                int base = (int) Math.pow(10, i);
                ans += base - Math.pow(10, i - 1);
            }
        }
        System.out.println(ans);
    }
}