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;

class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int count = 0;
    int remit[] = {9,900,90000};
    int judge = 0;

    if(n >= 10000){
        judge = 3;
    }else if(n >= 100){
        judge = 2;
    }else if(n < 99){
        judge = 1;
    }

    if(judge == 3){
        if(n == 100000){
            count += remit[0] + remit[1] + remit[2];
        }else{
            count += (n - 9999) + remit[0] + remit[1];
        }

    }else if(judge == 2){
        if(n >= 1000){
            count += remit[0] + remit[1];
        }else{
            count += (n - 99) + remit[0]; 
        }

    }else if(judge == 1){
        if(n >= 9){
            count += remit[0];
        }else{
            count = n;
        }
    }

    System.out.println(count);
  }
}