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 scan = new Scanner(System.in);
	        String N  = scan.next();

	        //12345
	        //123456
	        long sum = 0L;
	        if(N.length() ==1) {
	        	System.out.println(N);
	        	return;
	        }
	        if(N.length()%2 ==1) {
	        	//sum += Integer.parseInt( N.substring(1, N.length()))+1;
	        	sum += Long.parseLong( N.substring(1, N.length()))+1;
	        }
	        long n = Long.parseLong(N)/10;

	        while(n>0) {
	        	int digit = String.valueOf(n).length();
	        	if(digit % 2 ==1) {
	        		sum+= (long)(Math.pow(10, digit) - Math.pow(10, digit-1) );
	        	}
	        	n /=10;

	        }

	       	System.out.println(sum);
	       	scan.close();
	}
}