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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val N = readLine()!!.toInt()
    val d = Math.log10(N.toDouble()).toInt() + 1
    val ans = when (d) {
        1 -> N
        2 -> 9
        3 -> 9 + (N - 99)
        4 -> 9 + 900
        5 -> 9 + 900 + (N - 9999)
        6 -> 9 + 900 + 90000
        else -> -1
    }
    println(ans)
}