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)

import java.util.*

fun main(args: Array<String>) {
    var n = IO.readInt()
    var ans = 0
    for (i in 1..n) {
        if (i.toString().length % 2 == 1) { ans = ans + 1}
    }   
    println(ans)
}

object IO {
    fun readInt(): Int {
        return readLine()!!.toInt()
    }
    fun readIntList(): List<Int> {
        return readLine()!!.split(" ").map(String::toInt)
    }
}