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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

"use strict"
const create = () => {
    const input = require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n");
    const res = {
        "list": input, "index": 0, "max": input.length,
        "hasNext": function() {return this.index < this.max},
        "next": function() { if (!this.hasNext()) {throw "ArrayIndexOutOfBoundsException";}else{return this.list[this.index++];}}
    };
    return res;
}
const o = create();
const next = () => o.next()
const nextInt = () => parseInt(o.next())
const nextStrArray = () => o.next().split(" ")
const nextIntArray = () => o.next().split(" ").map(el => parseInt(el))
const nextCharArray = () => o.next().split("")
const hasNext = () => o.hasNext()
const myout = (...x) => console.log(...x)
const Main = () => {
    let n = nextInt()
    const digit = (x) => String(x).split("").length
    let cnt = 0
    for(let i = 1; i<= n; i++){
        let tmp = digit(i)
        if(tmp % 2 !== 0) cnt++
    }
    console.log(cnt)
}
Main()