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 v5.12) (judge verdict: Runtime Error)

'use strict'

function main(input) {
  const Num = input.trim()
  const L = Num.length
  const n = ~~Num
  let _mod = 0
  if (L % 2) {
    _mod = (n % Math.pow(10, L - 1)) + 1
  }
  const ll = Math.floor(L / 2)
  let _tmp = ''
  for (let i = 1; i <= ll * 2; i++) {
    _tmp = i % 2 ? '9' + _tmp : '0' + _tmp
  }

  _tmp = parseInt(_tmp, 10)
  console.log(_tmp + _mod)
}
main(require('fs').readFilexSync('/dev/stdin', 'utf8'))