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: Accepted)

'use strict'
function main(inp) {
  inp = inp.trim()
  const N = inp - 0
  const L = inp.length - 1
  
  let ans = 0
  for (let i = 0; i < L; i++) {
    if (i%2 === 0) ans += Math.pow(10, i) * 9
  }
  if (L%2 === 0) ans+= N - (Math.pow(10,L) - 1)
  console.log(ans)
}
main(require('fs').readFileSync('/dev/stdin', 'utf8'));