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: Wrong Answer)

'use strict';
function main(input) {
  const arr = input.trim().split('\n');
  const a = Number(arr[0]);
  let count = 0;

  for (let i = 0; i < a; i ++) {
    if (String(i).length % 2 !== 0) {
      count ++;
    }
  }
  
  console.log(count);
}

main(require("fs").readFileSync("/dev/stdin", "utf8"));