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)

function main(input) {
    var args = input.trim().split(" ");
    var N = args[0];

    var total = 0;
    if (N.length % 2 === 1) {
        var diff = (10**(N.length - 1));
        total = N - diff + 1;
    }

    var validDigit = Math.ceil((N.length - 1) / 2);
    for(var i = 1; validDigit + 1 > i; i++) {
        total += 10**(i * 2 - 2) * 9;
    }

    console.log(total);
}

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