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)

function keta(num) {
    for(var i=1; i<=6; i++) {
        if(num < Math.pow(10, i)) return i%2;
    }
}

function Main(input) {
    input = parseInt(input);
    var ans = 0;
    var cnt = 1;
    while(cnt <= input) {
        ans += keta(cnt);
        cnt++;
    }
    console.log(ans);
}

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