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)

// inputに入力データ全体が入る
function Main(input) {
    // 1行目がinput[0], 2行目がinput[1], …に入る
    input = input.split("\n");
    tmp0 = input[0].split(" ");
    N = Number(tmp0[0])

    var ans = 0

    for(var i=1; i <= N; i++)
    {
      if(String(i).length % 2 === 1){
        ans++
      }
    }
    console.log(ans)
}
  //*この行以降は編集しないでください（標準入出力から一度に読み込み、Mainを呼び出します）
Main(require("fs").readFileSync("/dev/stdin", "utf8"));
// Main("136\n");