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)

function Main(input){
var N = new Number(input);
var Np = input.split("");
var count = 0;
switch(Np.length){
  case 1:
    count += new Number(Np[0]);
    console.log(count);
  break;
  case 2:
    count += 9;
    console.log(count);
  break;
  case 3:
    count += new Number(N)-90;
    console.log(count);
  break;
  case 4:
    count += 909;
    console.log(count);
  break;
  case 5:
    count += new Number(N)-9000;
    console.log(count);
  break;
  case 6:
    count += 90909;
    console.log(count);
  break;
  default:
    console.log("error");
  break; 

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