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 Main(input) {
  var n = parseInt(input);
  var k = Math.floor(Math.log10(n)); // 10^k≦nを満たす最大のk
  var sum = 0;
  for(var i=1; i<=k; i+=2){
  	sum += Math.pow(10, (i-1))*9;
  }
  if(k%2==0){
  	sum += n-Math.pow(10, Math.floor(Math.log10(n))) + 1;
  }
  console.log(sum);
}
Main(require("fs").readFileSync("/dev/stdin", "utf8"));