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) {
	
	input = input.split(" ");
	
	//文字列から10進数に変換するときはparseIntを使います
	var N= parseInt(input[0], 10);
	
	var n = 0;
	
	if (N <= 9){var n = N;}
	   else if ( N >=10 && N <= 99){n = 9;}
	         else if ( N >= 100 && N<= 999 ){n = N-99+9;}
	          else if (N >= 1000 && N<=9999){n = 909;}
	                                             
	               else if (N >=10000 && N <=99999){ n = N-9999  + 909;}
	               else{var n = 90909;}

   console.log(n);               
               
     }
    
	
//*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"));