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 Go (1.6) (judge verdict: Wrong Answer)

package main

import "fmt"

func main() {
	var a int
	fmt.Scan(&a)
	var count int
	if a == 100000 {
		fmt.Println(90909)
		return
	}
	if a >= 10 {
		count += 9
	} else {
		count += a
	}
	if a >= 100 {
		count += a - 99
	}

	if a >= 10000 {
		count = 9 + 999 - 100 + 1 + a - 9999
	}
	fmt.Println(count)
}