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 N int
	fmt.Scan(&N)
	// 1-9
	// 100-999
	// 10000-99999
	if N < 10 {
		fmt.Println(N)
	} else if 10 <= N && N < 100 {
		fmt.Println(9)
	} else if 100 <= N && N < 1000 {
		fmt.Println(9 + (N - 100 + 1))
	} else if 1000 <= N && N < 10000 {
		fmt.Println(9 + 900)
	} else if 10000 < N && N < 100000 {
		fmt.Println(9 + 900 + (N - 10000 + 1))
	} else {
		fmt.Println(9 + 900 + 90000)
	}

}