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)

	keta := 6
	warukazu := 100000
	for N/warukazu < 1 {
		warukazu /= 10
		keta--
	}

	var count int
	if keta <= 1 {
		count = 1
	} else if keta <= 2 {
		count = 9
	} else if keta <= 3 {
		count = N - 90
	} else if keta <= 4 {
		count = 909
	} else if keta <= 5 {
		count = N - 9090
	} else if keta <= 6 {
		count = 90909
	}

	fmt.Printf("%d\n", count)

}