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: Accepted)

package main

import (
	"fmt"
	"strconv"
	"strings"
)

func digits(n int) int {
	s := strconv.Itoa(n)
	return len(strings.Split(s, ""))
}

func isOdd(n int) bool {
	if n%2 == 1 {
		return true
	}
	return false
}

func main() {
	var n int
	fmt.Scan(&n)

	var result int
	for ; 0 < n; n-- {
		if isOdd(digits(n)) {
			result++
		}
	}
	fmt.Println(result)
}