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 (
	"bufio"
	"fmt"
	"math"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func read() string {
	sc.Scan()
	return sc.Text()
}

func main() {
	sc.Split(bufio.ScanWords)
	N, _ := strconv.Atoi(read())
	order := int(math.Trunc(math.Log10(float64(N)+0.1)) + 1)
	sub := 0
	for i := 0; i <= order; i++ {
		if i%2 == 0 {
			sub += int(math.Pow10(i))
		}
	}
	fmt.Printf("%v\n", N-sub)
}