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"
	"bufio"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)
func scanInt() int {
	sc.Scan()
	a,_ := strconv.Atoi(sc.Text())
	return a
}

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

func main() {
	sc.Split(bufio.ScanWords)
	
	n := scanInt()

	ans := 0
	d := 1
	for i := 1; i < n+1; i*=10 {
		if d%2 == 0 {
			d++
			continue
		}
		
		if n/i/10 == 0 {
			ans += n-i+1
		} else {
			ans += i*10-i
		}
		d++
	}
	
	fmt.Println(ans)
}