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 (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func main() {
	scanInit()

	n := nextInt()

	cnt := 0
	for i := 1; i <= n; i++ {
		if keta(i)%2 == 1 {
			cnt++
		}
	}

	fmt.Println(cnt)

}

// 正数の桁数を返す
func keta(n int) int {
	keta := 0
	if n == 0 {
		return 1
	}
	for n > 0 {
		keta++
		n /= 10
	}
	return keta
}

// ---- readfunc
func scanInit() {
	const cap = 200 * 1024 // Default=64
	var buf = make([]byte, cap)
	sc.Buffer(buf, cap)
	sc.Split(bufio.ScanWords)
	return
}
func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}
func nextStr() string {
	sc.Scan()
	return sc.Text()
}