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 (
  "fmt"
)

func num(n int) int {
   count := 1
  for n >= 10 {
    n = n / 10
    count += 1
  }
  return count
}

func main(){
  var N int
  fmt.Scan(&N)
  
  result := 0
  for i := 1; i <= N; i++ {
    if num(i) % 2 != 0 {
      result += 1
    }
  }
  fmt.Println(result)
}