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"
)
 
func main() {
 
  var n int
  c := 0
  fmt.Scan(&n) 
  
  for i := 1; i <= n; i++ {
    if (i < 10) || (99 < i && i < 1000) || ( 9999 < i && i < 100000) {
      c++
    } 
  }
 
  fmt.Println(c)
}