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
        fmt.Scanf("%d", &n)
        fmt.Printf("%d\n", numOfOddDigitNum(n))
}

func numOfOddDigitNum(n int) (count int) {
        for i := 1; i <= n; i++ {
                if digit(i)%2 == 1 {
                        count++
                }
        }
        return count
}
func digit(n int) int {
        if n == 0 {
                return 0
        }
        digit := 0
        for i := 1; i <= n; i *= 10 {
                digit++
        }
        return digit
}