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: Wrong Answer)

package main

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

func min64(a,b int64) int64 {
	if a < b {
		return a
	}
	return b
}

func max64(a,b int64) int64 {
	if a < b {
		return b
	}
	return a
}

func answer(n int64) int64 {
	if (n < 10) {
		return n;
	}
	if (n < 100) {
		return 9;
	}
	if (n < 1000) {
		return (n - 99)  +  9;
	}
	if (n < 10000) {
		return 900 +  9;
	}
	if (n < 100000) {
		return (n - 9000) + 900 +  9;
	}
	return 90000 + 900 + 9;
}

func main() {
	scanner := bufio.NewScanner(os.Stdin)

	for scanner.Scan() {
		text := scanner.Text()
		N, _ := strconv.ParseInt(text, 10, 0)


		result := answer(N);
		fmt.Printf("%d", result)
	}
}