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 Python (3.4.3) (judge verdict: Wrong Answer)

# -*- coding:utf-8 -*-

N = int(input())

num = len(str(N))

if num == 1:
    print(N)
    exit(0)

odd = 0
for x in range(1, num):
    if x % 2 == 1:
        odd += (10 ** x)
        if x != 1:
            odd -= 10 ** (x-1)

if num % 2 == 1:
    odd += N % (10**(num-1)) + 1

odd -= 1

print(odd)