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())
d = len(str(n))
ans = 0

if d == 1:
  print(n)
else:
  for i in range(1, d+1):
    if i % 2 == 0:
      continue

    if i == d:
      ans += n - (10**(i-1))
      break;

    if i == 1:
      ans += 9
    else:
      ans += (10**i) - (10**(i-1))

  print(ans)