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: Accepted)

def digitCount(num):
  digits = 0
  while num > 0:
    digits += 1
    num = num // 10
  return digits

def main():
  n = int(input())
  ans = 0
  for i in range(1,n+1):
    if digitCount(i) % 2 != 0:
      ans += 1
  print(ans)

if __name__ == '__main__':
  main()