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 main():
    N = input()
    cnt = 9
    len_n = len(N)
    if len_n == 1:
        cnt = int(N)
    elif len_n % 2 == 0:
        len_n -= 1
    else:
        cnt += int(N[1:]) + 1
        if int(N[0]) > 1:
            cnt += (int(N[0])-1) * 10 ** (len_n-1)
        len_n -= 2
    while len_n > 1:
        cnt += 10 ** (len_n - 1) * 9
        len_n -= 2
    print(cnt)


if __name__ == '__main__':
    main()