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)

N = int(input())
i = 1
count = 0

if N < 10:
    for i in range(1,N):
        count += 1

elif 10 <= N < 1000:
    for i in range(1,10):
        count += 1
    if N == 100:
        count += 1
    i = 100
    for i in range(100,N) :
        count += 1

elif 1000 <= N <= 100000:
    for i in range(1,10):
        count += 1
    i = 100
    for i in range(100, 1000):
        count += 1
    if N == 10000:
        count += 1
    i = 10000
    for i in range(10000, N):
        count += 1
print(count)