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)

import numpy as np

N = int(input())

n = N

count = 0

num = 0

while n >= 10:
    n = n / 10
    count = count + 1

if(int(n) != 0):
	n = N - (10 ** count) + 1

for i in range(count + 1):
    if(i % 2 == 1):
        num = num + (0.9 * (10 ** i))

if(count % 2 == 0):
    num = num + n

print(int(num))