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)

a= int(input())
result = 0
for i in range(10):
    if a >= 10 ** (i-1) and a < 10 ** i :
        rank = i 
        break

for i in range(rank+1):
    
    if i == 1 and rank is not i:
        result += 10 ** i -1
    elif i == 3 and rank is not i:
        result += 10 ** i  - 10** (i-1)
    elif i == 5 and rank is not i:
        result += 10 ** i  - 10** (i-1)
    elif i == rank and i %2 is not 0:
        result += a - 10 ** (i-1) + 1
    
        
        
print(result)