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)

n=int(input())
m=str(n)
minus1=10**(len(m)-2)*9
minus=minus1
if len(m)==1:
    print(n)
elif len(m)%2==0:
    for i in range(len(m)//2):
        minus1+=minus//(100**(len(m)//2-i))
    print(minus1)
else:
    for i in range(len(m)//2):
        minus1+=minus//(100**(len(m)//2-i))
    print(minus1//10+n-10**(len(m)-1)+1)