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)

s=input()
s_int=int(s)
ans=0

for i in range(1,len(s)):
    if i%2==1:
        ans+=9*(10**(i-1))
if len(s)%2==0:
    print(ans)
elif len(s)%2==1 and len(s)>=2:
    ans+=s_int-10**(len(s)-1)+1
    print(ans)
elif len(s)==1:
    print(s_int-1)