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 (3.8.2) (judge verdict: Wrong Answer)

n    = int(input())
mx   = 10**5
li   = [0]*mx
cnt  = 0
unit = 10

for i in range(mx):
    chk = (i+1)//unit

    if chk < 1:
        cnt+=1
    elif chk < 10:
        pass
    else:
        cnt+=1
        unit*100

    li[i] = cnt

print(li[n-1])