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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

import sys
from operator import itemgetter
import math
stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  # ignore trailing spaces

n = ni()
keta = len(str(n))
rs = 0
if keta % 2 == 1:
    rs += n % (10 ** (keta - 1)) + 1
li = [0, 9, 0, 900, 0, 90000]
keta -= 1
while keta > 0:
    rs += li[keta]
    keta -= 1
print(rs)