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 LuaJIT (2.0.4) (judge verdict: Accepted)

n=io.read("*n")

function digit(n)
  local dig=1
  local c=n
  while math.floor(c/10)~=0 do
    dig=dig+1
    c=math.floor(c/10)
  end
  return dig
end

ans=0
for i=1,n do
  if digit(i)%2==1 then
    ans=ans+1
  end
end

print(ans)