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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"

using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < n; i++)
#define out(ans) cout << ans << endl;
const long long mod = 1e9 + 7;

int GetDigit(ll num)
{
  ll digit = 0;
  while (num != 0)
  {
    num /= 10;
    digit++;
  }
  return digit;
}

int main()
{
  ll n;
  cin >> n;
  ll count = 0;
  for (ll i = 1; i <= n; i++)
  {
    if(GetDigit(i)%2==1)
      count++;
  }
  out(count);

  return 0;
}