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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(int i = 0; i < (n); i++)
#define all(v) v.begin(), v.end()

int count(int x) {
  int y = 0;
  while(x > 0) {
    x /= 10;
    y++;
  }
  return y;
}

int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  ll n;
  cin >> n;
  ll cnt = 0;
  
  rep (i, n) {
    if (count(i) %2 == 1)
      cnt++;
  }
  
  cout << cnt << "\n";

}