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)

#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;

int n, ans;

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        int t = i;
        int cnt = 1;
        while (t/=10) ++cnt;
        ans += cnt & 1;
    }
    cout << ans << '\n';
}