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 <iostream>

using namespace std;

class Solution {
public:
    int Count(int N) {
        int ret = 0;
        for (int i = 1; i <= N; i++) {
            int cnt = 0;
            int tmp = i;
            while (tmp) {
                cnt++;
                tmp /= 10;
            }
            if ((cnt & 1))
                ret++;
        }
        return ret;
    }
};

int main() {
    int N;
    cin >> N;
    Solution S;
    cout << S.Count(N) << endl;
}