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>
#include <cmath>

using namespace std;

int main() {

    int N, Ncopy;
    cin >> N;
    Ncopy = N;

    int c = 0;
    for (int i=0; i<10; i++) {
        N = N / 10;
        if (N != 0) {
            c += 1;
        }
        else {
            break;
        }
    }

    int d;
    int count = 0;
    if (c%2 == 0) { 
        d = c/2;
        for (int i=0; i<d;i++) {
            count += 9 * pow(10, (i*2));
        }
        count += (Ncopy - pow(10, c) + 1);  
    } else {
        d = c/2 + 1;
        for (int i=0; i<d;i++) {
            count += 9 * pow(10, (i*2));
        }
    }

    cout << count;


    return 0;
}