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>

#define MIN_INT -2147483648
#define MAX_INT 2147483647
#define MIN_LONG -9223372036854775808L
#define MAX_LONG 9223372036854775807L

#define long long long int

using namespace std;

// @author: pashka

int main() {
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    int c = 1;
    int t = 10;
    int res = 0;
    for (int i = 1; i <= n; i++) {
        if (i >= t) {
            t *= 10;
            c++;
        }
        if (c % 2 == 1) {
            res++;
        }
    }
    cout << res;

    return 0;
}