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 <string>
#include <vector>
#include <algorithm>
#include <cmath>
#include <stdlib.h>
using namespace std;

int main()
{
	double n;
	cin >> n;

	int count = 0;
	double temp = n;
	while (temp / 10 >= 1) {
		count++;
		temp = temp / 10;
	}

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

	cout << answer << endl;

	return 0;
}