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 <deque>
#include <list>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <array>
#include <queue>
#include <stack>
#include <algorithm>
#include <tuple>
#include <numeric>
#include <type_traits>
#include <limits>

int main()
{
	int N;

	std::cin >> N;

	if (N < 10)
		std::cout << N << "\n";
	else if (N < 100)
		std::cout << 9 << "\n";
	else if (N < 1000)
		std::cout << (9 + (N-99)) << "\n";
	else if (N < 10000)
		std::cout << 909 << "\n";
	else if (N < 100000)
		std::cout << (909 + (N - 9999)) << "\n";
	else
		std::cout << 90909 << "\n";
}