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<list>
#include<vector>
#include<algorithm>
#include<string>
#include<stdio.h>
#include<math.h>
#include<unordered_map>
#include<map>

using namespace std;
using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using lst = list<ll>;


ll MOD = 1000000007;


int main()
{
	int N;
	cin >> N;
	int M = N;

	int digit=0;
	int ans = 0;
	int ten = 1;
	while (M != 0)
	{
		++digit;
		ten *= 10;

		if (digit % 2 == 1)
		{
			if (M < 10)
			{
				ans += N - ten / 10 + 1;
			}
			else
			{
				ans += ten - ten/10;
			}
		}
		M /= 10;
	}

	cout << ans << endl;


	return 0;
}