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 rep(i,n)  for(int i = 0; i < n; i++)
#define all(cont) cont.begin(), cont.end()
#define rall(cont) cont.rbegin(), cont.rend()
#define unique(a)       sort((a).begin(), a.end()), (a).erase(unique((a).begin(), (a).end()),(a).end())
#define fastio          ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define PB push_back
#define endl "\n"

using namespace std;

typedef long long LL;
typedef long double LD;
typedef vector<int> VI;

LL gcd(LL a, LL b)
{
	if (!a)return b; return gcd(b%a, a);
}
double intlog(double base, double x) {
	return (double)((double)log(x) / (double)log(base));
}

int main()
{
	int n;
	cin >> n;
	int count = 0;
	for (int i = 1; i <= n; i++)
	{
		string s = to_string(i);
		if (s.length() & 1)
			count++;

}
	cout << count << endl;
}