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> 
using namespace std;
 
typedef long long int ll;
typedef pair<int,int> pii;

#define ff first
#define ss second
#define vi vector<int>
#define fo(n) for(int i = 0;i < int(n);i++)
#define foo(a,b) for(int i = int(a);i <= int(b);i++)
#define mk make_pair
#define pb push_back
#define all(x) x.begin(),x.end()
#define fastio ios_base:: sync_with_stdio(false)


const int MAXN = 1e6;
const int mod = 1e9 + 7;
const int inf = 1e9;

int nd(int n){
	int d = 0;

	while(n){
		d++;
		n /= 10;
	}

	return d;
}


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

	int cnt = 0;
	for(int i = 1;i <= n;i++){
		if(nd(i)%2 == 1) cnt++;
	}

	cout << cnt << endl;
	return 0;
 }