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)

/*In the name of Allah*/
#include <bits/stdc++.h>
using namespace std;
int sz(int x){
	int c = 0;
	while(x){
		x/=10;
		c++;
	}
	return c;
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	int n;
	cin>>n;
	int ans= 0;
	for(int i = 1; i <= n ; i++){
		if(sz(i)&1)ans++;
	} 
	cout<<ans;
	return 0;
}