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 int long long
#define f(i,n) for(int i=0;i<(n);i++)
#define all(x) (x).begin(),(x).end()
using namespace std;
signed main(){
	string s;
	cin>>s;
	if(s.size()%2){
		int a=0;
		f(i,s.size()-1){
			a*=10;
			if(i%2)a+=9;
		}
		a+=stoi(s)+1-pow(10,s.size()-1);
		cout<<a<<endl;
	}
	else{
		f(i,s.size()-1){
			if(i%2==0)cout<<9;
			else cout<<0;
		}
	}
	return 0;
}