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 <cstdlib>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;


int main(){
    int n;
    cin >> n;
    
    if(n < 10){
        cout << n << endl;
    }
    else if (n < std::pow(10,2)){
        cout << 9 << endl;
    }
    else if (n < std::pow(10,3)){
        cout << n - 100 + 1 + 9 << endl;
    }
    else if (n < std::pow(10,4)){
        cout << 900 + 9 << endl;
    }
    else if (n < std::pow(10,5)){
        cout << n - 10000 + 1 + 900 + 9 << endl;
    }
    else if (n == std::pow(10,5)){
        cout << n - 10000 + 900 + 9 << endl;
    }
   
}