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 ll;

int main(){
    ll n;
    cin >> n;
    ll count = 0;
    ll c = 0;
    if(n==100000)
        count -= 1;
    if(n>=10000)
    {
        count += n - 10000 +1;
    }
    if(n>=100)
    {
        if(n>=1000)
            count += 900;
        else
        {
            count += n - 100 + 1;
        }
        
    }
    if(n>=10)
        count += 9;
    else
        count += n;
    cout << count;
}