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;

int func(int num){
    int k = 0;
    while(num != 0){
        num /= 10;
        k++;
    }
    return k;
}

int main(){
    int N;
    cin >> N;
    int ans = 0;
    for(int i = 1; i <= N; i++){
        if(func(i) % 2 == 1){
            ans++;
        }
        else{
            continue;
        }
    }
    cout << ans << endl;
    return 0;
}