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 <vector>
#include <algorithm>
#include <stdio.h>
#include <cmath>
#include <map>
#include <utility>
#include <queue>
#define ull unsigned long long
#define ll long long
#define mod 998244353
using namespace std;

int main()
{
    int n;
    cin>>n;

    int br = 0;
    for (int i=1;i<=n;i++){
        int d = i;
        int b = 0;
        while ( d ) d/=10, b++;
        if ( b & 1 ) br++;
    }

    cout<<br<<endl;
    return 0;
}