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 <fstream>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>

unsigned gcd(unsigned a, unsigned b) {
  if(a < b) return gcd(b, a);
  unsigned r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}

#define LOCAL  1
#define UPLOAD 2

using namespace std;

int main()
{
#if ENVIRONMENT == LOCAL
    std::ifstream in("input.txt");
    std::cin.rdbuf(in.rdbuf());
#endif

    int N;
    cin >> N;

    int count = 0;
    for(int i=0;i<=N;i++){
        if((int)floor(log10(i)+1) % 2 == 1){
            count++;
        }
    }

    cout << count << endl;

    return 0;
}