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>
#define rep(i,n) for(int i = 0; i < (n); i++)
using namespace std;
using ll = long long;
using P = pair<int, int>;

bool dig(int n) {
  int res = 0;
  while(n > 0) {
    res++;
    n /= 10;
  }
  if(res%2 == 1) return true;
  else return false;
}

int main() {
  int n;
  cin >> n;
  int count = 0;
  for(int i = 1; i <= n; i++) {
    if(dig(i)) count++;
  }
  cout << count << endl;
  return 0;
}