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 <iomanip>
#include <cstdio>
#include <math.h>
#include <vector>
#include <map>
#include <queue>
#include <utility>
#include <algorithm>
#include <string>
using namespace std;
int main() {
  int N;
  cin >> N;
  bool dig  = true;
  int digit = 1;
  int ans   = 0;
  for (int i = 1; i <= N; i++) {
	  if (dig) {
		  ans++;
		}
	  if (i == 10 * digit - 1) {
		  digit *= 10;
		  dig    = !dig;
		}
	}
  cout << ans << endl;
  return 0;
}