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 <math.h>
using namespace std;

int main() {
  int N, n, digit=0, pow=1, num=0;
  cin >> N;
  for(n=N; n>0; n/=10) {
    pow *= 10;
    ++digit;
  }
  if(digit%2) {
    num = -(pow-1-N);
    pow *= 10;
  }
  num += (pow-1)/11;
  cout << num << endl;
  return 0;
}