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: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main() {
  int N, N_;
  cin >> N;
  N_ = N;
  int ans = 0;
  int i;
  for (i = 0; N > 0; N /= 10) {
    i++;
  }
  if (i > 1) ans += 9;
  if (i > 3) ans += 999 - 100 + 1;
  if (i > 5) ans += 99999 - 10000 + 1;
  if (i % 2 == 1) ans += N_ - ((int)pow(10, i - 1) + 1) + 1;
  cout << ans;
}