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>
using namespace std;

int main() {
  int N;
  cin >> N;
  int a;
  int count = 0;

  for(int i = 0; i < N; i++){
      if(N < 10){
          count = N;
      }
      if(N >= 10 && N < 100){
          count = 9;
      }
      if(N >= 100 && N < 1000){
          count = N - 90;
      }
      if(N >= 1000 && N < 10000){
          count = N - 90;
      }
      if(N >= 10000 && N < 100000){
          count = N - 9090;
      }
      if(N >= 100000 && N < 1000000){
          count = N - 9091;
      }
      if(N == 1000000){
          count = N - 9092;
      }
    }
    cout << count;
}