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;
  if(N==10000){
    cout <<90909<<endl;
  }
  if(N<10){
    cout <<N<<endl;
  }
  else if(N<100){
    cout <<9<<endl;
  }
  else if(N<1000){
    cout <<N-90<<endl;
  }
  else if(N<9999){
    cout <<909<<endl;
  }  
  else if(N<99999){
    cout << N-9090 <<endl;
  }
}