Count all possible N digit numbers that satisfy the given condition in C++

In this tutorial, we will be discussing a program to find the number of possible N digit numbers that satisfy the given condition.

For this we will be provided with an integer. Our task is to check which one of number having N digits follow

Number + Reverse(Number) = 10N -1

Example

#include <bits/stdc++.h>
using namespace std;
//returning the count of numbers
string count_num(int N){
   if (N % 2 == 1)
      return 0;
   string result = "9";
   for (int i = 1; i <= N / 2 - 1; i++)
      result += "0";
   return result;
}
int main(){
   int N = 4;
   cout << count_num(N);
   return 0;
}