Find unique numbers in array

Question

Well, I have to find how many different numbers are in an array.

For example if array is: 1 9 4 5 8 3 1 3 5

The output should be 6, because 1,9,4,5,8,3 are uniqu

Answer 1

Since you've stated that you cannot use the standard library and must use loops, let's try this solution instead.

#include 

using namespace std; // you're a bad, bad boy!

int main() 
{
    int r = 0, a[50], n;

    cout << "How many numbers will you input? ";
    cin >> n;

    if(n <= 0)
    {
        cout << "What? Put me in Coach. I'm ready! I can do this!" << endl;
        return -1;
    }

    if(n > 50)
    {
        cout << "So many numbers! I... can't do this Coach!" << endl;
        return -1;
    }   

    cout << "OK... Enter your numbers now." << endl;

    for (int i = 0; i < n; i++)
        cin >> a[i];


    cout << "Let's see... ";

    // We could sort the list but that's a bit too much. We will choose the
    // naive approach which is O(n^2), but that's OK. We're still learning!

    for (int i = 0; i != n; i++) 
    { // Go through the list once.      
        for (int j = 0; j != i; j++)
        { // And check if this number has already appeared in the list:
            if((i != j) && (a[j] == a[i]))
            { // A duplicate number!        
                r++; 
                break;
            }
        }
    }

    cout << "I count " << n - r << " unique numbers!" << endl;

    return 0;
}

I urge you to not submit this code as your homework - at least not without understanding it. You will only do yourself a disservice, and chances are that your instructor will know that you didn't write it anyways: I've been a grader before, and it's fairly obvious when someone's code quality magically improves.