Kickstart 2022 interesting numbers

Question

The question is to find the number of interesting numbers lying between two numbers. By the interesting number, they mean that the product of its digits is divisible by the sum of its digits.

For example: 459 => product = 4 * 5 * 9 = 180, and sum = 4 + 5 + 9 = 18; 180 % 18 == 0, hence it is an interesting number.

My solution for this problem is having run time error and time complexity of O(n2).

#include<iostream>
using namespace std;
int main(){
    int x,y,p=1,s=0,count=0,r;
    cout<<"enter two numbers"<<endl;
    cin>>x>>y;
    for(int i=x;i<=y;i++)
     {
            r=0;
         while(i>1)
           {
            r=i%10;
            s+=r;
            p*=r;
            i/=10;
           }
        if(p%s==0)
        {
            count++;
        }
    }
    cout<<"count of interesting numbers are"<<count<<endl;
    return 0;
}

Answer 1

If s is zero then if(p%s==0) will produce a divide by zero error.

Inside your for loop you modify the value of i to 0 or 1, this will mean the for loop never completes and will continuously check 1 and 2.

You also don't reinitialise p and s for each iteration of the for loop so will produce the wrong answer anyway. In general limit the scope of variables to where they are actually needed as this helps to avoid this type of bug.

Something like this should fix these problems:

#include <iostream>

int main()
{
    std::cout << "enter two numbers\n";
    int begin;
    int end;
    std::cin >> begin >> end;
    int count = 0;
    for (int number = begin; number <= end; number++) {
        int sum = 0;
        int product = 1;
        int value = number;
        while (value != 0) {
            int digit = value % 10;
            sum += digit;
            product *= digit;
            value /= 10;
        }
        if (sum != 0 && product % sum == 0) {
            count++;
        }
    }
    std::cout << "count of interesting numbers are " << count << "\n";
    return 0;
}

I'd guess the contest is trying to get you to do something more efficient than this, for example after calculating the sum and product for 1234 to find the sum for 1235 you just need to add one and for the product you can divide by 4 then multiply by 5.