CODEWITHSUNDEEP
c++
David Choi
David Choi

Reputation: 3

Finding Sum of Square of Digits Beginner Bug C++

So, I started learning C++ recently. This code is trying to add the sum of the squares of each numbers digits. For example: 243: 2*2 + 4*4 + 3*3 = 29.

int sumOfSquareDigits(int n) //BUG WITH INPUT OF 10, 100, 1000, etc.
{
    int digits = findDigits(n);
    int number;
    int remainder;
    int *allDigits = new int[digits];
    for (int i = 0; i < digits; i++) { //assigns digits to array
        if (i + 1 == digits){ //sees if there is a ones value left
            allDigits[i] = n;
        }
        else {
            remainder = (n % findPower10(digits - (i + 1)));
            number = ((n - remainder) / findPower10(digits - (i + 1)));
            allDigits[i] = number; //records leftmost digit
            n = n - (allDigits[i] * findPower10(digits - (i + 1))); //gets rid of leftmost number and starts over
        }
    }
    int result = 0;
    for (int i = 0; i < digits; i++) { //finds sum of squared digits
        result = result + (allDigits[i] * allDigits[i]);
    }
    delete [] allDigits;
    return result;
}

int findDigits(int n) //finds out how many digits the number has
{
int digits = 0;
int test;
    do {
        digits++;
        test = findPower10(digits);
    } while (n > test);
    return digits;
}

int findPower10(int n) { //function for calculating powers of 10
    int result = 1;
    for (int i = 0; i < n; i++)
        result = result * 10;
    return result;
}

And after running the code, I've figured out that it (barely) mostly works. I've found that whenever a user inputs a value of 10, 100, 1000, etc. it always returns a value of 100. I'd like to solve this only using the iostream header.

Sorry if my code isn't too readable or organized! It would also be helpful if there are any shortcuts to my super long code, thanks!

Upvotes: 0

Views: 1000

Answers (2)

Ricardo Gallego
Ricardo Gallego

Reputation: 36

The problem is in the findDigits function. For the values 10, 100, 1000 etc, it calculates the number of the digits minus one. This happens because of the comparison in the loop, you are stopping when n is less or equal to test, but in these cases n is equal test and you should run the next iteration.

So, you should change the line 33: 

} while (n > test);

to:

} while (n >= test);

Now, it should work just fine. But it will not work for negative numbers (I don't know this is required, but the solution bellow works for that case too).

I came up with a much simpler solution:

int sumOfSquareDigits(int n)
{
    // Variable to mantain the total sum of the squares 
    int sum = 0;
    // This loop will change n until it is zero
    while (n != 0) {
        /// The current digit we will calculate the square is the rightmost digit,
        //   so we just get its value using the mod operator 
        int current = n % 10;
        // Add its square to the sum 
        sum += current*current;
        // You divide n by 10, this 'removes' one digit of n 
        n = n / 10;
    }
    return sum;
}

Upvotes: 2

Dimitrios Bouzas
Dimitrios Bouzas

Reputation: 42929

I found the problem challenging managed to reduce your code to the following lines:

long long sumOfSquareDigits(long long i) {  
  long long sum(0L);
  do {
    long long r = i % 10;
    sum += (r * r);    
  } while(i /= 10);

  return sum;
}

Haven't test it thoroughly but I think it works OK.

Upvotes: 2

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