Python 程序使用递归 function 求数字平方和

Question

I want to make a function that gets the sum of the squares of its each digits. Although, I have seen some solutions in the internet, the one I have seen is "getting the sum of the squares of its digits" but given a list. For example, instead of starting at the integer 133, they use [1,3,3] as an input. I tested this function and it works great, but I want to use an integer as an input and not a list.

Edit: For example the given is 123. So the function must return 14. (1^2 + 2^2 + 3^2)

My idea is:

    def squarer(x):                           # where x is an integer (not a list)
        # <insert a recursive formula>
        # return the sum

Answer 1

One simple way to make a recursive function is, to always square the first digit and then call the same function with the remaining digits.

def squarer(x: int) -> int:
    if x < 10:  # trivial case, when x has only one digit
        return x**2
    first_digit = int(str(x)[0])  # extract the first digit
    return first_digit**2 + squarer(int(str(x)[1:]))  # recursive call

Here, I used the functions int() and str() to always access the individual digits and cast the the variable back to an integer.

Answer 2

You can use an integer as an input and transform it into a list.

arr  = [int(x) for x in input("Enter a number: ")]

and then you pass your list into your function.

def squarer(arr):                           # where arr is a list
        # <insert a recursive formula>
        # return the sum

Answer 3

Try to convert your number to string so you can iterate by each digit. And after you can convert string to list or just convert each digit to int before you apply other logic

Answer 4

If I understood the question correctly, this code should solve your problem

It is not of the highest quality but it works:)

def sq(x):

    sum = 0
    while x > 0:
        y = x/10
        r = y*10
        r = x-r
        r = r*r
        sum = sum+r
        x = y

    return sum


print(sq(12))

Answer 5

Recursive and without any transformations:

def squarer(n):
    if n <= 0:
        return 0
    rest, first_digit = divmod(n, 10)
    return first_digit**2 + squarer(rest)