n=1
Sum=0
while n<=20:
    for i in range(1,n 1):
        if(n%i==0):
            Sum =i
        if(Sum==n):
            print(Sum)
    n =1

I need to print all the perfect numbers between 1 to 20.

CodePudding user response：

Perfect number is defined here

To achieve that, I would isolate the perfect number check in a function as follows:

def is_perfect(n):
    s = 0
    for i in range(1,n//2 1):
        if(n%i==0):
            s  = i
    return n == s    

n=20
print([x for x in range(1, n 1) if is_perfect(x)])

CodePudding user response：

One problem is that you never reset the sum but it should be reset for every new number you try. Otherwise Sum will grow over n.

Another issue is, that according to the definition of a perfect number, the number itself is excluded in the division test. But since you run up to n 1, the number itself will always be added and thus Sum is always greater than n.

Third issue is the indentation of the second if. It will print more often than you expect.

Fixing the code is left as an exercise for the reader. If you want a copy&paste solution, there are probably plenty enough on the Internet.

Besides these primary issues which prevent you from the correct results, the code could be improved in several aspects

• Python coding style: remove the redundant parentheses
• Clean code: naming of the variables
• Reuse: define a method is_perfect(number)
• Use type hints: def is_perfect_number(number:int)->bool:
• Performance: division check needs to run until sqrt(n) only.
• Performance: division check could run in steps of 2 once 2 itself has been tested