CODEWITHSUNDEEP
python
Nilou
Nilou

Reputation: 9

how to write a code about perfect numbers in python?

I was trying to write a code including a function in order to find numbers that are complete(perfect number:the number which is = its denominator's summation) but I had some problem with it...would you help me?

L=[]
def denominator(n):

    k=1
    while(int(k) < int(n)):
        s=int(n%k)
        if(s==0):
            L+=[s]
    k+1
    return L

def isPerfect(n):

    return denominator(n)
    a=sum(L)
    return(a==n)   

n=input("enter a number:")

if(isPerfect(n)):

    print("it is perfect.")
else:
    print("it's not perfect.")

Upvotes: 0

Views: 452

Answers (3)

Jithin Pavithran
Jithin Pavithran

Reputation: 1351

I think, this is more pythonic code for this problem:

if sum([x for x in range(1,n) if n%x==0])==n and n>0:
     print "complete"
else:
     print " Not complete"

Upvotes: 1

Serenity
Serenity

Reputation: 36695

You have a lot of mistakes in your code. L has to be defined in denominator; you has to add k not s to L; in while you has to increment k; denominator has to return list of dividers; you has to convert n to int after input and it is enough to check dividers until [n/2+1]. After removing these misprints your code will work:

def denominator(n):
    L = []
    k=1
    while(k < int(n/2.+1)):
        if(n%k==0):
            L+=[k,]
        k=k+1
    return L

def isPerfect(n):
    L = denominator(n)
    a=sum(L)
    return(a==n)   

n=input("enter a number:")

if(isPerfect(int(n))):
    print("it is perfect.")
else:
    print("it's not perfect.")

However you may reduce your code like:

def isPerfect(n):  
    if (n < 1): return False
    sum = 0  
    for k in range(1, int(n/2+1)):  
        if n % k == 0:  
            sum += k  
    return sum == n  
print(isPerfect(6), isPerfect(28))

Upvotes: 0

Anis
Anis

Reputation: 3094

With @Serenity's answer and the comments to your post I think you have all that you need to correct the errors in your code. Let me also give you this nearly one-liner that does what you want:

def is_perfect(n):
    return False if n <= 0 else (sum([s for s in range(1, n) if n%s == 0]) == n)

Upvotes: 1

Related Questions