Python- Armstrong Number-CodePudding

# Armstrong Number
num= int(input("Enter a number: "))
a=[]
b=[]
while num>0:
  digit=num   #Taking the last digit
  a.append(digit) # creating a list with the individual digits
  num=num//10

for i in a:
  numb=i**len(a) #Calculating the power
  b.append(numb) # Creating a new list with the numbers powered to length of the digits

summ=sum(b)      #Sum of all the digits in the new list "b"
print("Sum is:", summ)

if summ == num:
  print("Yes Armstrong No")
else:
  print("Not Armstrong")

The last summ==num "if" condition is always returning the else condition.

Eg if my number (num) is 371, then 3^3 7^3 1^3 is also 371 which is original number = sum and hence it's an Armstrong number so it should return "Yes Armstrong No" but it returning "No" (else condition)..I am unable to identify the error as summ==num (is true here).

CodePudding user response：

You are overwriting the number.

You can just write:

num = int(input("Enter a number: "))
sum = 0
for digit in str(num):
    sum  = int(digit)**len(str(num))

if sum == num:
  print("Yes Armstrong No")
else:
  print("Not Armstrong")

And it will work.

CodePudding user response：

You should make a copy of your original input:

original_num = int(input("Enter a number: "))
num = original_num

... and check that instead, as you overwrite num a few times:

if summ == original_num: