如何让我的 function 处理负数？

Question

repeat = "y"
while repeat == "y":
    #First get the two integers from the user
    a = int(input("Enter the first integer: "))
    b = int(input("Enter the second integer: "))
    #Start the answer with 0
    answer = 0

    print("A", "B")
    print("---")
    print(a, b)
    #run loop until b is not zero
    while b != 0: 
        #loop while 'b' is odd number
        if (b % 2 != 0):
            answer = answer + a 
            print(a*2, b//2) 
            a = a*2 #double every 'a' integers
            b = b//2 #halve the 'b' integers
        #loop while 'b' is even number
        elif (b % 2 == 0):
            print(a*2, b//2)
            a = a*2 #double every 'a' integers
            b = b//2 #halve the 'b' integers
    print("The product is {}.".format(answer))
    repeat = input("Would you like to repeat? (y/n)")
print("Goodbye!")

I am writing a program that uses Ancient Egyptian method to multiply. My program works for positive numbers but not negative. How do I fix it so that if both inputted values of user are negative. My result should give the product of any two positive, negative or one negative and positive number. My current program gives the product for any two positive values, or negative a value and positive b value. However, when user enters a negative b value, it produces infinite outputs.

Answer 1

The issue is with the floor division b//2 , when b is negative the result will be the lower integer, so -0.5 will be rounded to -1 . To avoid it cast to an int a regular division b = int(b / 2) .

After removing duplicate code the while loop looks like that

while b != 0:
    if b % 2 != 0:
        answer = answer + a
    print(a * 2, int(b / 2))
    a = a * 2
    b = int(b / 2)

Edit

To get the correct sign you can check the expected sign after getting the numbers and multiple by 1 or -1 at the end. Since you only check a for the answer you need to work on a positive a

....
answer = 0
sign = 1 if a * b > 0 else -1
a = abs(a)

while b != 0:
    ....

print("The product is {}.".format(answer * sign))

Answer 2

Welcome to Stack Overflow!!

This is what the program returned when I used -2 on the second integer ( b ).

A B
---
1 -2
2 -1
4 -1
8 -1
16 -1

If you notice, B stays on -1, while A starts increasing *2 and it does indifinitely because the while loop doesn't break. Let's look at the code

while b != 0: 
        #loop while 'b' is odd number
        if (b % 2 != 0):
            answer = answer + a 
            print(a*2, b//2) 
            a = a*2 #double every 'a' integers
            b = b//2 #halve the 'b' integers
        #loop while 'b' is even number
        elif (b % 2 == 0):
            print(a*2, b//2)
            a = a*2 #double every 'a' integers
            b = b//2 #halve the 'b' integers

So on the if section, you do a=a*2 and b=b//2 since b%2.=0 (-1%2 is -1), However, for the while to break, you need to get b=0. and you get that by doing b=b//2, The problem, as Guy said before, is that when you get b=-1 (as you saw on my example), doing -1//2 will give you -1. instead of the supposed 0 that you'll get if you do 1//2, Since you won't get b=0. the program never stops multiplying.

The solution is simple, as guy mentioned, use b=int(b/2)

EDIT for the negative numbers

The reason it gives you the wrong sign when you are using multiplication is this instruction

    while b != 0: 
        #loop while 'b' is odd number
        if (b % 2 != 0):
            answer = answer + a #<--------------- THE INSTRUCTION
            a = a*2 #double every 'a' integers
            # b = b//2 #halve the 'b' integers
            b = int(b/2)
            print(a, b) 

answer = answer + a #<--------------- THE INSTRUCTION

since you get your answer by just adding up A, you will have these two wrong scenarios ----> a positive and b negative ->gives you positive (the sign of A) when it should be negative ----> a negative and b negative -> gives you negative (the sign of A) when it should be positive

I tried to find on google how you handled negative numbers with the egyptian method but i didn't find anything, so I guess you can handle that issue with whichever method you prefer. An alternative to Guy method is multiplying the sign (1 or -1) to the result depending of b, not the multiplication

#------------------Your code-------------------------
    #Start the answer with 0
    answer = 0

    print("A", "B")
    print("---")
    print(a, b)
    #run loop until b is not zero

#------------------The fix proposed-------------------------
    #Start the answer with 0
    answer = 0
    sign = 1 if b > 0 else -1          #---------> the fix
    print("A", "B")
    print("---")
    print(a, b)
    #run loop until b is not zero 

and at the end, you multiply the sign with the general answer

#------------------Your code-------------------------
    elif (b % 2 == 0):
            print(a*2, b//2)
            a = a*2 #double every 'a' integers
            b = b//2 #halve the 'b' integers
    print("The product is {}.".format(answer))
    repeat = input("Would you like to repeat? (y/n)")
print("Goodbye!")

#------------------The fix proposed-------------------------
    elif (b % 2 == 0):
            a = a*2 #double every 'a' integers
            b = int(b/2) #---->the previous fix to your first problem
            print(a,b)
            # b = b//2 #halve the 'b' integers
    answer=answer*sign   #---->The another fix: the multiplication i mentioned
    print("The product is {}.".format(answer))
    repeat = input("Would you like to repeat? (y/n)")
print("Goodbye!")

With that, you should have the signs working properly

btw, the reason I changed the print insruction that you used to just print (a,b) at the end of the operation of the if sentence is to avoid redundant operations on the program.