How do you take a negative value in python?

I am a student in a concepts of programming class. The lab is run by a TA and today in lab he gave us a real simple little program to build. It was one where it would multiply by addition. Anyway, he had us use absolute to avoid breaking the prog with negatives. I whipped it up real quick and then argued with him for 10 minutes that it was bad math. It was, 4 * -5 does not equal 20, it equals -20. He said that he really dosen't care about that and that it would be too hard to make the prog handle the negatives anyway. So my question is how do I go about this.

here is the prog I turned in:

#get user input of numbers as variables

numa, numb = input("please give 2 numbers to multiply seperated with a comma:")

#standing variables
total = 0
count = 0

#output the total
while (count< abs(numb)):
    total = total + numa
    count = count + 1

#testing statements
if (numa, numb <= 0):
    print abs(total)
else:
    print total

I want to do it without absolutes, but every time I input negative numbers I get a big fat goosegg. I know there is some simple way to do it, I just can't find it.

asked Mar 16, 2010 at 2:37

3

Perhaps you would accomplish this with something to the effect of

text = raw_input("please give 2 numbers to multiply separated with a comma:")
split_text = text.split(',')
a = int(split_text[0])
b = int(split_text[1])
# The last three lines could be written: a, b = map(int, text.split(','))
# but you may find the code I used a bit easier to understand for now.

if b > 0:
    num_times = b
else:
    num_times = -b

total = 0
# While loops with counters basically should not be used, so I replaced the loop 
# with a for loop. Using a while loop at all is rare.
for i in xrange(num_times):
    total += a 
    # We do this a times, giving us total == a * abs(b)

if b < 0:
    # If b is negative, adjust the total to reflect this.
    total = -total

print total

or maybe

a * b

answered Mar 16, 2010 at 2:52

Mike GrahamMike Graham

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1

Too hard? Your TA is... well, the phrase would probably get me banned. Anyways, check to see if numb is negative. If it is then multiply numa by -1 and do numb = abs(numb). Then do the loop.

answered Mar 16, 2010 at 2:50

2

The abs() in the while condition is needed, since, well, it controls the number of iterations (how would you define a negative number of iterations?). You can correct it by inverting the sign of the result if numb is negative.

So this is the modified version of your code. Note I replaced the while loop with a cleaner for loop.

#get user input of numbers as variables
numa, numb = input("please give 2 numbers to multiply seperated with a comma:")

#standing variables
total = 0

#output the total
for count in range(abs(numb)):
    total += numa

if numb < 0:
    total = -total

print total

answered Mar 16, 2010 at 3:03

slackerslacker

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Try this on your TA:

# Simulate multiplying two N-bit two's-complement numbers
# into a 2N-bit accumulator
# Use shift-add so that it's O(base_2_log(N)) not O(N)

for numa, numb in ((3, 5), (-3, 5), (3, -5), (-3, -5), (-127, -127)):
    print numa, numb,
    accum = 0
    negate = False
    if numa < 0:
        negate = True
        numa = -numa
    while numa:
        if numa & 1:
            accum += numb
        numa >>= 1
        numb <<= 1
    if negate:
        accum = -accum
    print accum

output:

3 5 15
-3 5 -15
3 -5 -15
-3 -5 15
-127 -127 16129

answered Mar 16, 2010 at 4:13

John MachinJohn Machin

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3

How about something like that? (Uses no abs() nor mulitiplication)
Notes:

• the abs() function is only used for the optimization trick. This snippet can either be removed or recoded.
• the logic is less efficient since we're testing the sign of a and b with each iteration (price to pay to avoid both abs() and multiplication operator)

def multiply_by_addition(a, b):
""" School exercise: multiplies integers a and b, by successive additions.
"""
   if abs(a) > abs(b):
      a, b = b, a     # optimize by reducing number of iterations
   total = 0
   while a != 0:
      if a > 0:
         a -= 1
         total += b
      else:
         a += 1
         total -= b
   return total

multiply_by_addition(2,3)
6
multiply_by_addition(4,3)
12
multiply_by_addition(-4,3)
-12
multiply_by_addition(4,-3)
-12
multiply_by_addition(-4,-3)
12

answered Mar 16, 2010 at 3:13

mjvmjv

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4

Thanks everyone, you all helped me learn a lot. This is what I came up with using some of your suggestions

#this is apparently a better way of getting multiple inputs at the same time than the 
#way I was doing it
text = raw_input("please give 2 numbers to multiply separated with a comma:")
split_text = text.split(',')
numa = int(split_text[0])
numb = int(split_text[1])

#standing variables
total = 0

if numb > 0:
    repeat = numb
else:
    repeat = -numb

#for loops work better than while loops and are cheaper
#output the total
for count in range(repeat):
    total += numa


#check to make sure the output is accurate
if numb < 0:
    total = -total


print total

Thanks for all the help everyone.

answered Mar 16, 2010 at 23:42

dman762000dman762000

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5

Try doing this:

num1 = int(input("Enter your first number: "))
num2 = int(input("Enter your second number: "))
ans = num1*num2


if num1 > 0 or num2 > 0:
    print(ans)

elif num1 > 0 and num2 < 0 or num1 < 0 and num1 > 0:
    print("-"+ans)

elif num1 < 0 and num2 < 0:
    print("Your product is "+ans)
else:
    print("Invalid entry")

answered Dec 2, 2021 at 2:46

import time

print ('Two Digit Multiplication Calculator')
print ('===================================')
print ()
print ('Give me two numbers.')

x = int ( input (':'))

y = int ( input (':'))

z = 0

print ()


while x > 0:
    print (':',z)
    x = x - 1
    z = y + z
    time.sleep (.2)
    if x == 0:
        print ('Final answer: ',z)

while x < 0:
    print (':',-(z))
    x = x + 1
    z = y + z
    time.sleep (.2)
    if x == 0:
        print ('Final answer: ',-(z))

print ()  

answered Nov 10, 2014 at 20:34

KitiyoKitiyo

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1