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I was doing a basic exercise for the function While, he asked to create a program that found the MMC between 2 numbers... I managed to do, but the program continues to print the answer non-stop on loop. Where I went wrong?
num1 = int(input("Digite um número inteiro:"))
num2 = int(input("Digite outro número inteiro:"))
if num1 > num2:
maior = num1
else:
maior = num2
while True:
if maior % num1 == 0 and maior % num2 == 0:
print(maior)
else:
maior += 1
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By the way a lazy and inefficient version: calculate the common multiples in [a.. a*b] and select the first...
[x for x in range(a,a*b+1) if x%b==0 and x%a==0][0]
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You can simply add one break after printing its "major" variable, so it will interomper the while. That way:
num1 = int(input("Digite um número inteiro:"))
num2 = int(input("Digite outro número inteiro:"))
if num1 > num2:
maior = num1
else:
maior = num2
while True:
if maior % num1 == 0 and maior % num2 == 0:
print(maior)
break
else:
maior += 1
I hope I’ve helped.
EDIT:
Like you said you haven’t seen break try to do this way with a bow for that will make your code even more "clean".
num1 = int(input("Digite um número inteiro:"))
num2 = int(input("Digite outro número inteiro:"))
if num1 > num2:
maior = num1
else:
maior = num2
for i in range(maior):
aux = num1 * i
if (aux % num2) == 0:
mmc = aux
print(mmc)
I hope I helped friend.
It works that way, but there is no way to do it without the break? The book that did the exercise had not taught 'break', so must have.
@Matthew has, I’ve done.
@Matthew made the changes, take a look.
@Victornogueira , this iterative solution seemed to force the break. Does this actually work in Python? As I recall, for Python works on generators, so it should simply ignore changes to the value of i
Yes, @Jeffersonquesado was a mistake, that line is irrelevant. I’ll edit.
To the entrance 4,6, the output should be 12. Check?
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I think this code is wrong, but if you’re gonna do it like this, then do it like this:
num1 = int(input("Digite um número inteiro:"))
num2 = int(input("Digite outro número inteiro:"))
maior = num1 if num1 > num2 else num2
while maior % num1 != 0 or maior % num2 != 0:
maior += 1
print(maior)
Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.
In case, the condition denied.
Test, input 2 and 3; expected result 6; entering loop, 3 % 2 will be non-zero, then the value of maior will not be changed, will be returned 3
@Jeffersonquesquesado como eu disse, this code is too weird, I answered only what was the focus of the question that is to end the loop endlessly and without use break that is horrible.
I also think the parole hearing is weird... wouldn’t it maior = num1 if num1>num2 else num2?
I would do the @Mrghost solution, but this incremental, although inefficient, would generate the right result.
I fixed those slips.
If it were a simple solution maybe, it seems to be on the right track, but I don’t think it needs all that.
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Really what was causing this loop was the condition while True: so there is no parameter that tells Python that this condition can be false and will always run the script. Follow an example here without using the break I hope it serves your purpose:
mdc = num1
i = num2
resto = None
while resto is not 0:
resto = mdc % i
mdc = i
i = resto
resp = (mdc * i) / mdc
print("Resultado: "+str(resp))
Flame x of mdc and everything gets less obscure
Opa worth friend, good observation.
For those who do not know, this is the adaptation of Euclid’s algorithm to calculate the MMC. This works because num1 * num2 = mmc * mdc
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I saw this question and did this function, the advantage is that it receives a list of numbers, and the MMC can be calculated beyond 2 numbers.
def mmc(nums):
max_ = max(nums)
i = 1
while True:
mult = max_ * i
if all(mult%nr == 0 for nr in nums):
return mult
i += 1
>>> mmc([5, 6, 8])
>>> 120
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Maybe a break when finding the desired number? Right after the print?
– Jefferson Quesado
@Jeffersonquesado Yes, it works, but the book still hasn’t taught this break: so there must be some way to do without it
– Mateus
So boot as stop condition, instead of making endless loop
– Jefferson Quesado