Question is to find two numbers within 1 to num that multiply to become num

Question

t=int(input())
while t > 0:
    num=int(input())
    lst=[]
    n=0
    for i in range(1,math.floor(math.sqrt(num))+1):
        if num%i==0:
            lst.append(i)
    n=len(lst)
    if n == 1:
        print(1,num)
    else:
        print(lst[1],int(num/lst[1]))
    t-=1

Question is to find two numbers within 1 to num that multiply to become num(for prime it is only 1, num and for rest any 2 factors). The time limit exceeded error is the thing that I am seeing now. Can anyone help me in optimizing it?

That's kind of a weird problem. If you only need to return two numbers, can't you just return `1` and `num`? — Samwise, Dec 19 '20 at 16:47
noooooo for prime numbers 1 and num and for rest, other factors — Chandrachud Pati, Dec 19 '20 at 16:49
Maybe only iterate up to `floor(sqrt(num))` since anything after that point is going to be a combination you've already tried? — Samwise, Dec 19 '20 at 16:50

PILLI VINEETH · Answer 1 · 2020-12-19T17:46:28.780

0

You can optimise the code by changing the limit inside the range to sqrt(num).I hope this solves your problem.

You can try like this:

 import math
 t= int(input())
 while(t>0):
     num = int(input())
     lst = []
     for i in range(1,(math.sqrt(num))+1):
          if(num%i == 0):
              lst.append((i,num/i))
              break
     print(lst[0])
     t= t-1

edited Dec 19 '20 at 17:46

answered Dec 19 '20 at 17:05

PILLI VINEETH

21
2

Can you tell me which part is wrong in this? – PILLI VINEETH Dec 19 '20 at 17:45
```lst.append``` part, it is showing append() takes 2 positional arguments but 3 were given-this error – Chandrachud Pati Dec 19 '20 at 18:16
Yeah,I actually forgot to close one bracket.That is why this error is coming.Now I have fixed it.Can you try once again? – PILLI VINEETH Dec 19 '20 at 18:27

Tomerikoo · Answer 2 · 2020-12-19T17:12:53.720

0

If you just need to find any 2 factors, there is no need to use a list. Just break after the first one found.
As suggested in the comments, there is no need to iterate until num, as the factors will start to repeat. It is enough to iterate until the square.
Lastly, you can take advantage of the built-in divmod function to avoid calculating the modulus and then the division.

from math import sqrt

num = int(input())
for n in range(2, int(sqrt(num)) + 1):
    div, mod = divmod(num, n)
    if mod == 0:
        print(n, div)
        break
else:
    print(1, num)

This uses the for/else construct so in case no factors were found, just 1, num will be printed which is always true.

edited Dec 19 '20 at 17:12

answered Dec 19 '20 at 17:06

Tomerikoo

18,379
16
47
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try this code for ```num=45``` sir,you won't get any result – Chandrachud Pati Dec 19 '20 at 17:19
I don't know then, why it isn't showing in my case. Though all things are correct here – Chandrachud Pati Dec 19 '20 at 17:21
@ChandrachudPati make sure you are running correctly. There is no way to don't get any output here. As long as the loop didn't break, the `else` is executed. If the loop `break`s, then there is a print right before it... – Tomerikoo Dec 19 '20 at 17:22
Have a look sir, in my answer – Chandrachud Pati Dec 19 '20 at 17:29
Well, it is not equivalent to mine.... The `break` should be at the same level of the print... Please delete the answer as well – Tomerikoo Dec 19 '20 at 17:31
Sir, I have tried your code even, but it is showing time limit error, the exact one, break just after print and same identation – Chandrachud Pati Dec 19 '20 at 17:31
Well I really don't know how to optimize it further... I am not sure it is possible. You do have to go over all those numbers to check them in the end... – Tomerikoo Dec 19 '20 at 17:33

Question is to find two numbers within 1 to num that multiply to become num

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