Product of three numbers, if there is any, else print NO (python)-CodePudding

I have been given the following homework to solve in python.

You are given one integer number n. Find three distinct integers a,b,c such that 2≤a,b,c, and a⋅b⋅c=n or say that it is impossible to do it.

If there are several answers, you can print any.

You have to answer t independent test cases. The first line of the input contains one integer t (1≤t≤100) — the number of test cases.

The next n lines describe test cases. The i-th test case is given on a new line as one integer n.

Input: 5=t, 64, 32, 97, 2, 12345

Output: YES, 2 4 8 , NO, NO, NO, YES, 3 5 823

I have tried and it works except for the last example output. I am not sure what is wrong and how to solve this. Any help would be appreciated. (I know it would be better to use 'or' but I wanted to make sure I include every condition to see better whats wrong.)

import math
def divisors(t,n):
  for k in range(t,0,-1):
    a=0
    b=0
    c=0
    k=0
    l=n
    for i in (2,math.sqrt(n)):
         if n%i==0 :
           if k==0:
              a=i
              n =n/a
           elif k==1:
              b=i
           k=k 1
         if k==2:
           break
    if (a==1 or b==1):
      print('NO')
      break
    if a*b==0:
      print('NO')
      break
    if a==b:
      print('NO')
      break
    c=l/(a*b)
    if c==b:
      print('NO')
      break
    if a==c:
      print('NO')
      break
    if(c<2):
      print('NO')
      break
    else:
      print('YES', int(a ),int(b ), int(c))
      break

CodePudding user response：

My attempt:

import numpy as np
def div(cases): #Python can work without indexing (No of cases).
    for i in cases:
        r = [] #Here, I store all the factors. Is reset every case.
        for j in range(2,i 1): #Loop through all the possible divisors.
            if i % j == 0: #Check if the case has no remainder with the current factor.
                r  = [j] #If factor found, add to list
                i = i//j #Divide the case by the found factor
        if len(r) < 3:
            print('NO')
        else:
            print(f'YES: {[r[0], r[1], np.prod(r[2:])]}') #Print 3 and 3 numbers only
            #print(f'YES: {r}')

Input:

div([64, 32, 97, 2, 12345])

Output:

YES: [2, 4, 8]
NO
NO
NO
YES: [3, 5, 823]

Run it here.