I have to write a code to get 10 numbers at input. Then, find the number with higher number of prime factors. If two number have the same amount of prime factor, then print the bigger one.

For instance, for sample input of : input values:

123
43
54
12
76
84
98
678
543
231

the Output values should be like this:

678 3

Here is my code, however the output is not correct:

import math

# Below function will print the
# all prime factor of given number
def prime_factors(num):
    # Using the while loop, we will print the number of two's that divide n
    added1 = 0
    added2 = 0
    while num % 2 == 0:
        added1 = 1
        print(2, )
        num = num / 2

    for i in range(3, int(math.sqrt(num))   1, 2):
        # while i divides n , print i ad divide n
        while num % i == 0:
            added2 = 1
            print(i, )
            added2  = 1
            num = num / i
    if num > 2:
        print(num)

    added = added1   added2
    print(added)
    return added
    # calling function


input_numbers = []
for i in range(0,10):
    input_numbers.append(int(input()))

contest = 1
for item in input_numbers:
    if prime_factors(item) > contest:
        contest  = 1
        contest2 = item


print(contest2)

CodePudding user response：

Using an algorithm from this answer I wrote one to solve your problem.

def prime_factors(n):
    i = 2
    factors = []
    while i * i <= n:
        if n % i:
            i  = 1
        else:
            n //= i
            factors.append(i)
    if n > 1:
        factors.append(n)
    return factors

input_numbers = []
for i in range(0,10):
    input_numbers.append(int(input()))

highest_num = 0
highest_amount = 1
for num in input_numbers:
    factors = list(dict.fromkeys(prime_factors(num)))
    factors_amount = len(factors)
    if factors_amount > highest_amount:
        highest_num = num
        highest_amount = factors_amount
    elif factors_amount == highest_amount:
        if num > highest_num:
            highest_num = num

print(highest_num,highest_amount)