如何比较 Python 中素数最多的数字？

Question

我必须编写代码才能在输入时获得 10 个数字。 然后，找出质因数较多的数。 如果两个数具有相同数量的质因数，则打印较大的一个。

例如，对于示例输入： 输入值：

123
43
54
12
76
84
98
678
543
231

Output 值应该是这样的：

678 3

这是我的代码，但是 output 不正确：

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)

Answer 1

使用此答案中的算法，我写了一个算法来解决您的问题。

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)