此文為延續 Python 找質數 一文，在解此題之前，請先看質因數分解的影片講解。
要解這題，有幾種作法：
方法一：
1. 除數從 2 開始，若此被除數可以被除數 2 整除，就被除數就一直除以 2；若不行除數換用 3。
2. 一直重複到 除數的平方大於或小於 被除數。
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| def prime_factoring(n): | |
| divisor = 2 | |
| factors = [] | |
| while divisor * divisor <= n: | |
| if n % divisor: | |
| divisor += 1 | |
| else: | |
| n //= divisor | |
| factors.append(divisor) | |
| if n > 1: | |
| factors.append(n) | |
| return factors | |
| aNumber = (int)(input("Enter a number:")); | |
| print(prime_factoring(aNumber)); |
方法二：
1. 求出小於此被除數開根號後的所有質數。
2. 在用這些質數去除被除數算出每個質因數出現的次數。
2. 在用這些質數去除被除數算出每個質因數出現的次數。
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| import math; | |
| def eratosthenes(n): | |
| P = [i for i in range(2, (int)(math.sqrt(n)+1))] | |
| p = 0 | |
| while True: | |
| for i in P[p + 1:]: | |
| if i % P[p] == 0: | |
| P.remove(i) | |
| if P[p]**2 >= P[-1]: | |
| break | |
| p += 1 | |
| return P | |
| aNumber = (int)(input("Enter a number:")); | |
| factors = eratosthenes(aNumber); | |
| num = aNumber; | |
| for divisor in factors: | |
| while( num % divisor == 0 ): | |
| num //= divisor | |
| print(divisor, ' ', end = ''); |
方法二好像不怎麼好XD。
除了以上方法外，讀者您還有想到什麼方法呢？
除了以上方法外，讀者您還有想到什麼方法呢？