python打破嵌套循环与多个if功能

Question

我有一个代码，我需要找到一个数字的主要因素。在发现主要因素后，我需要知道如何破解我的代码，所以当它达到1.我刚刚开始使用python进行编码，并且我还没有和其他馆藏一起使用familiair，所以我想知道是否可以制作一个只打破常规的Python代码。 NUM达到1

num = int(input("Give me a number:")) 
priemgetallen = [] 

for reeks in range(2, num+1): 
    print(reeks) 
    for priemgetal in range(2,reeks): 
     if reeks % priemgetal != 0: 
      print(priemgetal) 
      if num%priemgetal == 0: 
       print("This is the old num", num) 
       num = num/priemgetal 
       print("This is the new num", num) 
       priemgetallen.append(priemgetal) 
       if num > 1: 
        if num%priemgetal == 0: 
         print("This is the new num", num) 
         num = num/priemgetal 
         print("This is the old num", num) 
         priemgetallen.append(priemgetal) 
       else: 
        print(priemgetallen) 
        break 
      else: 
       print("Num stays old") 

print(priemgetallen) 

Answer 1

后实际具有if声明任何金额并不重要，应立即停止，break将退出循环你，而不是if语句。

所以你只需要处理嵌套循环，我可以通过将你的代码放入一个函数并使用return退出它来解决这个问题。就像这样：

def compute_priemgetallen(num): 
    priemgetallen = [] 
    for reeks in range(2, num+1): 
     print(reeks) 
     for priemgetal in range(2,reeks): 
      if reeks % priemgetal != 0: 
       print(priemgetal) 
       if num%priemgetal == 0: 
        print("This is the old num", num) 
        num = num/priemgetal 
        print("This is the new num", num) 
        priemgetallen.append(priemgetal) 
        if num > 1: 
         if num%priemgetal == 0: 
          print("This is the new num", num) 
          num = num/priemgetal 
          print("This is the old num", num) 
          priemgetallen.append(priemgetal) 
        else: 
         return priemgetallen # here we break out of all loops 
       else: 
        print("Num stays old") 


num = int(input("Give me a number:")) 
print(compute_priemgetallen(num))