Question

我正在尝试在循环中使用break语句。我正在尝试构建一个函数，该函数用户userInput找到小于给定数字的最大素数。这是我的代码：

def primeFinder():
    aVar = 2 # This is a possible factor
    divisorList = []  # This is a factor list of userInout, the inputted number
    while aVar <= userInput: # Checking until the variable cannot be a factor
            if userInput % aVar == 0: # If userInput divided by the variable has a remainder of zero, it means it is a factor
                    divisorList.append(aVar) # Adding the factor to th factor list
                    aVar+=1 # Making the possible factor numeral one bigger to test again
            else:
                    aVar +=1 # Making the possible factor numeral one bigger to test again

    print divisorList
    if len(divisorList) > 1:  # If there are more than one factor...
            print "Your input",userInput, " is not prime" # It is not prime
    else:   
            print "Your input", userInput," is prime."  # It is prime

当我找到最大的素数时，我需要知道在哪里放置break语句来停止程序。我把它放在哪里，为什么？返回功能不起作用。这是发生的事情：

[2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, 500, 1000]
Your input 1000  is not prime
[3, 9, 27, 37, 111, 333, 999]
Your input 999  is not prime
[2, 499, 998]
Your input 998  is not prime
[997]
Your input 997  is prime.
[2, 3, 4, 6, 12, 83, 166, 249, 332, 498, 996]
Your input 996  is not prime
[5, 199, 995]
Your input 995  is not prime
[2, 7, 14, 71, 142, 497, 994]
Your input 994  is not prime
[3, 331, 993]
Your input 993  is not prime
[2, 4, 8, 16, 31, 32, 62, 124, 248, 496, 992]
Your input 992  is not prime
[991]
Your input 991  is prime.
[2, 3, 5, 6, 9, 10, 11, 15, 18, 22, 30, 33, 45, 55, 66, 90, 99, 110,     165,     198, 330, 495, 990]
Your input 990  is not prime

Answer 1

我认为你最好从上到下迭代以获得最大的素数因子。像这样：

primeFactor=1
for aVar in range(userInput-1,1,-1): #loop from userInput-1 to 2
    if userInput % aVar == 0:
        primeFactor=aVar
        break;

print "The largest prime factor is "+str(primeFactor).

Answer 2

到目前为止你的代码看起来很不错。你所描述的功能是找到一个给定的（单个）整数是素数。你现在要做的就是从1000开始直到找到素数为止。

根据数字是否为素数，我认为返回True或False更清晰。接受userInput作为你的主要发现功能的论据也是一个小清洁。所以，你可以改变你的功能：

def primeFinder(userInput):
    aVar = 2 # This is a possible factor
    divisorList = []  # This is a factor list of userInout, the inputted number
    while aVar <= userInput: # Checking until the variable cannot be a factor
            if userInput % aVar == 0: # If userInput divided by the variable has a remainder of zero, it means it is a factor
                    divisorList.append(aVar) # Adding the factor to th factor list
                    aVar+=1 # Making the possible factor numeral one bigger to test again
            else:
                    aVar +=1 # Making the possible factor numeral one bigger to test again

    #print divisorList
    if len(divisorList) > 1:  # If there are more than one factor...
            return False
    else:
            return True

然后，要找到最大值（根据您的问题使用休息），您可以这样做：

for num in range(1000, -1, -1): # count down from 1000
    if primeFinder(num):
        print "The largest prime under 1000 is:"
        print num
        break # Break out of loop after we've found largest result, as per question

Answer 3

当n太大时，尝试使用xrange以获得更好的性能：

from math import sqrt


def is_prime(n):
    if n in (2, 3):
        return True

    for i in xrange(2, int(sqrt(n)) + 1):
        if n % i == 0:
            return False

    return True


def find_largest_prime(n):
    for i in xrange(n - 1, 1, -1):
        if is_prime(i):
            return i

print(find_largest_prime(1000))

如何在这个特定情况下使用break子句

3 个答案: