SPOILER ALERT！这可能会影响你的回答＃3
我设法获得了一段代码，但由于我正在分析的数量很多，因此计算解决方案需要花费很长时间。
我觉得蛮力不是正确的方法......
有助于提高此代码的效率吗？
# What is the largest prime factor of the number 600851475143
# Set variables
number = 600851475143
primeList = []
primeFactorList = []
# Make list of prime numbers < 'number'
for x in range(2, number+1):
isPrime = True
# Don't calculate for more than the sqrt of number for efficiency
for y in range(2, int(x**0.5)+1):
if x % y == 0:
isPrime = False
break
if isPrime:
primeList.append(x)
# Iterate over primeList to check for prime factors of 'number'
for i in primeList:
if number % i == 0:
primeFactorList.append(i)
# Print largest prime factor of 'number'
print(max(primeFactorList))
答案 0 :(得分：1)
您可以使用用户定义的功能，例如
<head>
<script>
function add() {
var lookupType = document.getElementById("lookupType");
var typeID = -1;
if (lookupType != null) typeID = lookupType.value;
document.forms[0].action = "ManageLookupSupport!add.action?" + encodeURI("keyValue.keyType=" + typeID);
document.forms[0].submit();
}
</script>
</head>
<div class="container-fluid" id="page">
<div id="page-wrapper">
<s:form theme="simple">
<div class="panel-body" align="center">
<s:submit cssClass="btn btn-primary" value="Add"
onclick="javascript:add();return false"
data-toggle="modal" data-target="#lookupModal"/>
</div>
</s:form>
</div>
</div>
它将返回布尔值。您可以使用此功能来验证素因子。
OR
你可以连续将数字除以2
def isprime(n):
if n < 2:
return False
for i in range(2,(n**0.5)+1):
if n % i == 0:
return False
return True
n必须是奇数现在跳过2 in for循环，然后打印每个除数
n = 600851475143
while n % 2 == 0:
print(2),
n = n / 2
此时，n将等于1，除非n是素数。所以
for i in range(3,n**0.5+1,2):
while n % i== 0:
print(i)
n = n / i
将自己打印为主要因素。
玩得开心
答案 1 :(得分：1)
我首先要解决您尝试的特定算法中的一些基本问题：
您不需要预先生成素数。在您需要的时候即时生成它们 - 您还会看到您生成的素数超出了您的需求（您只需要尝试素数到sqrt(600851475143)）
# What is the largest prime factor of the number 600851475143
# Set variables
number = 600851475143
primeList = []
primeFactorList = []
def primeList():
# Make list of prime numbers < 'number'
for x in range(2, number+1):
isPrime = True
# Don't calculate for more than the sqrt of number for efficiency
for y in range(2, int(x**0.5)+1):
if x % y == 0:
isPrime = False
break
if isPrime:
yield x
# Iterate over primeList to check for prime factors of 'number'
for i in primeList():
if i > number**0.5:
break
if number % i == 0:
primeFactorList.append(i)
# Print largest prime factor of 'number'
print(max(primeFactorList))
使用生成器（请参阅yield？）PrimeList()甚至可以通过将质数更改为永久返回素数：
def primeList():
x = 2
while True:
isPrime = True
# Don't calculate for more than the sqrt of number for efficiency
for y in range(2, int(x**0.5)+1):
if x % y == 0:
isPrime = False
break
if isPrime:
yield x
x += 1
虽然我无法帮助，但稍微优化它以跳过大于2的偶数：
def primeList():
yield 2
x = 3
while True:
isPrime = True
# Don't calculate for more than the sqrt of number for efficiency
for y in range(2, int(x**0.5)+1):
if x % y == 0:
isPrime = False
break
if isPrime:
yield x
x += 2
如果你放弃了枚举素数的初步想法并一次针对number尝试一个素数，那么还有另一种选择：相反直接处理number并将其排除在外 - 即，做什么botengboteng suggests并直接分解数字。
这会快得多，因为我们现在检查的数字要少得多：
number = 600851475143
def factors(num):
factors = []
if num % 2 == 0:
factors.append(2)
while num % 2 == 0:
num = num // 2
for f in range(3, int(num**0.5)+1, 2):
if num % f == 0:
factors.append(f)
while num % f == 0:
num = num // f
# Don't keep going if we're dividing by potential factors
# bigger than what is left.
if f > num:
break
if num > 1:
factors.append(num)
return factors
# grab last factor for maximum.
print(factors(number)[-1])
答案 2 :(得分：0)
首先计算你的数字的sqrt。
number = 600851475143
number_sqrt = (number**0.5)+1
然后在您的最外层循环中，仅搜索值小于您的数字的sqrt根的素数。您将不需要任何高于此的素数。 （您可以根据您的列表推断任何更大的内容。）
for x in range(2, number_sqrt+1):
要推断最大因素，只需将您的数字除以因子列表中的项目，包括它们之间的任何组合，并确定结果是否为素数。
无需重新计算素数列表。但是要定义一个函数来确定数字是否为素数。
我希望我很清楚。祝好运。非常有趣的问题。
答案 3 :(得分：0)
我制作了这段代码，如果您有任何问题可以自由评论，一切都会被解释
def max_prime_divisor_of(n):
for p in range(2, n+1)[::-1]: #We try to find the max prime who divides it, so start descending
if n%p is not 0: #If it doesn't divide it does not matter if its prime, so skip it
continue
for i in range(3, int(p**0.5)+1, 2): #Iterate over odd numbers
if p%i is 0:
break #If is not prime, skip it
if p%2 is 0: #If its multiple of 2, skip it
break
else: #If it didn't break it, is prime and divide our number, we got it
return p #return it
continue #If it broke, means is not prime, instead is just a non-prime divisor, skip it
如果你不知道：它在range(2, n+1)[::-1]中的作用与reversed(range(2, n+1))相同，所以这意味着不是以 2 < / strong>，它以 n 开头，因为我们正在搜索最大素数。 （基本上它会颠倒列表以便以这种方式开始）
def max_prime_divisor_of(n): #Decompose by its divisor
while True:
try:
n = next(n//p for p in range(2, n) if n%p is 0) #Decompose with the first divisor that we find and repeat
except StopIteration: #If the number doesn't have a divisor different from itself and 1, means its prime
return n
如果你不知道：它在next(n//p for p in range(2, n) if n%p is 0)中的作用是获得第一个除数为n的数字