blocks|key|1661994|text|我想我明白了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1661995|def+sum_divisors(n):
++++return+sum([i+for+i+in+range(1,+n)
++++++++++++++++if+n+%25+i+==+0])

print(sum_divisors(0))
print(sum_divisors(3))+#+Should+sum+of+1
print(sum_divisors(36))+#+Should+sum+of+1%2B2%2B3%2B4%2B6%2B9%2B12%2B18
print(sum_divisors(102))+#+Should+be+sum+of+2%2B3%2B6%2B17%2B34%2B51|code-block|syntax|javascript|1661996|输出|1661997|0
1
55
114|1661998|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I think I got it:

<pre class="lang-py prettyprint-override"><code>def sum_divisors(n):
 return sum([i for i in range(1, n)
 if n % i == 0])

print(sum_divisors(0))
print(sum_divisors(3)) # Should sum of 1
print(sum_divisors(36)) # Should sum of 1+2+3+4+6+9+12+18
print(sum_divisors(102)) # Should be sum of 2+3+6+17+34+51
</code></pre>

Output

<pre><code>0
1
55
114
</code></pre>

blocks|key|2563434|text|我们需要先找到数字的除数，然后再求和。找出一个数是否是另一个数的除数非常简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2563435|if+x+%25+y+==+0:
++is_divisor+=+True|code-block|syntax|javascript|2563436|Modulo+-%25-返回除法的余数。9%252=1.10%253=1。但是，我们需要知道要检查哪些数字。我们可以遍历每个数字直到我们的输入：|2563437|for+i+in+range(n):
++if+n+%25+i+==+0:
++++is_divisor+=+True|2563438|...but，这太多了。例如，我们知道n的最高可能因子不是n本身，如果n是偶数，则是1/2n，如果不是偶数，则更低。相反，让我们专注于获取每一对，然后检查是否应该继续。我们还需要存储所有这些数字，所以让我们创建一些列表：|2563439|lower_divisors+=+[]
upper_divisors+=+[]|2563440|我们想知道我们当前最高的下除数和最低的最高除数是什么，也知道从哪里开始。我们可以用1填充较低的除数列表，用n填充最高的除数列表。我们可以考虑稍后删除n。|2563441|lower_divisors+=+[1]
upper_divisors+=+[n]

continu+=+True

while+continu:
++#+Let's+get+the+last+item+in+each+list
++highest_low+=+lower_divisors[-1]
++lowest_high+=+upper_divisors[-1]

++for+i+in+range(highest_low+%2B+1,+lowest_high):
++++if+n+%25+i+==+0:
++++++lower_divisors.append(i)
++++++upper_divisors.append(n+//+i)
++++++break
++continu+=+False

#+If+we've+left+the+loop,+we+have+all+of+our+divisors.
lower_sum+=+sum(lower_divisors)
higher_sum+=+sum(upper_divisors)+-+n+++#Gotta+take+it+back+out!
sm+=+lower_sum+%2B+higher_sum
return+sm|2563442|这应该是一种快速实现目标的方法，而不会让你的计算机工作更困难-但是...这需要更多的步骤来编写。我写了更多的步骤，以清楚我正在做什么，并使其具有可读性，但它肯定可以修剪很多。|2563443|在运行代码时，当n超过8位时，循环遍历从1到n的每个数字的简单方法开始真正积累起来。在我的机器上，9位数大约需要12-16秒。上述较长的方法在不到1/10秒的时间内返回32位的n值。|2563444|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|16|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|17|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|18|8|@]|9|@]|A|$]]]|X|$]]

We need to find the divisors of the number first, and then sum them. Finding out if a number is a divisor of another number is pretty simple:

<pre><code>if x % y == 0:
 is_divisor = True
</code></pre>

Modulo - % - returns the remainder of a division. 9 % 2 = 1. 10 % 3 = 1. However, we'll need to know which numbers to check. We could just loop through every number up to our input:

<pre><code>for i in range(n):
 if n % i == 0:
 is_divisor = True
</code></pre>

...but that's too much. For example, we know that the highest possible divisor of n that isn't n itself, is 1/2n if n is even, and lower if it's not. Instead, let's focus on getting each pair, and then check if we should continue. We also need to store all of these numbers, so lets create some lists:

<pre><code>lower_divisors = []
upper_divisors = []
</code></pre>

We'll want to know what our current highest lower divisor and lowest highest divisor are, and also know where to start. We can populate the lower divisor list with 1 and the highest with n. We can worry about removing n later.

<pre><code>lower_divisors = [1]
upper_divisors = [n]

continu = True

while continu:
 # Let's get the last item in each list
 highest_low = lower_divisors[-1]
 lowest_high = upper_divisors[-1]

 for i in range(highest_low + 1, lowest_high):
 if n % i == 0:
 lower_divisors.append(i)
 upper_divisors.append(n // i)
 break
 continu = False

# If we've left the loop, we have all of our divisors.
lower_sum = sum(lower_divisors)
higher_sum = sum(upper_divisors) - n #Gotta take it back out!
sm = lower_sum + higher_sum
return sm
</code></pre>

That should be a fast way to achieve your goal, without making your computer work harder - but... it's a lot more steps to write. I wrote it in far more steps to make it clear what I was doing and to be readable, but it could definitely be trimmed up a lot.

Upon running the code, the simple method of looping through each number from 1 to n starts to really add up when n is over 8 digits. 9 digit numbers were taking around 12-16 seconds on my machine. The longer approach above returns 32-digit values for n in under 1/10th of a second.

blocks|key|1662045|text|您可以使用组合的filter()和lambda()函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1662046|num+=+36
div_sum+=+sum(list(filter(lambda+div+:+num+%25+div+==+0+,+range(1,+num))))

#+divisors+:++[1,+2,+3,+4,+6,+9,+12,+18]
#+divisors+sum+:++55+

print('divisors+sum+:+',+div_sum)|code-block|syntax|javascript|1662047|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could use a combined filter() and lambda() function

<pre class="lang-py prettyprint-override"><code>num = 36
div_sum = sum(list(filter(lambda div : num % div == 0 , range(1, num))))

# divisors : [1, 2, 3, 4, 6, 9, 12, 18]
# divisors sum : 55 

print('divisors sum : ', div_sum)
</code></pre>

Finding the sum of all the divisors of a number, without including that particular number that is being divided.
For example if we want to find the divisors of number 6 they would be 1,2,3. Number 6 should be included as it is a divisor, but for this scenario we are not considering it.
Note:- number 3 and 4 are not included since, when you divide number 6 by number 3, there would be a remainder.
A divisor is a number that divides into another without a remainder.
<pre class="lang-py prettyprint-override"><code>def sum_divisors(n):
 sum = 0
 # Return the sum of all divisors of n, not including n
 return sum

print(sum_divisors(0)) # 0
print(sum_divisors(3)) # Should sum of 1 # 1
print(sum_divisors(36)) # Should sum of 1+2+3+4+6+9+12+18 # 55
print(sum_divisors(102)) # Should be sum of 2+3+6+17+34+51 # 114
</code></pre>

How to find the sum of all the divisors of a number, without including it

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

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求一个数的所有因子的和，不包括被除的那个特定的数。例如，如果我们想要找到数字6的除数，它们应该是1,2,3。数字6应该包括在内，因为它是一个除数，但对于这种情况，我们不考虑它。注意：-数字3和4不包括在内，因为当你将数字6除以数字3时，会有一个余数。除数是一个除以另一个没有余数的数。def sum_divisors(n...

问如何求一个数的所有因子的和，而不包含它
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问如何求一个数的所有因子的和，而不包含它EN