[英]Combinations for three numbers to sum up to 1000
I need every combination of three positive integers with the sum of 1000. 我需要三个正整数的每个组合,总和为1000。
This was my attempt but I'm unsure if this is correct since I have no way to validate it. 这是我的尝试,但我不确定这是否正确,因为我无法验证它。
def getSum():
l = []
for x in range(1, 999):
total = 1000-x
for y in range(1, 999):
total = total-y
if total>0:
l.append([x, y, total])
return l
print len(getSum())
I get 28776 different combinations. 我得到了28776种不同的组合。 Is that correct?
那是对的吗?
Since 1+998+1
and 1+1+998
are not the same thing, there are some incredible amount of combinations: 由于
1+998+1
和1+1+998
不是一回事,因此有一些令人难以置信的组合:
This line can generate them all: 这一行可以生成所有:
[(i, 1000-i-k, k) for i in range(1,999) for k in range(1,1000-i)]
Results: 结果:
[...
(1, 4, 995),
(1, 3, 996),
(1, 2, 997),
(1, 1, 998),
(2, 997, 1),
(2, 996, 2),
...]
The length of this list is: 这个清单的长度是:
498501
No, that number is not correct. 不,这个数字不正确。 The problem with your code is this line:
你的代码的问题是这一行:
total = total-y
Here, you decrease total
further and further with each value of y
that you try, never resetting it to the value after just subtracting x
. 在这里,您可以降低
total
越走越用的每个值y
你尝试,它只是减去后从未重置价值x
。 To fix it, create a new variable, eg total2
, and use that in the inner loop. 要修复它,请创建一个新变量,例如
total2
,并在内部循环中使用它。
total2 = total-y
This way, you get 498501
combinations. 这样,您就获得了
498501
组合。 Also, you can break
from the inner loop as soon as total2 < 0
. 此外,只要
total2 < 0
,就可以从内循环break
。
If you need just the number of combinations: Note that there are N-1
combinations to sum two numbers to N
, eg for N==4
: 1+3
, 2+2
, 3+1
(assuming you consider 1+3
and 3+1
different). 如果你只需要组合的数量 :注意有
N-1
组合将两个数字相加到N
,例如对于N==4
: 1+3
2+2
3+1
(假设你考虑1+3
和3+1
不同)。 You can extend this to the case of three numbers as partitioning the number in two parts two times. 您可以将此扩展为三个数字的情况,将数字分为两部分两次。 This way, you only need a single loop.
这样,您只需要一个循环。 And this can be simplified further to an O(1) formula.
并且这可以进一步简化为O(1)公式。
Example, with naive approach using product
as reference: 例如,使用
product
作为参考的天真方法:
>>> N = 100 # to make reference faster
>>> sum(1 for t in product(range(1, N+1), repeat=3) if sum(t)==N)
4851
>>> sum(N-1-i for i in range(1, N-1))
4851
>>> ((N-2)*(N-1))//2
4851
Of course, also works for N = 1000
(or much, much larger): 当然,也适用于
N = 1000
(或更多,更大):
>>> N = 1000
>>> sum(N-1-i for i in range(1, N-1))
498501
>>> ((N-2)*(N-1))//2
498501
If you treated [1,1,998] and [1,998,1] the same (no unique integers): 如果你对[1,1,998]和[1,998,1]的处理方式相同(没有唯一的整数):
def getSum():
l = []
for x in range(1, 999):
total = 1000-x
for y in range(1, 999):
total = total-y
if total>0:
z = [x, y, total]
z.sort()
if z not in l:
l.append(z)
return l
a = getSum()
print(len(a))
If you want 3 unique integers: 如果你想要3个唯一整数:
def getSum():
l = []
for x in range(1, 999):
total = 1000-x
for y in range(1, 999):
total = total-y
if total>0:
z = [x, y, total]
z.sort()
if (z not in l) and (not((len(set(z)) < len(z)))):
l.append(z)
return l
a = getSum()
print(len(a))
Otherwise your code is (in my sense) ok. 否则你的代码(在我看来)是好的。 I haven't check your answer yet...
我还没检查你的答案......
EDIT : I have checked it using brutal force. 编辑:我用残酷的力量检查了它。 The correct answer is actually 498501 if you treated (1,1,998) and (998,1,1) differently.
如果您对(1,1,998)和(998,1,1)的处理方式不同,那么正确的答案实际上是498501。 Currently I don't know why...
目前我不知道为什么......
Try this: 尝试这个:
def getSum():
l = []
for x in range(1, 6):
for y in range(1, 6):
total = 6-(y+x)
if total>0:
s = set([x, y, total])
if s not in l:
l.append(s)
print(x, y, total)
return l
print (len(getSum()))
This is my algorithm, Although there is better ways. 这是我的算法,虽然有更好的方法。 In this case I wrote code for number 6 and printed all combinations to show how it works.
在这种情况下,我编写了6号代码并打印了所有组合以显示它是如何工作的。 You can set 1000 or any number instead of 6 in this code(in 3 position) and ignore print() line.
您可以在此代码中设置1000或任何数字而不是6(在3位置)并忽略print()行。
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