我正在尝试找到一种方法来显示加起来给定整数的所有可能的X整数集.例如,要获得5个我将拥有的所有2个整数集:
1, 4
2, 3
Run Code Online (Sandbox Code Playgroud)
或者对于产生6的3个整数:
1, 1, 4
1, 2, 3
2, 2, 2
Run Code Online (Sandbox Code Playgroud)
我只需要整数不包括0,它只需要在一组中最多可以工作15次,最多可以工作30次.数.
我甚至不确定这是否有一个数学术语.我觉得它与分解相似吗?
Jas*_*rff 19
这是解决此问题的一种方法:
def sum_to_n(n, size, limit=None):
"""Produce all lists of `size` positive integers in decreasing order
that add up to `n`."""
if size == 1:
yield [n]
return
if limit is None:
limit = n
start = (n + size - 1) // size
stop = min(limit, n - size + 1) + 1
for i in range(start, stop):
for tail in sum_to_n(n - i, size - 1, i):
yield [i] + tail
Run Code Online (Sandbox Code Playgroud)
你可以像这样使用它.
for partition in sum_to_n(6, 3):
print partition
[2, 2, 2]
[3, 2, 1]
[4, 1, 1]
Run Code Online (Sandbox Code Playgroud)
有一个片段在这里:
from itertools import combinations, chain
def sum_to_n(n):
'Generate the series of +ve integer lists which sum to a +ve integer, n.'
from operator import sub
b, mid, e = [0], list(range(1, n)), [n]
splits = (d for i in range(n) for d in combinations(mid, i))
return (list(map(sub, chain(s, e), chain(b, s))) for s in splits)
Run Code Online (Sandbox Code Playgroud)
像这样使用它:
for p in sum_to_n(4):
print p
Run Code Online (Sandbox Code Playgroud)
输出:
[4] [1, 3] [2, 2] [3, 1] [1, 1, 2] [1, 2, 1] [2, 1, 1] [1, 1, 1, 1]