python分解接收数字的数字函数和分解方法

这是一个非常自然的递归公式:

from copy import copy
def recurse_find(decomposed,remaining,valid_numbers):
    #base case
    if remaining == 0:
        return decomposed
    #find all valid subtractions
    else:
        ans = []
        for number in valid_numbers:
            if remaining - number >= 0:
                new_decomposed = copy(decomposed)
                new_decomposed.append(number)
                cand = recurse_find(new_decomposed,remaining- 
                              number,valid_numbers)
                if cand:
                    ans.append(cand)
        if len(ans) > 0:
            return ans

print(recurse_find([],5,[1,2,5]))-->与您请求的输出匹配。

试用使用 itertools.product 对于列表理解中的不同重复,则在列表理解之外进行检查:

import itertools
def func(n,l):
   return [a for i in [itertools.product(l,repeat=x) for x in range(1,6)] for a in i if sum(a) == n]

print(func(5,[1,2,5]))

输出:

[(5,), (1, 2, 2), (2, 1, 2), (2, 2, 1), (1, 1, 1, 2), (1, 1, 2, 1), (1, 2, 1, 1), (2, 1, 1, 1), (1, 1, 1, 1, 1)]

如果需要列表:

print(list(map(list,func(5,[1,2,5]))))

输出:

[[5], [1, 2, 2], [2, 1, 2], [2, 2, 1], [1, 1, 1, 2], [1, 1, 2, 1], [1, 2, 1, 1], [2, 1, 1, 1], [1, 1, 1, 1, 1]]