例如，10,[1,2,3]
輸出類似：
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
2 + 2 + 2 +2 + 2
3 + 3 + 3 + 2
3 + 2 + 2 + 2 + 1
注意：是小于等于，list 内的正整數有可能并不能正好等于 a.
通過itertools.combinations_with_replacement我們寫短一點的代碼：
def solve2(lst, bound):
max_length = bound // min(lst)
for n in range(1, max_length+1):
for c in itertools.combinations_with_replacement(lst,n):
if sum(c) <= bound:
print('+'.join(map(str, c)))
solve2([1,2,3], 10)
假設該問題符合下列假設:
清單內元素可重複使用
隻要是能滿足小於等於上限值的組合都可接受, 就算遠小於上限值甚至是零也可以
以下是暴力法:
# code for python3
from itertools import combinations
def solve(lst, upperbound):
candidates = []
for n in lst:
for count in range(upperbound//n):
candidates.append(n)
allcomb = set()
for l in range(1, len(candidates)+1):
for comb in combinations(candidates, l):
if not comb in allcomb:
allcomb.add(comb)
if sum(comb) <= upperbound:
print('+'.join([str(n)for n in comb]))
solve([1,2,3], 10)
我回答過的問題: Python-QA