from itertools import combinations, product

def func(numbers, m, s):
    def determinate(args):
        return sum(args) == s
    return filter(determinate, product(numbers, repeat = m))

print(func(range(10), 2, 5))

输出:
    [(0, 5), (1, 4), (2, 3), (3, 2), (4, 1), (5, 0)]

【 在 wenti (question) 的大作中提到: 】
: n个数，任取m个，使得和为s
: m不定，可重复取