Question

我想找到一种有效的算法，将整数除以最大，最小范围内的某个值。应该尽可能少的值。

例如： max = 7，min = 3 然后

8 = 4 + 4
9 = 4 + 5
16 = 5 + 5 + 6 (not 4 + 4 + 4 + 4)

修改

为了更清楚，让我们举一个例子。假设你有一堆苹果，你想把它们装进篮子里。每个篮子可以包含3到7个苹果，并且您希望使用的篮子数量尽可能少。

**我提到价值应该均分，但这并不重要。我更担心的是篮子数量减少。

Answer 1

这让我觉得这是一个有趣的问题，所以我去了解快速解决方案。我认为这可能是一个有趣的起点，它会为您提供尽可能少的数字或尽可能相似的数字的有效解决方案，所有这些都在min_bound定义的范围内和max_bound number = int（输入（＆＃34;数字：＆＃34;）） min_bound = 3 max_bound = 7

def improve(solution):
    solution = list(reversed(solution))
    for i, num in enumerate(solution):
        if i >= 2:
            average = sum(solution[:i]) / i
            if average.is_integer():
                for x in range(i):
                    solution[x] = int(average)
                break
    return solution

def find_numbers(number, division, common_number):
    extra_number = number - common_number * division
    numbers_in_solution = [common_number] * division
    if extra_number < min_bound and \
    extra_number + common_number <= max_bound:
        numbers_in_solution[-1] += extra_number
    elif extra_number < min_bound or extra_number > max_bound:
        return None
    else:
        numbers_in_solution.append(extra_number)
    solution = improve(numbers_in_solution)
    return solution

def tst(number):
    try:
        solution = None
        for division in range(number//max_bound, number//min_bound + 1):  # Reverse the order of this for numbers as close in value to each other as possible.
            if round (number / division) in range(min_bound, max_bound + 1):
                solution = find_numbers(number, division, round(number / division))
            elif (number // division) in range(min_bound, max_bound + 1):  # Rarely required but catches edge cases
                solution = find_numbers(number, division, number // division)
            if solution:
                print(sum(solution), solution)
                break
    except ZeroDivisionError:
        print("Solution is 1, your input is less than the max_bound")

tst(number)
for x in range(1,100):
    tst(x)

此代码只是为了展示一个想法，我确信可以调整它以获得更好的性能。

将值拆分为max，min范围内的值

1 个答案: