K产品阵列

Question

我正在研究一个算法问题。您有一个数组编号、数组大小 t 、编号 number_of_elements 和编号 multiplication_value。您必须找到数组元素的任意一组 number_of_elements 索引，其乘积将等于 multiplication_value。保证存在这样的一组索引

这个问题看起来像 2 个总和，但我无法将其推断到我的情况。我已经尝试过 O(n) 的朴素算法，但是当你在数组中有错误的第一个数字时它失败了。我认为这里有一种使用递归的方法。我想这是众所周知的问题，但我找不到解决方案

示例在：

t = 7
number_of_elements = 2
multiplication_value = 27
numbers = [9,1,1,27,3,27,3]

示例：

1 3

我的代码思路：


    def return_index_values(numbers,multiplication_value,number_of_elements):
    cur_number = int(multiplication_value)
    list_of_indexes = []
    values = []
    for i in range(len(numbers)):
        if ((cur_number == 1) and (len(values) == number_of_elements)):
            print(values)
            #finishing if everything worked
            break
        else:               
            if (cur_number % int(numbers[i]) == 0):
                if(len(values) < number_of_elements):
            #pushing values if possible
                    values.append(int(numbers[i]))
                    list_of_indexes.append(i)
                    cur_number = int(cur_number / int(numbers[i]))
                    print(cur_number)
                else:
                    pass
            if(len(values) == number_of_elements):
                if mult_check(values,int(multiplication_value)):
               #mult_check checks if the array's element multiplication gives a value
                    break
                else:
               #started dealing with bad cases, but it doesn't work properly
                    values.sort()
                    val_popped = values.pop()
                    cur_number = cur_number * val_popped

我的代码不合适

 numbers = [9,3,1,27,3,27,3]

Answer 1

您可以使用 itertools.combinations() (https://www.geeksforgeeks.org/itertools-combinations-module-python-print-possible-combinations/) 以所有可能的方式从您的列表中 select number_of_elements 条目，然后检查它们是否乘以所需数量。

Answer 2

这是一种实现方式。不一定是最好的解决方案，但它可以让您了解如何完成。

它首先根据保留索引信息的元素对 numbers 进行排序。然后它执行递归调用。

number_of_elements = 2
multiplication_value = 27
numbers = [9,1,1,27,3,27,3]

def preprocess(numbers, multiplication_value, number_of_elements):
    l = []
    for i, num in enumerate(numbers):
        l.append((num, i))
    return sorted(l, key = lambda tup: tup[0])

def subroutine(numbers, multiplication_value, number_of_elements, idx_start, result):
    if idx_start >= len(numbers):
        return False
    if number_of_elements == 0:
        return True if multiplication_value == 1 else False
    for i in range(idx_start, len(numbers)):
        num = numbers[i][0]
        if num <= multiplication_value:
            if multiplication_value % num == 0:
                idx = numbers[i][1]
                result.append(idx)
                found = subroutine(numbers, multiplication_value / num, number_of_elements - 1, i + 1, result)
                if not found:
                    del result[-1]
                else:
                    return True
        else:
            return False
    return False

result = []
processed_numbers = preprocess(numbers, multiplication_value, number_of_elements)
subroutine(processed_numbers, multiplication_value, number_of_elements, 0, result)
print(result)

K产品阵列

K product array

algorithm

divide-and-conquer