如何解决以下贪婪/动态问题？

Question

我必须检查是否有可能从给定的数字中得到总和 > 严格大于给定的总和

例如：

如果给定的总和是 9 并且给定的数字是 [3,5,7] 那么正确答案是 1*3 +1*7=10>9 所以输出应该是 [1,0,1]

例子2：

如果给定的总和是 13 并且数字是 [1,3,5] 那么正确答案是 0*1+3*3+1*5=14>13 但 9*1+0*3+5=14 是>错误的所以正确的输出是 [0,3,1] 如何解决这个问题

Answer 1

可以使用无界背包算法的一个版本来获得解决方案。

这里我们从Python Unbound Knapsack修改代码

修改有两个目标：

1. 最小化数组中使用的值的计数

2. 创建小于限制的最大总和（即给定总和 + 1）

def knapsack_unbounded_dp(arr, C):
    C += 1  # actual limit

    # Form index value pairs (to keep track of the original indexes after sorting)
    items = [(i, v) for i, v in enumerate(arr)]

    # Sort values in descending order
    items = sorted(items, key=lambda item: item[1], reverse=True)

    # Sack keeps track of:
    #  max value so i.e. sack[0] and
    # count of how many each item are in the sack i.e. sack[1]
    sack = [(0, [0 for i in items]) for i in range(0, C+1)]   # value, [item counts]

    for i,item in enumerate(items):
        # For current item we check if a previous entry could have done better
        # by adding this item to the sack
        index, value = item
        for c in range(value, C+1):   # check all previous values
            sackwithout = sack[c-value]  # previous max sack to try adding this item to
            trial = sackwithout[0] + value  # adding value to max without using it
            used = sackwithout[1][i]        # count of i-them item
            if sack[c][0] < trial:
                # old max sack with this added item is better
                sack[c] = (trial, sackwithout[1][:])
                sack[c][1][i] +=1   # use one more

    value, bagged = sack[C]

    # index and count of each array value
    new_bagged = [(i, v) for (i, _), v in zip(items, bagged)]

    # Re-sort based upon original order of array indexes
    new_bagged.sort(key=lambda t: t[0])

    # counts based upon original array order
    cnts = [v for i, v in new_bagged]

return sum(cnts), value, cnts

测试代码

for t in [([1, 3, 5], 8), ([3, 5, 7], 9), ([1, 3, 5], 13)]:
  result = knapsack_unbounded_dp(t[0], t[1])
  print(f'Test: {t[0]}, Given Sum {t[1]}')
  print(f'Result counts {result[2]}, with max sum {result[1]}, Total Count {result[0]}\n')

输出

Test: [1, 3, 5], Given Sum 8
Result counts [0, 3, 0], with max sum 9, Total Count 3

Test: [3, 5, 7], Given Sum 9
Result counts [0, 2, 0], with max sum 10, Total Count 2

Test: [1, 3, 5], Given Sum 13
Result counts [0, 3, 1], with max sum 14, Total Count 4