Question

这个任务旨在实现一个原始计算器的动态编程方法，该计算器只能加1，乘以2并乘以3.因此，输入n确定达到n的最小操作数。我已经实现了一个非常天真的dp或我认为是dp方法。它不起作用。我没有其他人要问。对于n = 5的输入，以下输出为：（[0,1,2,2,3,4]，[1,1,2,3,4,5]）而有两个正确的输出列表编号= [1,2,4,5]或[1,3,4,5]。一些帮助将不胜感激。

def DPmin_operations(n):

numbers = []
minNumOperations = [0]*(n+1)
numOps = 0
numbers.append(1)

for k in range(1,n+1):
    minNumOperations[k] = 10000

    # for *3 operator
    if k % 3 == 0:
        numOps = minNumOperations[k//3] + 1
        if numOps < minNumOperations[k]:
            minNumOperations[k] = numOps
            numbers.append(k)
    # for *2 operator
    elif k % 2 == 0:
        numOps = minNumOperations[k//2] + 1
        if numOps < minNumOperations[k]:
            minNumOperations[k] = numOps
            numbers.append(k)
    # for + 1 operator 
    elif k >= 1:
        numOps = minNumOperations[k - 1] + 1
        if numOps < minNumOperations[k]:
            minNumOperations[k] = numOps
            numbers.append(k)

return (minNumOperations, numbers)

Answer 1

请注意，elif块应该是if块。目前，您正在使用总是试图除以3的贪婪算法;如果失败了，那就试着除以2;如果失败，则减去1.可能数字可被6整除，因此所有三个选项都是可能的，但除以2则更优，然后除以3.

至于获取你的数字列表，最后这样做。存储所有可能的父母，然后从目标向后工作，看看你是如何到达那里的。

def dp_min_ops(n):
    all_parents = [None] * (n + 1)
    all_min_ops = [0] + [None] * n

    for k in range(1, n + 1):
        curr_parent = k - 1
        curr_min_ops = all_min_ops[curr_parent] + 1

        if k % 3 == 0:
            parent = k // 3
            num_ops = all_min_ops[parent] + 1
            if num_ops < curr_min_ops:
                curr_parent, curr_min_ops = parent, num_ops

        if k % 2 == 0:
            parent = k // 2
            num_ops = all_min_ops[parent] + 1
            if num_ops < curr_min_ops:
                curr_parent, curr_min_ops = parent, num_ops

        all_parents[k], all_min_ops[k] = curr_parent, curr_min_ops

    numbers = []
    k = n
    while k > 0:
        numbers.append(k)
        k = all_parents[k]
    numbers.reverse()

    return all_min_ops, numbers

print(dp_min_ops(5))   # ([0, 1, 2, 2, 3, 4], [1, 3, 4, 5])
print(dp_min_ops(10))  # ([0, 1, 2, 2, 3, 4, 3, 4, 4, 3, 4], [1, 3, 9, 10])

动态编程 - 原始计算器Python

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