Question

我正试图找出一种方法来解决（在c ++中），给定一个价格不变的物品清单，如果他们有X金额，一个人可以购买多少种X物品。

到目前为止，我已经尝试使用嵌套for循环来尝试强制解决方案，但是，我觉得我可能会错过一个我似乎无法看到的非常简单的解决方案。

非常感谢任何帮助。感谢。

Answer 1

这与常见的编程问题非常类似：“你可以用多少种方式将Y种类型的硬币与Z值组合起来制作X美元”，即用Y币类型改变X美元。

这是一个可以移植到C ++的通用解决方案：

I = list of items SORTED from highest to lowest price
N = number of items bought so far
M = money left
S = money to start

function shop(I, N, M, S):
  if M < 0: return 0 //we bought something illegally! 
  if M == 0:
    //If we have no money, we've bought all we could. 
    //Check our condition that num items N = starting money S
    return 1 if N == S, else 0 
  if I is empty:
    return 0 //no more item combos, no way to buy things.
  else:
    newI = I with first element removed 
    //Every time we want to buy stuff, we have two options: 
    //1. buy something of highest value and subtract money
    //2. Shop ignoring the next highest value item
    return shop(newI, N, M, S) + shop(I, N+1, M-(cost of first item), S)

使用X美元，您可以从电话开始：

shop(I, 0, X, X)

如何购买物品的组合

1 个答案: