原题描述如下

The magic shop in Mars is offering some magic coupons. Each coupon has an integer N printed on it, meaning that when you use this coupon with a product, you may get N times the value of that product back! What is more, the shop also offers some bonus product for free. However, if you apply a coupon with a positive N to this bonus product, you will have to pay the shop N times the value of the bonus product... but hey, magically, they have some coupons with negative N's!
For example, given a set of coupons { 1 2 4 −1 }, and a set of product values { 7 6 −2 −3 } (in Mars dollars M$) where a negative value corresponds to a bonus product. You can apply coupon 3 (with N being 4) to product 1 (with value M$7) to get M$28 back; coupon 2 to product 2 to get M$12 back; and coupon 4 to product 4 to get M$3 back. On the other hand, if you apply coupon 3 to product 4, you will have to pay M$12 to the shop.
Each coupon and each product may be selected at most once. Your task is to get as much money back as possible.

大致意思就是，我手上有魔法券，这个券上有正数或负数的数值，其魔法的点就在于，用这个券的数值和商品的价值相乘，自己所得就是相乘后的结果，即负负得正，正负为负，所以想要获得最大值，那就让最大的负数依次相乘，最大的正数依次相乘。
但是有一点，多余的负数呢？
那就不要呗，题目没有说一定要全部用完！！这是一道最优解的问题！！

问题点

如果把这道题理解成，全部组合后实现最大值，那么思路上就不太一样，就多考虑正负相乘的情况。但本质上还是先排序，在按正负数分别操作。

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