PAT 1037 Magic Coupon[dp]

1037 Magic Coupon（25 分）

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.

Input Specification:

Each input file contains one test case. For each case, the first line contains the number of coupons N​C​​, followed by a line with N​C​​ coupon integers. Then the next line contains the number of products N​P​​, followed by a line with N​P​​ product values. Here 1≤N​C​​,N​P​​≤10​5​​, and it is guaranteed that all the numbers will not exceed 2​30​​.

Output Specification:

For each test case, simply print in a line the maximum amount of money you can get back.

Sample Input:

4
1 2 4 -1
4
7 6 -2 -3

Sample Output:

43

题目大意：有n个勺子，每个勺子都有一个参数N，有m个积，每个勺子可以和一个积配对，那么求可以产生的最大正整数。

//就是将两者排序，从大到小，但是有负数怎么办呢？

代码转自： https://www.liuchuo.net/archives/2253

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
    int m, n, ans = , p = , q = ;
    scanf("%d", &m);
    vector<int> v1(m);
    for(int i = ; i < m; i++)
        scanf("%d", &v1[i]);
    scanf("%d", &n);
    vector<int> v2(n);
    for(int i = ; i < n; i++)
        scanf("%d", &v2[i]);
    sort(v1.begin(), v1.end());//从小到大排序。
    sort(v2.begin(), v2.end());
    while(p < m && q < n && v1[p] <  && v2[q] < ) {
        ans += v1[p] * v2[q];//都小于0的相加。
        p++; q++;
    }
    p = m - , q = n - ;
    while(p >=  && q >=  && v1[p] >  && v2[q] > ) {//这里将0算上了。
        ans += v1[p] * v2[q];
        p--; q--;
    }
    printf("%d", ans);
    return ;
}

//看完题解感觉真的是水题。

1.将其从大到小或者从小到大排序均可，假设从小到大拍；

2.那么将左边的负数分别相乘得到结果，右边的整数相乘得到结果即可。

//emmm，这么简单的吗