PAT 甲级 1037 Magic Coupon

https://pintia.cn/problem-sets/994805342720868352/problems/994805451374313472

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 − }, and a set of product values { 7 6 − − } (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, 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

代码：

#include <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 10;
int N, M;
int a[maxn], b[maxn], c[maxn], d[maxn];
int v1[maxn], v2[maxn];
int num1 = 0, num2 = 0, num3 = 0, num4 = 0;

bool cmp(int x, int y) {
    return x > y;
}

int main() {
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++) {
        scanf("%d", &v1[i]);
        if(v1[i] >= 0)
            a[num1 ++] = v1[i];
        else
            b[num2 ++] = v1[i];
    }
    scanf("%d", &M);
    for(int i = 1; i <= M; i ++) {
        scanf("%d", &v2[i]);
        if(v2[i] >= 0)
            c[num3 ++] = v2[i];
        else
            d[num4 ++] = v2[i];
    }

    sort(a, a + num1, cmp);
    sort(b, b + num2);
    sort(c, c + num3, cmp);
    sort(d, d + num4);

    int len1 = min(num1, num3);
    int len2 = min(num2, num4);

    int sum = 0;
    for(int i = 0; i < len1; i ++) {
        sum += a[i] * c[i];
    }
    for(int i = 0; i < len2; i ++) {
        if(b[i] * d[i] >= 0)
            sum += b[i] * d[i];
    }
    printf("%d\n", sum);
    return 0;
}