1037 Magic Coupon (25 point(s))

本文链接： https://blog.csdn.net/huaxuewan/article/details/101793111

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 ≤ 10⁵, and it is guaranteed that all the numbers will not exceed 2³⁰.

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_C 张优惠券，每张券有不同数字，有 N_P 种产品，每种产品有不同价值；优惠券数字乘产品价值即金币数；

可以从两个序列中选相同数目的元素，一对一相乘，求可得金币的最大值；

设计思路：

贪心算法：

分别对两个序列按相同顺序排序，这里是从小到大的顺序
前面都是负数的元素相乘，后面都是正数的元素相乘，累计求和，即为最大值

编译器：C (gcc)

#include <stdio.h>

int cmp(const void *a, const void *b)
{
        int *n1 = (int *)a, *n2 = (int *)b;
        return *n1 - *n2;
}

int main(void)
{
        int nc, np;
        int coupon[100000] = {0}, product[100000] = {0};
        int i, p, q, sum;

        scanf("%d", &nc);
        for (i = 0; i < nc; i++) {
                scanf("%d", &coupon[i]);
        }
        scanf("%d", &np);
        for (i = 0; i < np; i++) {
                scanf("%d", &product[i]);
        }

        qsort(coupon, nc, sizeof(coupon[0]), cmp);
        qsort(product, np, sizeof(product[0]), cmp);

        sum = 0;
        p = 0;
        q = 0;
        while (p < nc && q < np && coupon[p] < 0 && product[q] < 0) {
                sum += coupon[p] * product[q];
                p++;
                q++;
        }
        p = nc - 1;
        q = np - 1;
        while (p >= 0 && q >= 0 && coupon[p] > 0 && product[q] > 0) {
                sum += coupon[p] * product[q];
                p--;
                q--;
        }

        printf("%d", sum);
        return 0;
}