1037 Magic Coupon (25 分)(PAT甲级)

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.

Code

#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
int comp(const void *a, const void *b ){
    return (*(int*)b - *(int*)a);
}
int main()
{
    int c, p, k = 0, j = 0, res = 0;
    scanf("%d", &c);
    int coupons[c];
    for(int i=0; i<c; i++){
        scanf("%d", &coupons[i]);
    }
    scanf("%d", &p);
    int products[p];
    for(int i=0; i<p; i++){
        scanf("%d", &products[i]);
    }
    qsort(coupons, c, sizeof(int), comp);
    qsort(products, p, sizeof(int), comp);
    if(c > p){
        while(coupons[k]>0&&products[j]>0&&j<p){
            res += coupons[k]*products[j];
            k++; j++;
        }

        if(j != p){
            c--; p--;
            while(coupons[c]<0&&products[p]<0&&p>=j){
                res += coupons[c]*products[p];
                c--; p--;
            }
        }
    }else{
        while(coupons[k]>0&&products[j]>0&&k<c){
            res += coupons[k]*products[j];
            k++; j++;
        }
        if(k != c){
            c--; p--;
            while(coupons[c]<0&&products[p]<0&&c>=k){
                res += coupons[c]*products[p];
                c--; p--;
            }
        }
    }
    printf("%d\n", res);
    return 0;
}