PAT-A 1037. Magic Coupon

1037. Magic Coupon

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 NC, followed by a line with NC coupon integers. Then the next line contains the number of products NP, followed by a line with NP product values. Here 1<= NC, NP <= 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:

41 2 4 -147 6 -2 -3

Sample Output:

43

程序代码：

#include<stdio.h>#define MAX 100000int coupons[MAX],value[MAX];int comp(void* a,void* b);int main(){    int m,n,i;    scanf("%d",&m);    for(i=0;i<m;i++)        scanf("%d",&coupons[i]);    scanf("%d",&n);        for(i=0;i<n;i++)                scanf("%d",&value[i]);    qsort(coupons,m,sizeof(int),comp);    qsort(value,n,sizeof(int),comp);    int p=m-1,q=n-1;    long long sum = 0;    while(coupons[p]>0&&value[q]>0)    {        sum += coupons[p]*value[q];        p--;        q--;    }    p=0,q=0;    while(coupons[p]<0&&value[q]<0)    {        sum += coupons[p]*value[q];        p++;        q++;    }       printf("%d",sum);    return 0;}int comp(void* a,void* b){    int* m=(int*)a;    int* n=(int*)b;    return *m -*n;}