PAT A1037 Magic Coupon+For语句中多条执行

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

代码：

#include <iostream>
#include<algorithm>

using namespace std;

int main()
{
    int Nc,Np;
    cin >> Nc;
    int coupon[Nc];
    for(int i=0;i<Nc;i++){
        scanf("%d",&coupon[i]);
    }
    cin >> Np;
    int product[Np];
    for(int i=0;i<Np;i++){
        scanf("%d",&product[i]);
    }
    sort(coupon,coupon+Nc);
    sort(product,product+Np);
    long long sum=0;
    for(int i=0;i<Nc && i<Np && coupon[i]<0 && product[i]<0;i++){ //从前往后，最小的两负数相乘
        sum+=coupon[i]*product[i];
    }
    for(int i=Nc-1,j=Np-1;i>=0 && j>=0 && coupon[i]>0 && product[j]>0;i--,j--){ 
    //从后往前，最大的正数相乘，注意两个语句用,隔开
        sum+=coupon[i]*product[j];
    }
    printf("%lld",sum);
    return 0;
}