PAT (Advanced Level) Practice 1037 Magic Coupon （25 分）

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

 排序，先从前往后找，碰到相乘等于或小于0的情况退出，然后记下再从后往前找，遇到之前退出的位置或者相乘等于或小于0的情况退出，然后就能求出答案。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
const int maxn=1e5+5;
typedef long long ll;
ll a[maxn];
ll b[maxn];
int nc,np;
int compare(ll a,ll b)
{
    return a>b;
}
int main()
{
    scanf("%d",&nc);
    for (int i=0;i<nc;i++)
    {
        scanf("%lld",&a[i]);
    }
    sort(a,a+nc,compare);
    scanf("%d",&np);
    for (int i=0;i<np;i++)
    {
        scanf("%lld",&b[i]);
    }
    sort(b,b+np,compare);
    ll sum=0;
    int loc=0;
    for (;loc<nc&&loc<np;loc++)
    {
        if(a[loc]*b[loc]>0)
            sum+=a[loc]*b[loc];
        else
            break;
    }
    for (int i=nc-1,j=np-1;i>=0&&j>=0&&i>loc&&j>loc;i--,j--)
    {
        if(a[i]*b[j]>0)
            sum+=a[i]*b[j];
        else
            break;
    }
    printf("%lld\n",sum);
    return 0;
}