PAT Advanced Level 1037 Magic Coupon (25)

The magic shop in Mars is offering some magic coupons. Each coupon hasan integer N printed on it, meaning that when you use this coupon with aproduct, you may get N times the value of that product back! What ismore, the shop also offers some bonus product for free. However, if youapply a coupon with a positive N to this bonus product, you will have topay 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 productvalues {7 6 -2 -3} (in Mars dollars M\$) where a negative valuecorresponds 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 2to get M\$12 back; and coupon 4 to product 4 to get M\$3 back. On theother hand, if you apply coupon 3 to product 4, you will have to payM\$12 to the shop.

Each coupon and each product may be selected at most once. Your task isto get as much money back as possible.

Input Specification:

Each input file contains one test case. For each case, the first linecontains the number of coupons NC, followed by a line with NC couponintegers. Then the next line contains the number of products NP,followed by a line with NP product values. Here 1<= NC, NP <=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 moneyyou can get back.

Sample Input:

Sample Output:

要求：输入各个coupon的值和各个Product的值，得到两两乘积的正数和的最大值。

思路：想到负数与负数相互搭配，正数和正数相互搭配，

1、得到输入的值，存放在数组中co[],pro[]，根据从大到小排序

2、每个数组分别在头尾设置各一个int类型，指向首尾两个位置，c1,c2,p1,p2。

3、首和首相乘，尾和尾相乘，取大的整数，加到sum中，直到只有负数或者某个数组的值遍历结束。

代码如下：

#include <iostream>
#include <cstdio>
#include <algorithm>
#define Maxsize 100010
using namespace std;

int main()
{
    int nc,np;
    int i;
    long left,right;
    unsigned long sum=0;
    int co[Maxsize]={0},pro[Maxsize]={0};
    int c1,c2,p1,p2;
    scanf("%d",&nc);
    for(i=0;i<nc;i++)
    {
        scanf("%d",&co[i]);
    }
    scanf("%d",&np);
    for(i=0;i<np;i++)
    {
        scanf("%d",&pro[i]);
    }
    sort(co,co+nc);               //排序
    sort(pro,pro+np);
    c1=p1=0;
    c2=nc-1;
    p2=np-1;
        while(c1<=c2 && p2>=p1)          //当存在某个数组遍历结束，跳出循环
        {
            left=co[c1]*pro[p1];
            right= co[c2]*pro[p2];
            if(left> 0 && left >=right )       //假如左边得整数大   sum+=   c1,p1右移
            {
                sum+=left;
                c1++;
                p1++;
                continue;
            }
            if(right >0 && right>=left)       //假如右边的正数大，sum+=，  c1，p1左移
            {
                sum+=right;
                c2--;
                p2--;
                continue;                        //else 或者continue用一个，否则，left和right没有更新就比较。
            }
            if(right<=0 && left <=0)                
                break;
        }
    printf("%ld",sum);
    return 0;
}

PAT Advanced Level 1037 Magic Coupon (25)

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