1037 Magic Coupon （25 point(s)）

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:

4
1 2 4 -1
4
7 6 -2 -3

Sample Output:

43

经验总结：

这一题，使用的是贪心算法，两个数组排一下序（这里设为从小到大），然后按顺序，从左边两个数组都是负数的话就相乘并且加入收益，从右边两个都是正数就相乘并且加入收益，只要不满足条件就跳出，最终算出结果就行啦~注意最终结果要用long long 存储，毕竟题目说所有数不大于2^30，两个2^30相乘就2^60多了，注意一点就行~

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100010;
int c[maxn],p[maxn];
int main()
{
    long long ans=0;
    int nc,np;
    scanf("%d",&nc);
    for(int i=0;i<nc;++i)
        scanf("%d",&c[i]);
    scanf("%d",&np);
    for(int i=0;i<np;++i)
        scanf("%d",&p[i]);
    sort(c,c+nc);
    sort(p,p+np);
    int L=nc>np?np:nc;
    for(int i=0;i<L;++i)
    {
        if(c[i]<0&&p[i]<0)
            ans+=(long long)c[i]*p[i];
        else
            break;
    }
    int i=nc-1;
    int j=np-1;
    while(i>=0&&j>=0)
    {
        if(c[i]>0&&p[j]>0)
            ans+=(long long)c[i]*p[j];
        else
            break;
        --i;--j;
    }
    printf("%lld\n",ans);
    return 0;
}