本文共 2099 字，大约阅读时间需要 6 分钟。

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 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:

41 2 4 -147 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 
   
    #include 
    
     #include 
     
      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
      
       np?np:nc;	for(int i=0;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;}
       
      
     
    
   

转载地址：https://meteorrain.blog.csdn.net/article/details/86619397 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！