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
//题目大意是:在两个集合里,选择相同数量的数分别相乘再求和,要求找出和的最大值
//思想分析:典型的贪心。首先,显然两数要同号(如果异号的话,相乘的结果为负,只会拉低最后求得的和);其次,用绝对值大的数乘以绝对值大的数。下面以整数为例简单证明下:
因此呢,要分两部分求,结果 = 负数 * 负数 + 正数 * 正数
下面是我AC的代码:
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int C[100005], P[100005]; //C数组是第一个数组,P是第二个
int main() {
int n, m;
cin >> n;
for (int i = 0; i < n; i++)
cin >> C[i]; //读入第一个集合
cin >> m;
for (int i = 0; i < m; i++)
cin >> P[i]; //读入第二个集合
sort(C, C + n); //对两个数组从小到大排列
sort(P, P + m);
int ans = 0;
//正数 * 正数 部分,逆序遍历数组
for (int i = n - 1, j = m - 1; i >= 0, j >= 0; i--, j--)
if (C[i] < 0 || P[j] < 0) //若有一个为负则此部分结束,0对结果无影响,故不用考虑
break;
else
ans += C[i] * P[j];
//负数 * 负数 部分,顺序遍历数组
for (int i = 0, j = 0; i < n, j < m; i++, j++)
if (C[i] > 0 || P[j] > 0) //发现有一个为正数,则停止
break;
else
ans += C[i] * P[j];
cout << ans << endl; //输出结果
return 0;
}
//最后,先看后赞是好习惯噢!!