Description:
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
43
//NKW 甲级真题1012 NKW上的测试样例有问题 PTA上AC
#pragma warning(disable:4996)
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
const int maxn = 100010;
int a1[maxn], a2[maxn];
int main(){
int np, nc, i = 0, j = 0, sum = 0;
scanf("%d", &nc);
for (int i = 0; i < nc; i++)
scanf("%d", &a1[i]);
scanf("%d", &np);
for (int i = 0; i < np; i++)
scanf("%d", &a2[i]);
sort(a1, a1 + nc);
sort(a2, a2 + np);
while (i < nc&&i < np&&a1[i] < 0 && a2[j] < 0)
sum += a1[i++] * a2[j++];
i = nc - 1;
j = np - 1;
while (i >= 0 && j >= 0 && a1[i] > 0 && a2[j]>0)
sum += a1[i--] * a2[j--];
printf("%d\n", sum);
system("pause");
return 0;
}