PAT 1037 Magic Coupon

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≤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

思路

给你两个序列，两个序列的任意两个元素可以相乘，每个元素只能用一次，求元素乘积 和的最大值。

我们只要保证同号的从大到小，依次相乘即可。用两个数组分别存储coupon中大于0的部分，和小于0的部分（将其转化为正数）。同样用两个数组存储product的值，将四个数组从大到小排序，四个数组，两两相乘求和即可。

记得longlong存储

代码

#include <stdio.h>
#include <string>
#include <stdlib.h>
#include <iostream>
#include <vector>
#include <string.h>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
#include <stack>
#include <functional>
#include <limits.h> 
using namespace std;
int cou1[100010];
int cou2[100010];
int pro1[100010];
int pro2[100010];
bool cmp(int a, int b){
	return a > b;
}
int main() {
	int N1, N2;
	long long sum = 0;
	scanf("%d", &N1);
	int t = 0, n1 = 0, n2 = 0, n3 = 0, n4 = 0; 
	for(int i = 0; i < N1; i++){
		scanf("%d", &t);
		if(t > 0)			cou1[n1++] = t;
		else if (t < 0)		cou2[n2++] = -t;
	}
	scanf("%d", &N2);
	for(int i = 0; i < N2; i++){
		scanf("%d", &t);
		if(t > 0)			pro1[n3++] = t;
		else if (t < 0)		pro2[n4++] = -t;
	}
	sort(cou1, cou1 + n1, cmp);
	sort(cou2, cou2 + n2, cmp);
	sort(pro1, pro1 + n3, cmp);
	sort(pro2, pro2 + n4, cmp);
	int i = 0;
	while(i < n1 && i < n3){
		sum += cou1[i] * pro1[i];
		i++;
	}
	i = 0;
	while(i < n2 && i < n4){
		sum += cou2[i] * pro2[i];
		i++;
	}
	cout << sum << endl;
	return 0;
}