PAT (Advanced Level) Practise 1037 Magic Coupon (25)

1037. Magic Coupon (25)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题意：给出两个数字序列，从这两个序列中分别选取相同数量的元素进行一对一相乘，问能得到的乘积之和最大为多少

解题思路：把这两个序列的正数、负数分别扔到vector里，然后进行乘积相加

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int n, x, ans;
vector<int>g1[2], g2[2];

int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &x);
			if (x > 0) g1[0].push_back(x);
			else g1[1].push_back(x);
		}
		scanf("%d", &n);
		for (int i = 1; i <= n; i++)
		{
			scanf("%d", &x);
			if (x > 0) g2[0].push_back(x);
			else g2[1].push_back(x);
		}
		sort(g1[0].begin(),g1[0].end(),greater<int>());
		sort(g2[0].begin(), g2[0].end(), greater<int>());
		sort(g1[1].begin(),g1[1].end());
		sort(g2[1].begin(), g2[1].end());
		for (int i = 0; i < min(g1[0].size(), g2[0].size()); i++) ans += g1[0][i] * g2[0][i];
		for (int i = 0; i < min(g1[1].size(), g2[1].size()); i++) ans += g1[1][i] * g2[1][i];
		printf("%d\n", ans);
	}
	return 0;
}