PAT甲级真题 1037 Magic Coupon (25分) C++实现_the magic shop in mars is offering some magic coup-优快云博客

在火星上的魔法商店，提供带有整数N的魔法优惠券，使用这些优惠券可以获得产品价值N倍的回馈。任务是在有限的选择下，通过合理搭配优惠券和产品，最大化获得的回馈金额。

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题目

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

思路

将两个数组升序排序，把左侧都是负数的依次相乘，把右侧都是正数的依次相乘，累加结果即为所求。

代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int m, n;
    cin >> m;
    vector<int> a(m);
    for (int i=0; i<m; i++){
        cin >> a[i];
    }
    cin >> n;
    vector<int> b(n);
    for (int i=0; i<n; i++){
        cin >> b[i];
    }

    sort(a.begin(), a.end());
    sort(b.begin(), b.end());

    long long sum = 0;
    for (int i=0; i<m && i<n; i++){
        if (a[i]<0 && b[i]<0){
            sum += (a[i] * b[i]);
        }
        else{
            break;
        }
    }
    for (int i=m-1, j=n-1; i>=0&&j>=0; i--, j--){
        if (a[i]>0 && b[j]>0){
            sum += (a[i] * b[j]);
        }
        else{
            break;
        }
    }
    cout << sum << endl;
    return 0;
}