PAT 1037 Magic Coupon (25分) 贪心+排序+负负/正正得正

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.

加粗样式Sample Input:

Sample Output:

题目解读

英语不行看懂题目是真的费劲，完全明白之后才发现这又是加减乘除。

先看题目：

给出NC个优惠券，每个优惠券面额可正可负；给出NP个产品价格，也是可正可负
一个优惠券只能用在一个产品上，不能重复使用
面额为N的优惠券用在价格为P的产品上，能获得 N * P 的回扣；所以当N>0&&P>0 或者 N<0&&P<0时，得到的是N*P的回利，二者异号时，是N*P的付出。
问：如何使用优惠券，才能获得最大的回扣？

分析：

既然要回扣，所以肯定考虑正正，负负。
既然要最大，肯定最小的负数优惠券用于最低的负数价格产品；最大的整数面额优惠券用于价格最高的正数价格产品。

思路：

用两个数组分别存储NC个优惠券面额和NP个产品价格
对两数组分别进行排序（从小到大）
从左边开始处理卡券和商品均负整数的情况，从右边开始处理卡券和商品均正整数的情况，累加结果。

完整代码

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

int main() {
    // NC中优惠券
    int NC, NP;
    cin >> NC;
    int coupons[NC];
    // 每个优惠券的面额
    for (int i = 0; i < NC; ++i) cin >> coupons[i];
    // NP中产品
    cin >> NP;
    int products[NP];
    // 每个产品标价
    for (int i = 0; i < NP; ++i) cin >> products[i];
    // 优惠券从小到大排序
    sort(coupons, coupons + NC);
    // 产品从低到高排序
    sort(products, products + NP);
    int ans = 0,i = 0, j = 0;
    // 最小的负数优惠券，用于最小的负数价格产品，获得最大的正数回馈
    while (i < NC && j < NP && coupons[i] < 0 && products[j] < 0) {
        ans += coupons[i] * products[j];
        ++i;++j;
    }
    // 最大正数优惠券，用于最大的正数价格产品，获得最大的正数回馈
    i = NC - 1, j = NP - 1;
    while (i >= 0 && j >= 0 && coupons[i] > 0 && products[j] > 0) {
        ans += coupons[i] * products[j];
        --i;--j;
    }
    // 一正一负情况不考虑
    cout << ans;      
}

来源：oschina

链接：https://my.oschina.net/u/4418383/blog/4307841

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