1037. Magic Coupon (25)

题目例如以下：

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

题目非常easy，为了得到钱。我们仅仅能通过两种途径。①正优惠券相应正的商品。②负优惠券相应负的商品。

对于正数。我们依照降序排序以获得最大的优惠券和最高价的商品。

对于负数。我们依照升序排序以获得绝对值最大的优惠券和商品。

因此。我们须要把正、负分开统计。因此须要4个容器（coupon和product各+、-两个）。

排序结束后。从前到后累加优惠券数值和商品数值的乘积，注意遍历容器时设置结束条件为较小的容器的大小。

#include <iostream>
#include <vector>
#include <stdio.h>
#include <algorithm>

using namespace std;

int comparePos(int a, int b){
    return a > b;
}

int compareNeg(int a, int b){
    return a < b;
}

int main()
{
    vector<int> posCoupons,negCoupons;
    vector<int> posProducts,negProducts;
    int N;
    cin >> N;
    int buffer;
    for(int i = 0; i < N; i++){
        scanf("%d",&buffer);
        if(buffer >= 0){
            posCoupons.push_back(buffer);
        }else{
            negCoupons.push_back(buffer);
        }
    }
    cin >> N;
        for(int i = 0; i < N; i++){
        scanf("%d",&buffer);
        if(buffer >= 0){
            posProducts.push_back(buffer);
        }else{
            negProducts.push_back(buffer);
        }
    }
    sort(posCoupons.begin(),posCoupons.end(),comparePos);
    sort(negCoupons.begin(),negCoupons.end(),compareNeg);
    sort(posProducts.begin(),posProducts.end(),comparePos);
    sort(negProducts.begin(),negProducts.end(),compareNeg);

    // 首先组合正的coupons和正的products
    int thres = posCoupons.size() > posProducts.size() ? posProducts.size() : posCoupons.size();
    int sum = 0;
    for(int i = 0; i < thres; i++){
        sum += posCoupons[i] * posProducts[i];
    }
    // 然后组合负的coupons和负的products
    thres = negCoupons.size() > negProducts.size() ? negProducts.size() : negCoupons.size();
    for(int i = 0; i < thres; i++){
        sum += negCoupons[i] * negProducts[i];
    }
    cout << sum << endl;
    return 0;
}