1037 Magic Coupon （25 分)

2019-03-17

1037 Magic Coupon （25 分)

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:

Sample Output:

作者: CHEN, Yue

单位: 浙江大学

时间限制: 400 ms

内存限制: 64 MB

代码长度限制: 16 KB

题目大意

有一组coupon，和一组bonus，其中各取一个数字相乘，求乘积和的最大值。贪心法即可，两个vector排序，前面均为负的相乘，后面均为正的相乘，一正一负的很显然不取。

代码

#include <iostream>
#include <vector>
#include <algorithm>
#include <deque>
#include <map>
#include <set>
#include <cstring>
#include <cmath>
#include<sstream>

using namespace std;


int main() {
    int nc,np;
    vector<long int>coupon;
    vector<long int>bonus;
    cin>>nc;
    long int t;
    long int res=0;
    for(int i=0;i<nc;i++){
        cin>>t;
        coupon.push_back(t);
    }
    cin>>np;
    for(int i=0;i<np;i++) {
        cin >> t;
        bonus.push_back(t);
    }
    sort(coupon.begin(),coupon.end());
    sort(bonus.begin(),bonus.end());
    for(int i=0,j=0;i<coupon.size()&&j<bonus.size();i++,j++){
        if(coupon[i]<0&&bonus[j]<0)
            res+=coupon[i]*bonus[j];
        else
            break;
    }
    for(int i=coupon.size()-1,j=bonus.size()-1;i>=0,j>=0;i--,j--){
        if(coupon[i]>0&&bonus[j]>0)
            res+=coupon[i]*bonus[j];
        else
            break;
    }
    cout<<res;
}