PAT 甲级 1037 Magic Coupon (25 分) （较简单,贪心）

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 − }, and a set of product values { 7 6 − − } (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, and it is guaranteed that all the numbers will not exceed 2​30​​.

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：

给出两个集合，从这两个集合里面选出数量相同的元素进行一对一相乘，求能够得到的最大乘积之和。

题解：

对每个集合，将正数和负数分开考虑，将每个集合里的整数从大到小排序；将每个集合里的负数从小到大排序，然后同位置的正数与正数相乘，负数与负数相乘。

注意点：

输入为0的不要管，直接忽略，否则测试点1过不去

 

AC代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll na[100005];
ll pa[100005];
ll nb[100005];
ll pb[100005];
int kna=0,kpa=0,knb=0,kpb=0;
int n1,n2;
bool cmp(ll x, ll y){
    return x>y;
}
ll s=0;
int main(){
    cin>>n1;
    ll x;
    for(int i=1;i<=n1;i++){
        cin>>x;
        if(x>0){
            pa[++kpa]=x;
        }else if(x<0){//=0的不要管 
            na[++kna]=x;
        }
    }
    cin>>n2;
    for(int i=1;i<=n2;i++){
        cin>>x;
        if(x>0){
            pb[++kpb]=x;
        }else if(x<0){//=0的不要管 
            nb[++knb]=x;
        }
    }
    sort(pa+1,pa+1+kpa,cmp);
    sort(pb+1,pb+1+kpb,cmp);
    sort(na+1,na+1+kna);
    sort(nb+1,nb+1+knb);
    int min_l=min(kpa,kpb);
    for(int i=1;i<=min_l;i++){
        s+=pa[i]*pb[i];
    }
    min_l=min(kna,knb);
    for(int i=1;i<=min_l;i++){
        s+=na[i]*nb[i];
    }
    cout<<s<<endl;
    return 0;
}