1037. Magic Coupon (25)

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

//有一个测试点没过
#include<cstdio>
#include<algorithm>
using namespace std;

int main(){
    int n,m;
    int coupon[100010],product[100010];
    scanf("%d",&n);
    for(int i = 0; i < n; i++){
        scanf("%d",&coupon[i]);
    }
    scanf("%d",&m);
    for(int i = 0; i < m; i++){
        scanf("%d",&product[i]);
    }
    sort(coupon,coupon+n);
    sort(product,product+m);
    int ans = 0;
    for(int i = 0;coupon[i] < 0&& product[i] < 0&& i < n&&i < m;i++){
        ans+=coupon[i]*product[i];
    }
    for(int i=n-1,j=m-1;coupon[i] > 0&& product[i] >0&&i >= 0&& j >= 0;i--,j--){   //product【j】
        ans+=coupon[i]*product[j];
    }
    printf("%d",ans);
    return 0;
}

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

int main(){
    int n,m;
    int coupon[100010],product[100010];
    scanf("%d",&n);
    for(int i = 0; i < n; i++){
        scanf("%d",&coupon[i]);
    }
    scanf("%d",&m);
    for(int i = 0; i < m; i++){
        scanf("%d",&product[i]);
    }
    sort(coupon,coupon+n);
    sort(product,product+m);
    int i = 0,j, ans = 0;
    while(i < n && i < m && coupon[i] < 0 && product[i] < 0){
        ans += coupon[i]*product[i];
        i++;
    }
    i = n - 1;
    j = m - 1;
    while(i >= 0 && j >= 0 && coupon[i] > 0 && product[j] > 0){
        ans += coupon[i] * product[j];
        i--;
        j--;
    }
    printf("%d",ans);
    return 0;
}