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:

1 2 4 -1
4
7 6 -2 -3

Sample Output:

43

分析：

很简单的模拟题，两组数有正有负，问乘积和最大是多少。
注意两点：
1.优惠券可以不全用完
2.数据比较大，用longlong存储数据，scanf和printf输入/打印。

code：

//来自牛客网一位老哥，代码很简洁，原文cin会超时
#include <algorithm>
#include <iostream>
using namespace std;

bool cmp(long long a,long long b){
    return a>b;
}
long long a[100010],b[100010],sum=0;
int main(){
    int x,y;
    scanf("%d",&x);
    for(int i=0;i<x;i++) scanf("%lld",&a[i]);
    scanf("%d",&y);
    for(int i=0;i<y;i++) scanf("%lld",&b[i]);
    sort(a,a+x,cmp);
        sort(b,b+y,cmp);
        int num=0,j=0,k=0;
        while(a[j]>0&&b[k]>0){  //这里不用过多限定，因为输入本身就有一定的限定
    sum+=a[j]*b[k];
    j++;k++;
    }
    j=x-1;k=y-1;
    while(a[j]<0&&b[k]<0){
        sum+=a[j]*b[k];
        num++;
        j--;k--;
    }
    printf("%lld",sum);
    return 0;
}

柳婼code：

//用vector动态数组，不会有溢出危险
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;
int main() {
    int m, n, ans = 0, p = 0, q = 0;
    scanf("%d", &m);
    vector<int> v1(m);
    for(int i = 0; i < m; i++)
        scanf("%d", &v1[i]);
    scanf("%d", &n);
    vector<int> v2(n);
    for(int i = 0; i < n; i++)
        scanf("%d", &v2[i]);
    sort(v1.begin(), v1.end());
    sort(v2.begin(), v2.end());
    while(p < m && q < n && v1[p] < 0 && v2[q] < 0) {
        ans += v1[p] * v2[q];
        p++; q++;
    }
    p = m - 1, q = n - 1;
    while(p >= 0 && q >= 0 && v1[p] > 0 && v2[q] > 0) {
        ans += v1[p] * v2[q];
        p--; q--;
    }
    printf("%d", ans);
    return 0;
}