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

题目大意：给定两行数，从第一行取一个数与第二行取一个数相乘，循环几次，最后将每个乘的结果相加。问如何取得相加后最大的结果值！一种贪心的思想！

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
	long long a[100002],b[100002];
	long long sum=0;
	int nc,np;
	scanf("%d",&nc);
	int i,j;
	for(i=0;i<nc;i++){
		scanf("%lld",&a[i]);
	}
	scanf("%d",&np);
	for(i=0;i<np;i++){
		scanf("%lld",&b[i]);
	}
	sort(a,a+nc);
	sort(b,b+np);
	for(i=nc-1,j=np-1;b[j]>0&&a[i]>0&&j>=0&&i>=0;i--,j--){
		sum+=a[i]*b[j];
	}
	for(i=0,j=0;b[j]<0&&a[i]<0&&i<nc&&j<np;i++,j++){
		sum+=a[i]*b[j];
	}
	printf("%lld
",sum);
	return 0;
}