1037. Magic Coupon (25)-PAT

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

推荐指数：※

来源：http://pat.zju.edu.cn/contests/pat-a-practise/1037

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#include<iostream>
#include<stdio.h>
#include <stdlib.h>
#include<stdio.h>
using namespace std;
int compare(const void *a,const void *b){
	return *(int *)a-*(int *)b;
}
int main()
{
	int nc,np,i,j;
	scanf("%d",&nc);
	int *co=new int [nc];
	for(i=0;i<nc;i++)
		scanf("%d",&co[i]);
	scanf("%d",&np);
	int *po=new int [np];
	for(i=0;i<np;i++)
		scanf("%d",&po[i]);
	qsort(co,nc,sizeof(int),compare);
	qsort(po,np,sizeof(int),compare);
	long sum=0;
	for(i=0;i<np&&i<nc;i++){
		if(co[i]<0&&po[i]<0){
			sum+=co[i]*po[i];
		}
		else
			break;
	}
	for(i=nc-1,j=np-1;i>=0,j>=0;i--,j--){
		if(co[i]>0&&po[j]>0)
			sum+=co[i]*po[j];
		else
			break;
	}
	printf("%ld\n",sum);
	return 0;
}