4.4 PAT A1037 Magic Coupon (25分)(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 −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 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≤N​C​​,N​P​​≤10​5​​, 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

题意为从给定 的两组数中选择两个数求积所能获得的最大值。每个数只能选择一次，且正数加，负数要减。可以分别求出正数相乘的积的和和两个负数积的和，累加即可。

参考代码1：分别统计正负数

#include <cstdio>
#include <algorithm>
using namespace std;
int cou1[100020],cou2[100020];
int val1[100020],val2[100020];
bool cmp(int a,int b)
{
	return a>b;
}
int main()
{
	int nc,np;
	while(scanf("%d",&nc)!=EOF)
	{
		int tmp,c1=0,c2=0,v1=0,v2=0;
		for(int i=0;i<nc;++i)
		{
			scanf("%d",&tmp);
			if(tmp>0)
				cou1[c1++]=tmp; //正数 
			else
				cou2[c2++]=tmp;  //负数 
		}
		scanf("%d",&np);
		for(int i=0;i<np;++i)
		{
			scanf("%d",&tmp);
			if(tmp>0)
				val1[v1++]=tmp;
			else
				val2[v2++]=tmp; 
		}
		sort(cou1,cou1+c1,cmp); //正数从大到小排序 
		sort(cou2,cou2+c2);//负数从小到大 
		sort(val1,val1+v1,cmp);
		sort(val2,val2+v2);
		int ans=0;
		for(int i=0;i<c1&&i<v1;++i) //统计正数的积 
		{
			ans+=cou1[i]*val1[i];
		}	
		for(int i=0;i<c2&&i<v2;++i) //统计负数的积 
		{
			ans+=cou2[i]*val2[i];
		}
		printf("%d\n",ans); 
	}
	return 0;
}

 参考代码2：

#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int Nc[100010];
int Np[100010];
int main()
{
	int nc,np;
	while(scanf("%d",&nc)!=EOF)
	{
		for(int i=0;i<nc;++i)
			scanf("%d",&Nc[i]);
		scanf("%d",&np);
		for(int i=0;i<np;++i)
			scanf("%d",&Np[i]);
		sort (Nc,Nc+nc);
		sort (Np,Np+np);
		int i=0,j,ans=0;
		while(i<nc&&i<np &&Nc[i]<0&&Np[i]<0) //先从前到后统计负数的积 
		{
			ans+=Nc[i]*Np[i];
			++i;
		 } 
		i=nc-1;
		j=np-1;
		while(i>=0&&j>=0&&Nc[i]>0&&Np[j]>0) //从后往前统计正数 
		{
			ans+=Nc[i]*Np[j];
			--i;
			--j;
		}
		printf("%d\n",ans); 
	}
	return 0;
}