PAT 甲级 1037 Magic Coupon

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最新推荐文章于 2021-03-06 15:06:45 发布

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分类专栏： PAT甲级 PAT甲级/乙级机试经验文章标签： PAT甲级 Magic Coupon 1037

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探讨了在火星上一家魔法商店中如何利用特定规则的优惠券和产品价值获取最大利益的问题，通过贪心算法实现了最优策略，给出了AC代码实现。

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1037 Magic Coupon （25 point(s)）

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:

Sample Output:

经验总结：

这一题，使用的是贪心算法，两个数组排一下序（这里设为从小到大），然后按顺序，从左边两个数组都是负数的话就相乘并且加入收益，从右边两个都是正数就相乘并且加入收益，只要不满足条件就跳出，最终算出结果就行啦~注意最终结果要用long long 存储，毕竟题目说所有数不大于2^30，两个2^30相乘就2^60多了，注意一点就行~

AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=100010;
int c[maxn],p[maxn];
int main()
{
	long long ans=0;
	int nc,np;
	scanf("%d",&nc);
	for(int i=0;i<nc;++i)
		scanf("%d",&c[i]);
	scanf("%d",&np);
	for(int i=0;i<np;++i)
		scanf("%d",&p[i]);
	sort(c,c+nc);
	sort(p,p+np);
	int L=nc>np?np:nc;
	for(int i=0;i<L;++i)
	{
		if(c[i]<0&&p[i]<0)
			ans+=(long long)c[i]*p[i];
		else
			break;
	}
	int i=nc-1;
	int j=np-1;
	while(i>=0&&j>=0)
	{
		if(c[i]>0&&p[j]>0)
			ans+=(long long)c[i]*p[j];
		else
			break;
		--i;--j;
	}
	printf("%lld\n",ans);
    return 0;
}