【PAT】1037. Magic Coupon (25)

最新推荐文章于 2021-10-15 11:57:23 发布

realxuejin

最新推荐文章于 2021-10-15 11:57:23 发布

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分类专栏： PAT 文章标签： PAT

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本文介绍了一种在火星魔法店利用优惠券获取最大价值的方法。通过将正数优惠券与正价值商品匹配，并将负数优惠券与负价值商品搭配，以实现利益最大化。

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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:

分析：这道题目很简单啊，不知道为什么会25分的分值。步骤：（1）coupon分为正负两组，product分为正负两组；（2）coupon中正数最大的和product中正数最大的相乘；coupon中负数最小的和product中负数最小的相乘；最终将两组数据相加就可以了。

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

//从小到大排列 
bool cmp1(int a, int b){
	return a<b;
}
//从大到小排列 
bool cmp2(int a, int b){
	return a>b;
}

int main(int argc, char** argv) {
	int nc, np, i, c, v; 
	cin>>nc;
	vector<int> posC;
	vector<int> negC;
	for(i=0; i<nc; i++){
		cin>>c;
		if(c>0){
			posC.push_back(c);
		}else if(c < 0){
			negC.push_back(c);	
		}
	}
	sort(posC.begin(), posC.end(), cmp2);
	sort(negC.begin(), negC.end(), cmp1);
	
	cin>>np;
	vector<int> posP;
	vector<int> negP;
	for(i=0; i<np; i++){
		cin>>v;
		if(v>0){
			posP.push_back(v);
		}else if(v<0){
			negP.push_back(v);
		}	 
	}
	
	sort(posP.begin(), posP.end(), cmp2);
	sort(negP.begin(), negP.end(), cmp1);
	int total = 0;
	for(i=0; i<posC.size() && i<posP.size(); i++){
		total += posC[i]*posP[i];
	}
	for(i=0; i<negC.size() && i<negP.size(); i++){
		total += negC[i]*negP[i];
	}	
	cout<<total<<endl;
	
	return 0;
}