1037 Magic Coupon (25)

最新推荐文章于 2020-11-10 11:31:10 发布

ChenyutingZJU

最新推荐文章于 2020-11-10 11:31:10 发布

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本文介绍了一种算法，用于解决火星魔法店的优惠券问题，通过将优惠券与产品价值进行最优匹配，以获得最大化的回馈金额。该算法首先对优惠券和产品价值进行排序，然后分别从两端开始匹配，确保负值优惠券与负值产品价值相乘，正值优惠券与正值产品价值相乘，从而实现利益最大化。

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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 <= 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:

解题思路：
两数组中的值两两相乘使值最大。
1.先排序，小的在前。
2.不能使用同号来判断，那样可能会判断两遍。要明确地先判同负，再判同正。

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
bool cmp(int a,int b){
    return a<b;
}
int a[maxn],b[maxn];
int main(){
    int nc,np,ans=0;
    scanf("%d",&nc);
    for(int i=0;i<nc;i++)
        scanf("%d",&a[i]);
    scanf("%d",&np);
    for(int i=0;i<np;i++)
        scanf("%d",&b[i]);
    sort(a,a+nc,cmp);
    sort(b,b+np,cmp);
    int i=0,j=0,lena=nc-1,lenb=np-1;
    while(a[i]<0&&b[i]<0){
        ans+=(a[i]*b[i]);
        i++;
        j++;
    }
    while(a[lena]>0&&b[lenb]>0){
        ans+=(a[lena]*b[lenb]);
        lena--;
        lenb--;
    }
    printf("%d",ans);
    return 0;
}