PAT甲级 1037

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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 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:

题意：有n种优惠券和m种商品，若优惠券是正数，商品的价格也是正数，则可以得到两者乘积的金钱；换言之只要优惠券与商品的乘积是正数，则消费者可以得到成绩大小的金钱，否则，消费者就要向商家支付优惠券与商品乘积绝对值的金钱，问消费者最终能得到的最大金额。

总结：将优惠券、商品价格分别进行从小到大的排序，从前往后进行负数的乘积相加，从后往前进行正数的乘积相加，即能得出最大金额。

#include<iostream>
#include<bits/stdc++.h>
#define long 100001
using namespace std;
int a[100005],b[100005];
int main()
{
    int n1;
    cin>>n1;
    for(int i=0;i<n1;i++)
    {
        cin>>a[i];
    }
    sort(a,a+n1);
    int n2;
    cin>>n2;
    for(int j=0;j<n2;j++)
    {
        cin>>b[j];
    }
    sort(b,b+n2);
    int ans=0;
    int it1=0;
    int it2=0;
    while(it1<n1&&it2<n2)
    {
        if(a[it1]<0&&b[it2]<0)
        {
            ans+=a[it1]*b[it2];
            it1++;
            it2++;
        }
        else
        {
            break;
        }
    }
    it1=n1-1;
    it2=n2-1;
    while(it1>=0&&it2>=0)
    {
        if(a[it1]>0&&b[it2]>0)
        {
            ans+=a[it1]*b[it2];
            it1--;
            it2--;
        }
        else
        {
            break;
        }
    }
    cout<<ans<<endl;
}