PAT甲级 1037

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: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;
}

 

### 关于 PAT 甲级 1024 题目 PAT (Programming Ability Test) 是一项编程能力测试,其中甲级考试面向有一定编程基础的学生。对于 PAT 甲级 1024 题目,虽然具体题目描述直接给出,但从相似类型的题目分析来看,这类题目通常涉及较为复杂的算法设计。 #### 数据结构的选择与实现 针对此类问题,常用的数据结构包括但不限于二叉树节点定义: ```cpp struct Node { int val; Node* lchild, *rchild; }; ``` 此数据结构用于表示二叉树中的节点[^1]。通过这种方式构建的二叉树能够支持多种遍历操作,如前序、中序和后序遍历等。 #### 算法思路 当处理涉及到图论的问题时,深度优先搜索(DFS)是一种常见的解题策略。特别是当需要寻找最优路径或访问尽可能多的节点时,结合贪心算法可以在某些情况下提供有效的解决方案[^2]。 #### 输入输出格式说明 根据以往的经验,在解决 PAT 类型的问题时,输入部分往往遵循特定模式。例如,给定 N 行输入来描述每个节点的信息,每行按照如下格式:“Address Data Next”,这有助于理解如何解析输入并建立相应的数据模型[^4]。 #### 数学运算示例 有时也会遇到基本算术表达式的求值问题,比如分数之间的加减乘除运算。下面是一些简单的例子展示不同情况下的计算结果: - \( \frac{2}{3} + (-2) = -\frac{7}{3}\) -2) = -\frac{4}{3}\) - \( \frac{2}{3} ÷ (-2) = -\frac{1}{3}\) 这些运算是基于样例提供的信息得出的结果[^3]。
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