Magic Coupon (25)

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本文介绍了一道关于魔法优惠券的算法题目，旨在通过合理的策略选取优惠券与商品搭配，以实现收益最大化。文章提供了完整的代码实现，并采用了贪心算法进行求解。

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Magic Coupon (25)

时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB 判断程序 Standard (来自小小)
题目描述
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 M7) to get M28 back; coupon 2 to product 2 to get M12 back; and coupon 4 to product 4 to get M3 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

输入描述:
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.

输出描述:
For each test case, simply print in a line the maximum amount of money you can get back.

输入例子:
4
1 2 4 -1
4
7 6 -2 -3

输出例子:
43

题解：给出两个集合，然后选取元素使乘积最大，可以想到贪心策略，把两个数组从小到大排序，然后从左到右选取相同位置都是负数相乘，从右到左选取相同位置非负数相乘，然后便可以得到最大值。
平台是牛客，数据比较大，记得使用long long 类型。

#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
#define vi vector<int>
#define pii pair<int,int>
#define x first
#define y second
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define SZ(x) x.size()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define pi acos(-1)
#define mod 1000000007
#define inf 1000000007
#define ll long long
#define DBG(x) cerr<<(#x)<<"="<<x<<"\n";
#define N 200010
template <class U,class T> void Max(U &x, T y){if(x<y)x=y;}
template <class U,class T> void Min(U &x, T y){if(x>y)x=y;}
template <class T> void add(int &a,T b){a=(a+b)%mod;}
const int maxn=100001;
int n,m;
ll a1[maxn],a2[maxn];
int main()
{
    cin>>n;
    rep(i,0,n) scanf("%lld",&a1[i]);
    cin>>m;
    rep(i,0,m) scanf("%lld",&a2[i]);
    sort(a1,a1+n);
    sort(a2,a2+m);
    int  i=0,j=0;
    ll ans=0;
    while(a1[i]<0&&a2[i]<0&&i<n&&i<m)
    {
        ans+=a1[i]*a2[i];
        i++;
    }
    i=n-1,j=m-1;
    while(a1[i]>0&&a2[j]>0&&i>=0&&j>=0)
    {
        ans+=a1[i]*a2[j];
        i--,j--;
    }
    printf("%lld",ans);
    return 0;
}