PAT_A 1037 Magic Coupon

PAT_A 1037 Magic Coupon

分析

尽量增大总回报值即可得到结果。

PAT_A 1037 Magic Coupon

题目的描述

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 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

AC的代码

#include<bits/stdc++.h>

using namespace std;

bool cmp(int a,int b){
    return a>b;
}

int main(){
    int NC,NP;
    scanf("%d",&NC);
    vector<int> C(NC);
    for(int i=0;i<NC;i++){
        scanf("%d",&C[i]);
    }
    
    scanf("%d",&NP);
    vector<int> P(NP);
    for(int i=0;i<NP;i++){
        scanf("%d",&P[i]);
    }
    sort(C.begin(),C.end(),cmp);
    sort(P.begin(),P.end(),cmp);
    long long totlareward = 0;
    for(int i=0, j=0;i<NC&&j<NP&&C[i]>0&&P[j]>0;j++,i++){

        totlareward += C[i]*P[j];

    }
    for(int i=NC-1, j=NP-1;i>-1&&j>-1&&C[i]<0&&P[j]<0;i--,j--){

        totlareward += C[i]*P[j];


    }
    printf("%lld\n",totlareward);
    return 0;
}