PAT甲级 1037

题目 Magic Coupon

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 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.

解析

一组为折扣组，另一组为商品原价组。求折扣组与原价组相乘后最大的金额

代码

#include<bits/stdc++.h>

#define INF 1<<29

using namespace std;

bool cmp(int i, int j) {
    return i > j;
}

bool cmp1(int i, int j) {
    return i < j;
}

void pat1037() {
    int nc, np;
    int ncdata[100050], npdata[100050];
    int res = 0;
    cin >> nc;
    for (int i = 0; i < nc; ++i) {
        cin >> ncdata[i];
    }
    cin >> np;
    for (int i = 0; i < np; ++i) {
        cin >> npdata[i];
    }
    sort(ncdata, ncdata + nc, cmp);
    sort(npdata, npdata + np, cmp);

    for (int i = 0; i < nc && i < np; ++i) {
        if (ncdata[i] < 0 || npdata[i] < 0) {
            break;
        }
        res += ncdata[i] * npdata[i];
    }

    sort(ncdata, ncdata + nc, cmp1);
    sort(npdata, npdata + np, cmp1);

    for (int i = 0; i < nc && i < np; ++i) {
        if (ncdata[i] > 0 || npdata[i] > 0) {
            break;
        }
        res += ncdata[i] * npdata[i];
    }

    cout << res << endl;
}


int main() {
    pat1037();
    return 0;
}