PAT--1037. Magic Coupon

最新推荐文章于 2022-07-01 18:09:42 发布

原创最新推荐文章于 2022-07-01 18:09:42 发布 · 295 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#pat

PAT 专栏收录该内容

96 篇文章

订阅专栏

本文介绍了一种算法，用于解决如何使用不同数值的魔法优惠券购买产品以获得最大返利的问题。通过将优惠券和产品按正数、负数分类，并按绝对值大小排序，再配对求和，实现利益最大化。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:

4
1 2 4 -1
4
7 6 -2 -3

Sample Output:

题解

对nc[]和np[] 按照正数, 0, 负数并按绝对值降序的次序排序。
然后两个指针i, j扫描nc和np，符号相同的作乘积累加。

#include <bits/stdc++.h>
using namespace std;

const int maxn = 100010;
int nc, np;
int a[maxn], b[maxn];

int cmp(int x, int y){
    if(x >= 0 && y >= 0) return x > y;
    if(x <= 0 && y <= 0) return abs(x) > abs(y);
    if(x >= 0 && y <= 0) return 1;
    if(x <= 0 && y >= 0) return 0;
}

bool sign(int x, int y){
    return (x > 0 && y > 0) || (x < 0 && y < 0);
}

int main() {
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
#endif // ONLINE_JUDGE

    scanf("%d", &nc);
    for(int i = 0; i < nc; ++i) scanf("%d", a + i);
    scanf("%d", &np);
    for(int i = 0; i < np; ++i) scanf("%d", b + i);

    sort(a, a + nc, cmp);
    sort(b, b + np, cmp);

    long long ans = 0;
    int i = 0, j = 0;
    while(i < nc && j < np){
        if(sign(a[i], b[j])) {
            ans += a[i] * b[j];
            i++, j++;
        }else if(a[i] <= 0 && b[j] >= 0) j++;
        else i++;
    }

    printf("%lld\n", ans);

    return 0;
}