1037. Magic Coupon (25)【贪心】——PAT (Advanced Level) Practise

题目信息

1. Magic Coupon (25)

时间限制100 ms
内存限制65536 kB
代码长度限制16000 B

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:
43

解题思路

首尾同时向中间逼近进行贪心，选择大于0 的较大值

AC代码

#include <cstdio>#include <vector>#include <algorithm>using namespace std;int main(){    int n, t;    vector<int> a, b, c;    scanf("%d", &n);    while (n--){        scanf("%d", &t);        a.push_back(t);    }    scanf("%d", &n);    while (n--){        scanf("%d", &t);        b.push_back(t);    }    sort(a.begin(), a.end());    sort(b.begin(), b.end());    int ba = 0, bb = 0, ea = a.size() - 1, eb = b.size() - 1;    int s = 0;    while (ba <= ea && bb <= eb){        int f = a[ba] * b[bb];        int e = a[ea] * b[eb];        if (f >= e && f >= 0){            s += f;            ++ba;            ++bb;        }else if (e >= f && e >= 0){            s += e;            --ea;            --eb;        }else{            break;        }    }    printf("%d\n", s);    return 0;}