PAT 1037. Magic Coupon (25)（俩端乘法）

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1037. Magic Coupon (25)

时间限制
100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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)whereanegativevaluecorrespondstoabonusproduct.Youcanapplycoupon3(withNbeing4)toproduct1(withvalueM7) to get M28back;coupon2toproduct2togetM12 back; and coupon 4 to product 4 to get M3back.Ontheotherhand,ifyouapplycoupon3toproduct4,youwillhavetopayM12 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

题目大意

• 1.给出俩列数组，有正有负，问你分别在他们俩组中，取数字相乘，每个数字只能用一次，然后求最大的乘机和。
• 2.左边取最大的正数相乘，右边取绝对值最大的负数相乘。

AC代码

#include<iostream>#include<algorithm>#include<vector>using namespace std;bool cmp(int a,int b){    return a > b;}int main(int argc, char *argv[]){    int n1,n2;    cin >> n1;    vector<int>  have(n1);    for (int i = 0; i < n1; ++i) {        cin >> have[i];    }    cin >> n2;    vector<int> product(n2);    for (int i = 0; i < n2; ++i) {        cin >> product[i];    }    //排序    sort(have.begin(),have.end(),cmp);    sort(product.begin(),product.end(),cmp);    //左边取最大的正数相乘，右边取绝对值最大的负数相乘    int hl,hr,pl,pr,sum =0;    hl = 0,hr = n1-1,pl = 0,pr = n2 -1;    while (have[hl]>0&&product[pl]>0) {        sum += have[hl] * product[pl];        hl++;        pl++;    }    while (have[hr]<0&&product[pr]<0) {        sum += have[hr] * product[pr];        hr--;        pr--;    }    cout << sum << endl;    return 0;}