PAT (Advanced Level) 1037. Magic Coupon (25) 解题报告

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$) 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:

41 2 4 -147 6 -2 -3

Sample Output:

43

题意：从两组中各挑一个乘起来，求和最大

代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <cmath>using namespace std;int na, nb, a[101000], b[101000];int main(){    long long sum = 0;    scanf("%d", &na);    for(int i = 0; i < na; i++)        scanf("%d", &a[i]);    scanf("%d", &nb);    for(int i = 0; i < nb; i++)        scanf("%d", &b[i]);    sort(a, a + na);    sort(b, b + nb);    int t = 0;    for(int i = 0; i < na; i++)    {        if(a[i] > 0) break;        if(a[i] < 0 && b[t] < 0)        {            sum += a[i] * b[i];            t ++;        }    }    t = nb-1;    for(int i = na - 1; i >= 0; i--)    {        if(a[i] < 0) break;        if(a[i] > 0 && b[t] > 0)        {            sum += a[i] * b[t];            t --;        }    }    printf("%lld\n", sum);    return 0;}