PAT 甲级 1037. Magic Coupon (25)

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

ps：自己方法太挫了。。从大到小排序然后遍历，中间会遇到负数的问题。。。不进行优化还会超时。。

#include<iostream>#include<string>#include<cstdio>#include<vector>#include<algorithm>#include<queue>#include <cmath>#include <map>#include <set>#include <string.h>using namespace std;vector<int> v1, v2;bool cmp(int a, int b) {return a < b;}int main() {int n, m;cin >> n;v1.resize(n);for (int i = 0; i < n; i++) {cin >> v1[i];}cin >> m;v2.resize(m);for (int i = 0; i < m; i++) {cin >> v2[i];}sort(v1.begin(), v1.end(), cmp);sort(v2.begin(), v2.end(), cmp);int sum = 0, pos1 = 0, pos2 = 0;while (pos1 < n&&pos2 < m&&v1[pos1] < 0 && v2[pos2] < 0){sum += v1[pos1++] * v2[pos2++];}pos1 = n - 1, pos2 = m - 1;while (pos1 >= 0 && pos2 >= 0 && v1[pos1] > 0 && v2[pos2]>0) {sum += v1[pos1--] * v2[pos2--];}cout << sum<<endl;//cin >> n;return 0;}