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
题目大意:给出两个数字序列,从这两个序列中分别选取相同数量的元素进行一对一相乘,问能得到的乘积之和最大为多少~
分析:把这两个序列都从小到大排序,将前面都是负数的数相乘求和,然后将后面都是正数的数相乘求和~
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#include <cstdio> #include <vector> #include <algorithm> using namespace std; int main() { int m, n, ans = 0, p = 0, q = 0; scanf("%d", &m); vector<int> v1(m); for(int i = 0; i < m; i++) scanf("%d", &v1[i]); scanf("%d", &n); vector<int> v2(n); for(int i = 0; i < n; i++) scanf("%d", &v2[i]); sort(v1.begin(), v1.end()); sort(v2.begin(), v2.end()); while(p < m && q < n && v1[p] < 0 && v2[q] < 0) { ans += v1[p] * v2[q]; p++; q++; } p = m - 1, q = n - 1; while(p >= 0 && q >= 0 && v1[p] > 0 && v2[q] > 0) { ans += v1[p] * v2[q]; p--; q--; } printf("%d", ans); return 0; } |
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