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.
輸入描述：
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.
輸出描述：
For each test case, simply print in a line the maximum amount of money you can get back.
示例：
輸入：
4
1 2 4 -1
4
7 6 -2 -3
輸出：
43
題目分析：
在這道題中，它其實就是給你優惠券和產品，使用優惠券乘上你產品價值得到數，就是要返回你的金額；在題目中，開始的背景給你講了優惠券爲負，產品爲正，得到的結果就是負，那麼你就要給商店倒貼錢；反之，優惠券爲正，產品價值爲負，得到的結果也是負，也需要倒貼錢；但是在計算商店要返回給你的最大金額時，肯定不需要給店鋪返錢呀！這是生活中的常識，所以我們就懂得了題目想要你回答什麼。
你輸入了一組優惠券和一組產品價值，要計算其得到金額的最大值，我們用輸入示例來畫圖講解：
代碼如下：
import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
// 優惠券數量
int numNC = sc.nextInt();
// 優惠券價值
long[] NC = new long[numNC];
for(int i = 0; i < numNC; i++){
NC[i] = sc.nextLong();
}
// 產品數量
int numNP = sc.nextInt();
// 產品價值
long[] NP = new long[numNP];
for(int i = 0; i < numNP; i++){
NP[i] = sc.nextLong();
}
// 存放使用優惠券獲得的金額
long backMoney = 0;
// 先排序
Arrays.sort(NC);
Arrays.sort(NP);
// 遍歷二維數組
for(int i = 0, j = 0; i < numNC && j < numNP && NC[i] < 0 && NP[j] < 0; i++, j++){
backMoney += NC[i] * NP[j];
}
for(int i = numNC-1, j = numNP-1; i > 0 && j > 0 && NC[i] > 0 && NP[j] > 0; i--, j--){
backMoney += NC[i] * NP[j];
}
System.out.println(backMoney);
}
}
本地IDEA中運行：
牛客網在線OJ運行：
注意： 在線OJ上的測試數據太大，記得要用long類型。