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:

Sample Output:

分析：

抽象出来就是找到两个数列相应乘积的最大值，数仅仅能用一次，且能够不用。

要求优惠券组合的最大值，基于这样一个事实，把两边的优惠券都按递增顺序排好，然后选取都是绝对值最大的两个负数相乘，也能够选择绝对值最大的两个正数相乘。依次直到没有能够配对的，得到的累加和就是最大的。

特别要注意while中成立的条件，不然可能v_begin会越界导致V[v_begin]没有值

AC代码：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<vector>
using namespace std;
vector<long long>V1;
vector<long long>V2;
int main(){
 //freopen("F://Temp/input.txt", "r", stdin);
 int n, m;
 cin >> n;
 for (int i = 0; i < n; i++){
  long long tmp;
  cin >> tmp;
  V1.push_back(tmp);
 }
 cin >> m;
 if (n == 0 || m == 0){
  cout << 0 << endl;
  return 0;
 }
 for (int i = 0; i < m; i++){
  long long tmp;
  cin >> tmp;
  V2.push_back(tmp);
 }
 sort(V1.begin(), V1.end());
 sort(V2.begin(), V2.end());
 int v1_begin = 0, v1_end = V1.size() - 1 , v2_begin = 0, v2_end = V2.size() - 1;
 long long sum = 0;
 while (v1_begin <V1.size() && v2_begin < V2.size() && V1[v1_begin] <0 &&  V2[v2_begin] < 0){//找到各自中绝对值最大的负数进行相乘
  sum += V1[v1_begin] * V2[v2_begin];
  v1_begin++;
  v2_begin++;
 }
 while (v1_end >= 0 && v2_end >= 0 && V1[v1_end] > 0 && V2[v2_end] > 0){
  sum += V1[v1_end] * V2[v2_end];//找到各自中最大的正数进行相乘
  v1_end--;
  v2_end--;
 }
 cout << sum << endl;
 return 0;
}

截图：

——Apie陈小旭