1037 Magic Coupon （25 分）

1037 Magic Coupon （25 分）

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 N​C​​, followed by a line with N​C​​ coupon integers. Then the next line contains the number of products N​P​​, followed by a line with N​P​​product values. Here 1≤N​C​​,N​P​​≤10​5​​, and it is guaranteed that all the numbers will not exceed 2​30​​.

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


分析：1、从大到小排序，再将对应位置上乘积大于0的部分加起来。
      2、从小到大排序，再将对应位置上乘积大于0的部分加起来。
要注意过程中要控制两个数均为正数或负数，不然回重复加。。。

 1 /**
 2 * Copyright(c)
 3 * All rights reserved.
 4 * Author : Mered1th
 5 * Date : 2019-02-26-00.04.11
 6 * Description : A1037
 7 */
 8 #include<cstdio>
 9 #include<cstring>
10 #include<iostream>
11 #include<cmath>
12 #include<algorithm>
13 #include<string>
14 #include<unordered_set>
15 #include<map>
16 #include<vector>
17 #include<set>
18 using namespace std;
19 const int maxn=100010;
20 int a[maxn],b[maxn];
21 bool cmp1(int a,int b){
22     return a>b;
23 }
24 bool cmp2(int a,int b){
25     return a<b;
26 }
27 int main(){
28 #ifdef ONLINE_JUDGE
29 #else
30     freopen("1.txt", "r", stdin);
31 #endif
32     int n1,n2;
33     scanf("%d",&n1);
34     for(int i=0;i<n1;i++){
35         scanf("%d",&a[i]);
36     }
37     scanf("%d",&n2);
38     for(int i=0;i<n2;i++){
39         scanf("%d",&b[i]);
40     }
41     int ans=0;
42     sort(a,a+n1,cmp1);
43     sort(b,b+n2,cmp1);
44     for(int i=0;i<min(n1,n2);i++){
45         if(a[i]*b[i]>0&&a[i]>=0&&b[i]>0){
46             ans+=a[i]*b[i];
47         }
48         else break;
49     }
50     sort(a,a+n1,cmp2);
51     sort(b,b+n2,cmp2);
52     for(int i=0;i<min(n1,n2);i++){
53         if(a[i]*b[i]>0&&a[i]<0&&b[i]<0){
54             ans+=a[i]*b[i];
55         }
56         else break;
57     }
58     cout<<ans;
59     return 0;
60 }

 

转载于:https://www.cnblogs.com/Mered1th/p/10434875.html