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 NPproduct 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
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。。。
/** * Copyright(c) * All rights reserved. * Author : Mered1th * Date : 2019-02-26-00.04.11 * Description : A1037 */ #include<cstdio> #include<cstring> #include<iostream> #include<cmath> #include<algorithm> #include<string> #include<unordered_set> #include<map> #include<vector> #include<set> using namespace std; ; int a[maxn],b[maxn]; bool cmp1(int a,int b){ return a>b; } bool cmp2(int a,int b){ return a<b; } int main(){ #ifdef ONLINE_JUDGE #else freopen("1.txt", "r", stdin); #endif int n1,n2; scanf("%d",&n1); ;i<n1;i++){ scanf("%d",&a[i]); } scanf("%d",&n2); ;i<n2;i++){ scanf("%d",&b[i]); } ; sort(a,a+n1,cmp1); sort(b,b+n2,cmp1); ;i<min(n1,n2);i++){ &&a[i]>=&&b[i]>){ ans+=a[i]*b[i]; } else break; } sort(a,a+n1,cmp2); sort(b,b+n2,cmp2); ;i<min(n1,n2);i++){ &&a[i]<&&b[i]<){ ans+=a[i]*b[i]; } else break; } cout<<ans; ; }