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 ? }, and a set of product values { 7 6 ? ? } (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, 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
#include<iostream> #include<vector> #include<algorithm> #include<queue> #include<string> #include<map> #include<set> #include<stack> #include<string.h> #include<cstdio> #include<cmath> using namespace std; int main() { int nc; scanf("%d",&nc); vector<int> coupons(nc); for(int i=0; i<nc; i++) { scanf("%d",&coupons[i]); } int np; scanf("%d",&np); vector<int>products(np); //products.resize(np); for(int i=0; i<np; i++) { scanf("%d",&products[i]); } sort(coupons.begin(),coupons.end(),greater<int>()); sort(products.begin(),products.end(),greater<int>()); int sum=0; for(int i=0; i<nc&&i<np; i++) { // cout<<coupons[i]; if(coupons[i]>0&&products[i]>0) sum+=coupons[i]*products[i]; else break; } reverse(coupons.begin(),coupons.end()); reverse(products.begin(),products.end()); for(int i=0; i<nc&&i<np; i++) { if(coupons[i]<0&&products[i]<0) sum+=coupons[i]*products[i]; else break; } cout<<sum<<endl; return 0; }
原文地址:https://www.cnblogs.com/zhanghaijie/p/10324756.html