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:4 1 2 4 -1 4 7 6 -2 -3Sample Output:
43
题意:一个商店提供优惠券，使用此优惠券能使得商店返回此优惠券的倍数*商品的价格。注意这两个数字都可以是负数。求最大能返回的商品价值。
思路：典型的一道贪心的题目，把优惠券的倍数和商品的价格分别区分为正数和负数，再排序用正数乘正数，负数乘负数可得出结果。
#include<iostream>
#include<algorithm>
using namespace std;
int arry1[100005],arry2[100005],arry3[100005],arry4[100005];
bool cmp1(int x,int y)
{
return x<y;
}
bool cmp2(int x,int y)
{
return x>y;
}
int main()
{
int nc,np;
int c1=0,c2=0,p1=0,p2=0;
int i,k;
cin>>nc;
for(i=0;i<nc;i++)
{
cin>>k;
if(k>=0) arry1[c1++]=k;
else arry2[c2++]=k;
}
cin>>np;
for(i=0;i<np;i++)
{
cin>>k;
if(k>=0) arry3[p1++]=k;
else arry4[p2++]=k;
}
sort(arry1,arry1+c1,cmp2);
sort(arry2,arry2+c2,cmp1);
sort(arry3,arry3+p1,cmp2);
sort(arry4,arry4+p2,cmp1);
int sum=0;
//cout<<c1<<" "<<c2<<" "<<p1<<" "<<p2<<endl;
for(i=0;i<c1&&i<p1;i++)
{
sum=sum+arry1[i]*arry3[i];
}
for(i=0;i<c2&&i<p2;i++)
{
sum=sum+arry2[i]*arry4[i];
}
cout<<sum<<endl;
return 0;
}
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原文：http://blog.csdn.net/zju_ziqin/article/details/19964389