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
用了最土的办法，没用任何技巧，写了快100行，一次过，这种感觉～很美妙～
【五分钟后更新】原来大家只要写二十几行....
#include<cstdio>
#include <cstring>
#include<algorithm>
#include <iostream>
#include <cmath>
using namespace std;
const int maxn = 1e5 + 5;
int goods[maxn],coupon[maxn],coupon_f[maxn],coupon_z[maxn],goods_f[maxn],goods_z[maxn];
bool cmp(int a,int b)
{
return a > b;
}
int main()
{
int a,b,c,sum = 0;
int f1 = 1 ,f2 = 1;
int tmp1 = 0,tmp2 = 0,tmp11 = 0,tmp22 = 0;
cin>>a;
for(int i = 0 ; i < a; i++)
{
cin>>coupon[i];
if(coupon[i] < 0)
f1 = 1;
}
cin>>b;
for(int i = 0 ; i < b ;i ++)
{
cin>>goods[i];
if(goods[i] < 0)
f2 = 1;
}
if(f1 == 0 && f2 == 0)//都是正的
{
sort(coupon,coupon + a,cmp);
sort(goods,goods + b,cmp);
c = max(a,b);
for(int i = 0 ; i < c ;i++)
{
sum += coupon[i] * goods[i];
}
}
else if(f1 == 1 && f2 == 0)//券里有负的，商品都是正的
{
sort(coupon,coupon + a,cmp);
sort(goods,goods + b,cmp);
c = max(a,b);
for(int i = 0 ; i < c;i++)
{
if(coupon[i] < 0)
break;
sum += coupon[i] * goods[i];
}
}
else if(f1 == 0 && f2 == 1)//券都是正的,商品里有负的
{
sort(coupon,coupon + a,cmp);
sort(goods,goods + b,cmp);
c = max(a,b);
for(int i = 0 ; i < c;i++)
{
if(goods[i] < 0)
break;
sum += coupon[i] * goods[i];
}
}
else if(f1 == 1 && f2 == 1)
{
for(int i = 0 ; i < a; i++)
{
if(coupon[i] < 0)
coupon_f[tmp1 ++] = coupon[i];
if(coupon[i] > 0)
coupon_z[tmp2 ++] = coupon[i];
}
sort(coupon_f,coupon_f + tmp1);//负的小到大排，绝对值大到小
sort(coupon_z,coupon_z + tmp2,cmp);//正的大到小排
for(int i = 0 ; i < b; i++)
{
if(goods[i] < 0)
goods_f[tmp11 ++] = goods[i];
if(goods[i] > 0)
goods_z[tmp22 ++] = goods[i];
}
sort(goods_f,goods_f + tmp11);//负的小到大排，绝对值大到小
sort(goods_z,goods_z + tmp22,cmp);//正的大到小排
int d = max(tmp2,tmp22);
for(int i = 0;i < d;i++)//正的部分大到小分别相乘加起来就行
{
sum += goods_z[i] * coupon_z[i];
}
int f = max(tmp1,tmp11);
for(int i = 0;i < f;i++)//负的部分小到大分别相乘 再加起来
{
sum += goods_f[i] * coupon_f[i];
}
}
cout<<sum<<endl;
return 0;
}