时间限制
100 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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 -3
Sample Output:
43
解：思路题，我起初想开四个数组的，前两个数组分别存coupon里的正值和负值，后两个存product里的正值和负值，但写的时候觉得可以简化下，只开了两个数组，其他的自己写的时候控制下，完全可以实现四个数组一样的功能。这道题，我先存值，输入的过程中统计正负值的个数，得到coupon和product对应数组中正数个数的最小值和负数个数的最小值，排序，再正正相乘+负负相乘，最后得到结果。
代码：
#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n=0,m=0; //两数组对应的个数
int len1=0,len2=0; //统计两个数组中负值和正值个数的最小值
scanf("%d",&n);
int *a=new int[n+1];
int cnt1=0,cnt2=0;
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
if(a[i]>0) cnt1++;
if(a[i]<0) cnt2++;
}
scanf("%d",&m);
int *b=new int[m+1];
int dnt1=0,dnt2=0;
for(int i=0;i<m;i++)
{
scanf("%d",&b[i]);
if(b[i]>0) dnt1++;
if(b[i]<0) dnt2++;
}
len1=min(cnt1,dnt1);
len2=min(cnt2,dnt2);
sort(a,a+n);
sort(b,b+m);
int result=0;
for(int i=0;i<len2;i++)
{
if(a[i]<0&&b[i]<0)
{
result+=a[i]*b[i];
}
}
int j=0,en1=n-1,en2=m-1;
while(j<len1)
{
j++;
if(a[en1]>0&&b[en2]>0)
result+=a[en1--]*b[en2--];
}
printf("%d\n",result);
}
