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 NCcoupon integers. Then the next line contains the number of products NP, followed by a line with NP product values. Here 1≤NC,NP≤10^5, 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
题意:
给出两个集合,从这两个集合中分别选取相同数量的元素进行一对一相乘,问能够得到的成绩之和最大为多少。
1≤NC,NP≤10^5, all the numbers will not exceed 2^30
注意:
- 首先应该注意到,本题中两个集合中均存在正数、负数和零,所以要分开考虑。对于正数相乘,最大的两个正数相乘,成绩最大;最小的两个负数相乘,乘积最大;0 乘以任何数均为 0 ,不作为考虑。
- all the numbers will not exceed 2^30 表明:所有的数字以及结果可以设为 int 型
#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn = 100010;
int coupon[maxn], product[maxn];
int main(){
int n, m;
scanf("%d", &n);
for(int i = 0; i < n; i++){
scanf("%d", &coupon[i]);
}
scanf("%d", &m);
for(int i = 0; i < m; i++){
scanf("%d", &product[i]);
}
sort(coupon, coupon + n);
sort(product, product + m);
int i = 0, j, ans = 0;
while(i < n && i < m && coupon[i] < 0 && product[i] < 0){
ans += coupon[i] * product[i];
i++;
}
i = n - 1;
j = m - 1;
while(i >= 0 && j >= 0 && coupon[i] > 0 && product[j] > 0){
ans += coupon[i] * product[j];
i--;
j--;
}
printf("%d\n", ans);
return 0;
}