A1037. Magic Coupon

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

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
bool cmp(long long a, long long b){
    return a > b;
}
long long C[100001], P[100001], ans = 0;
int main(){
    int NC, NP;
    scanf("%d", &NC);
    for(int i = 0; i < NC; i++)
        scanf("%lld", &C[i]);
    scanf("%d", &NP);
    for(int i = 0; i < NP; i++)
        scanf("%lld", &P[i]);
    sort(C, C + NC, cmp);
    sort(P, P + NP, cmp);
    for(int i = 0; i < NC && i < NP && C[i] >= 0 && P[i] >= 0; i++)
        ans += C[i] * P[i];
    for(int i = NC - 1, j = NP - 1; i >= 0 && j >= 0 && C[i] <= 0 && P[j] <= 0; i--, j--)
        ans += C[i] * P[j];
    printf("%lld", ans);
    cin >> NP;
    return 0;
}

总结：

1、题意：给出两个集合，每个集合都分别有正数、负数或0。分别从两集合里选出相同个数的数，求他们的乘积之和最大为多少。需要注意的是，本题选取的乘积的组数是任意的，不需要用掉所有的数。因为正正相乘则正数越大积越大，负负相乘负数越小积越大。因此只需要将两个集合从大到小排序，从下标0开始依次正正相乘，直到遇见二者异号。同理从尾部向前负负相乘，直到异号。