pat-a1037. Magic Coupon (25)

简单贪心。题目大概意思就是两个数组的数，每次取两个数进行乘积进行累加。一直取到某个数组没数，求累加值的最大值

我做时把正数和负数分开了，正数从大到小，负数从小到大。。不想倒着乘

题目保证没超过int范围，但是乘积可能会。所以要用long long int..同时只定义结果是long long int应该是不可行的(没试过)。两个int乘积存到long long int里面应该不是正确的值

既然水题今天继续刷

#include<cstdio>#include<algorithm>#include<functional>using namespace std;typedef long long int LL;LL a[100010];LL a1[100010];LL b[100010];LL b1[100010];int main(){int n,m;long long int ans=0,t;int len1=0,len2=0,len3=0,len4=0;scanf("%d",&n);for(int i=0;i<n;++i){scanf("%lld",&t);if(t>0) a[len1++]=t;else if(t<0) a1[len2++]=t;}sort(a,a+len1,greater<int>());sort(a1,a1+len2);scanf("%d",&m);for(int i=0;i<m;++i){scanf("%lld",&t);if(t>0) b[len3++]=t;else if(t<0) b1[len4++]=t;}sort(b,b+len3,greater<int>());sort(b1,b1+len4);for(int i=0;i<len1&&i<len3;++i) ans+=a[i]*b[i];for(int i=0;i<len2&&i<len4;++i) ans+=a1[i]*b1[i];printf("%lld\n",ans);return 0;}

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:

41 2 4 -147 6 -2 -3

Sample Output:

43