PAT-A1037 Magic Coupon 题目内容及题解
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 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
题目大意
有一种神奇优惠券,使用后可以获得产品价格和优惠券相乘的价值,每个优惠券和每件产品只能计算一次,任务是获得尽可能多的返利。
解题思路
- 使用long long类型存储结果;
- 将产品和优惠券分别排序,正值与正值、负值与负值分别相乘;
- 计算结果并返回零值。
代码
#include<cstdio>
#include<algorithm>
#define maxn 100010
using namespace std;
bool cmp(long long a,long long b){
return a>b;
}
int main(){
int NC,NP;
int i,j;
long long C[maxn],P[maxn],fee=0;
scanf("%d",&NC);
for(i=0;i<NC;i++){
scanf("%lld",&C[i]);
}
sort(C,C+NC,cmp);
scanf("%d",&NP);
for(i=0;i<NP;i++){
scanf("%lld",&P[i]);
}
sort(P,P+NP,cmp);
i=0;
while(C[i]>0&&P[i]>0){
fee+=C[i]*P[i];
i++;
}
i=NC-1;
j=NP-1;
while(C[i]<0&&P[j]<0){
fee+=C[i]*P[j];
i--;
j--;
}
printf("%lld\n",fee);
return 0;
}
运行结果