pat1037. Magic Coupon (25)

1037. Magic Coupon (25)

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:

Sample Output:

提交代码

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 using namespace std;
 5 bool cmp(int a,int b){
 6     return a>b;
 7 }
 8 int main(){
 9     //freopen("D:\INPUT.txt", "r", stdin);
10     int nc,np,i,j;
11     scanf("%d",&nc);
12     long long *ncp=new long long[nc+5];
13     for(i=0;i<nc;i++){
14         scanf("%lld",&ncp[i]);
15     }
16     sort(ncp,ncp+nc,cmp);
17     scanf("%d",&np);
18     long long *npp=new long long[np+5];
19     for(i=0;i<np;i++){
20         scanf("%lld",&npp[i]);
21     }
22     sort(npp,npp+np,cmp);
23     long long sum=0;
24     i=0,j=0;
25     while(i<nc&&j<np){
26         long long mul=ncp[i]*npp[j];
27         if(mul>=0){
28             sum+=mul;
29             i++;
30             j++;
31             continue;
32         }
33         if(ncp[i]<0){
34             break;
35         }    
36         if(npp[j]<0){
37             break;
38         }
39     }
40     //跳出循环的可能：
41     //i或j越界 
42     //i指向一个负数
43     //j指向一个负数
44     int ii=nc-1,jj=np-1; 
45     for(;ii>=i&&jj>=j&&ncp[ii]*npp[jj]>=0;ii--,jj--){
46         sum+=ncp[ii]*npp[jj];
47     } 
48     printf("%lld
",sum);
49     return 0;
50 }