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:

4
1 2 4 -1
4
7 6 -2 -3

Sample Output:

43

---

提交代码

 

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
bool cmp(int a,int b){
    return a>b;
}
int main(){
    //freopen("D:\\INPUT.txt", "r", stdin);
    int nc,np,i,j;
    scanf("%d",&nc);
    long long *ncp=new long long[nc+5];
    for(i=0;i<nc;i++){
        scanf("%lld",&ncp[i]);
    }
    sort(ncp,ncp+nc,cmp);
    scanf("%d",&np);
    long long *npp=new long long[np+5];
    for(i=0;i<np;i++){
        scanf("%lld",&npp[i]);
    }
    sort(npp,npp+np,cmp);
    long long sum=0;
    i=0,j=0;
    while(i<nc&&j<np){
        long long mul=ncp[i]*npp[j];
        if(mul>=0){
            sum+=mul;
            i++;
            j++;
            continue;
        }
        if(ncp[i]<0){
            break;
        }    
        if(npp[j]<0){
            break;
        }
    }
    //跳出循环的可能：
    //i或j越界 
    //i指向一个负数
    //j指向一个负数
    int ii=nc-1,jj=np-1; 
    for(;ii>=i&&jj>=j&&ncp[ii]*npp[jj]>=0;ii--,jj--){
        sum+=ncp[ii]*npp[jj];
    } 
    printf("%lld\n",sum);
    return 0;
}