首页 > ACM题库 > HDU-杭电 > hdu 3794-magic coupon[解题报告]hoj
2015
04-13

hdu 3794-magic coupon[解题报告]hoj

Magic Coupon

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 570    Accepted Submission(s): 130



Problem Description
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
Your program must read test cases from standard input.
The input file consists of several test cases. 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 <= 10^6, and it is guaranteed that all the numbers will not exceed 10^7.
The input is finished by a negative NC.
 


Output
For each test case, your program must output to standard output. 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 4 3 2 6 1 3 2 6 3 -1
 


Sample Output
43 49
 


Source
 

#include "iostream"
#include "stdio.h"
#include "math.h"
#include "vector"
#include "queue"
#include "memory.h"
#include "algorithm"
#include "string"
using namespace std;
#define N 1000005
__int64 a1[N];
__int64 a2[N];

int main()
{
    int n1,n2,i,j;
    while(scanf("%d",&n1)!=EOF&&n1>=0)
    {
        for(i=0;i<n1;i++)
            scanf("%I64d",&a1[i]);
        scanf("%d",&n2);
        for(i=0;i<n2;i++)
            scanf("%I64d",&a2[i]);
        sort(a1,a1+n1);//或者降序可以用sort(a1,a1+n1,greater<int>()) 
        sort(a2,a2+n2);
        i=0;
        __int64 sum=0;
        while(a1[i]<0&&a2[i]<0&&i<n1&&i<n2)
            sum+=a1[i]*a2[i++];
        i=n1-1,j=n2-1;
        while(a1[i]>0&&a2[j]>0&&i>=0&&j>=0)
            sum+=a1[i--]*a2[j--];
        printf("%I64d\n",sum);
    }
}

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  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

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