1037. Magic Coupon (25)-PAT

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

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来源：http://pat.zju.edu.cn/contests/pat-a-practise/1037

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#include<iostream>#include<stdio.h>#include <stdlib.h>#include<stdio.h>using namespace std;int compare(const void *a,const void *b){return *(int *)a-*(int *)b;}int main(){int nc,np,i,j;scanf("%d",&nc);int *co=new int [nc];for(i=0;i<nc;i++)scanf("%d",&co[i]);scanf("%d",&np);int *po=new int [np];for(i=0;i<np;i++)scanf("%d",&po[i]);qsort(co,nc,sizeof(int),compare);qsort(po,np,sizeof(int),compare);long sum=0;for(i=0;i<np&&i<nc;i++){if(co[i]<0&&po[i]<0){sum+=co[i]*po[i];}elsebreak;}for(i=nc-1,j=np-1;i>=0,j>=0;i--,j--){if(co[i]>0&&po[j]>0)sum+=co[i]*po[j];elsebreak;}printf("%ld\n",sum);return 0;}