1037. Magic Coupon (25)

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)whereanegativevaluecorrespondstoabonusproduct.Youcanapplycoupon3(withNbeing4)toproduct1(withvalueM7) to get M28back;coupon2toproduct2togetM12 back; and coupon 4 to product 4 to get M3back.Ontheotherhand,ifyouapplycoupon3toproduct4,youwillhavetopayM12 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

算法分析：本题比较简单。利用两个数组存储coupon和product values，然后把它们由大到小排序，即正数在前负数在后。然后从头开始，把对应位置的两个较大的正数相乘，直到碰到负数为止。然后再从数组末端开始，把对应位置较小的负数相乘，直到碰到正数为止。

#include <stdio.h>#include <stdlib.h>int cmp(const void *a, const void *b){    return *(int*)b - *(int*)a;}int main(){    int nc, np, i, money = 0;    int coupons[100000] = {0}, pValues[100000] = {0};    scanf("%d", &nc);    for(i = 0; i < nc; i++)        scanf("%d", &coupons[i]);    scanf("%d", &np);    for(i = 0; i < np; i++)        scanf("%d", &pValues[i]);    qsort(coupons, nc, sizeof(int), cmp);    qsort(pValues, np, sizeof(int), cmp);    i = 0;    while(coupons[i] > 0 && pValues[i] > 0 && i < nc && i < np) //从头开始，直到碰到负数为止    {        money += coupons[i] * pValues[i];        i++;    }    while(coupons[--nc] < 0 && pValues[--np] < 0 && nc >= 0 && np >= 0) //从尾部开始，直到碰到正数为止        money += coupons[nc] * pValues[np];    printf("%d", money);    return 0;}