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原题
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≤10^5, and it is guaranteed that all the numbers will not exceed 2^30.
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 结尾无空行
Sample Output: 43 结尾无空行
我的思路
最初的想法是，每次找到最大的两个正数相乘，任何一组的正数找到最小的两个负数相乘，这种方法很简单，没有报告bug我跑了一次，但是测试用例中有一组超时操作(每次都要找到最大或最小的遍历，这么多循环能不超时吗？doge）。
所以改变想法，先把两组数据分开排序，然后按顺序乘以(正数从大到小乘以，负数从小到大乘以)。
通过代码
#include <bits/stdc .h> using namespace std; int main(){ long long *c,*p,total=0,result=0; int cn,pn; scanf("%d",&cn); c=(long long *)malloc(sizeof(long long)*(cn)); for(int i = 0;i<=cn-1;i ){ scanf("%lld",&c[i]); } scanf("%d",&pn); p=(long long *)malloc(sizeof(long long)*(pn)); for(int i = 0;i<=pn-1;i ){ scanf("%lld",&p[i]); } sort(c,c cn); sort(p,p pn); if(cn==0 || pn==0){ printf("%lld",total); return 0; } int i=0; while(i<cn&&i<pn&&c[i]<0&&p[i]<0){ result=c[i]*p[i]; total =result; i ; } i=0; while(i<cn&&i<pn&&c[cn-1-i]>0&&p[pn-1-i]>0){ result=c[cn-1-i]*p[pn-1-i]; total =result; i ; } printf("%lld",total); return 0; }总结
在我看来，我需要判断两个负数相乘的项目是否完成（正数相同），所以我在一开始就使用了以下简单的判断方法。我认为如果一组数据中的负数已经处理完毕，这一轮就会出现正数，而另一组负数仍然是剩余的，这一轮仍然是负数，所以我会得到负数（result<=此时退出循环就好了。
while(i<cn&&i<pn){ result=c[cn-1-i]*p[pn-1-i]; if(result<=0) break; else{ total =result; i ; } }在我自己使用的几个例子中，这种判断方法确实没有问题，例如：
4 1 2 4 -1 4 7 6 -2 -3但忽略了一种情况，即两组负数相同，同时轮换负数，没有0，此时不能退出上述判例，如：
5 -3 -2 1 2 3 4 -4 -3 2 5呜呜，暂时没想到下次怎么避免这种问题，先死记硬背。