PAT甲级 1037 Magic Coupon （测试点4超时）

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

分析：

其实思想很简单，就是把两个数组进行排序，正数从大到小排，负数从小到大排。测试点4超时就说排序函数的问题。在使用sort函数排序时，cmp函数不要有=。

代码： 

错误代码：

bool cmp(int a,int b){
	if(a<0&&b<0)
	return a<b; 
	return a>=b;
}

正确代码：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
int a[100005];
int b[100005];
bool cmp(int a,int b){
	if(a<0&&b<0)
	return a<b; 
	return a>b;
}
int main(){
	int nc,np;
	cin>>nc;
	for(int i=0;i<nc;i++){
		cin>>a[i];
	}
	cin>>np;
	for(int i=0;i<np;i++){
		cin>>b[i];
	}
	sort(a,a+nc,cmp);
	sort(b,b+np,cmp);
	int l1=0,l2=0;
	long long sum=0;
	while(l1<nc&&l2<np){
		if(a[l1]*b[l2]>0){
			sum+=a[l1]*b[l2];
			l1++;
			l2++;
		}else if(a[l1]>=0){
			l1++;
		}else if(b[l2]>=0){
			l2++;
		}
	}
	cout<<sum;
}