hdu 5504 GT and sequence

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5504

GT and sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2223    Accepted Submission(s): 510

Problem Description

You are given a sequence of N integers.

You should choose some numbers(at least one),and make the product of them as big as possible.

It guaranteed that **the absolute value of** any product of the numbers you choose in the initial sequence will not bigger than 263−1 .

 

Input

In the first line there is a number T (test numbers).

For each test,in the first line there is a number N ,and in the next line there are N numbers.

1≤T≤1000
1≤N≤62

You'd better print the enter in the last line when you hack others.

You'd better not print space in the last of each line when you hack others.

 

Output

For each test case,output the answer.

 

Sample Input

1

3

1 2 3

 

Sample Output

6

 

#include<stdio.h>
#include<string.h>
#include<stdio.h>
#include<stdlib.h>
#include<algorithm>
#define LL long long
#define MAX 1100
using namespace std;
int main()
{
	LL t,n,m,j,i,k;
	LL x,a1,b1,c;
    LL s[1100],a[MAX],b[MAX];
	scanf("%lld",&t);
	while(t--)
	{
		scanf("%lld",&m);
		a1=b1=c=0;
		for(i=0;i<m;i++)
		{
			scanf("%lld",&s[i]);
			if(s[i]<0)
			    a[a1++]=s[i];
			else if(s[i]>0)
			    b[b1++]=s[i];
			else
			    c++;
		}
		LL sum1=1;
		LL sum2=1; 
		if(c==m)//只输入0 
		{
			printf("0
");
			continue;
		}
		if(m==1)//只输入一个数 
		{
			printf("%lld
",s[0]);
			continue;
		}
		sort(a,a+a1);
		sort(b,b+b1);
		for(i=0;i<b1;i++)//正数和 
		    sum1*=b[i];
		    
	    if(a1%2==0)//偶数个负数的话，所有负数乘 
	    {
	    	for(i=0;i<a1;i++)
		        sum2*=a[i];
	    }
	    else //奇数个负数，最大的负数不乘 
	    for(i=0;i<a1-1;i++)
	        sum2*=a[i];
	        
		if(c!=0&&b1==0&&a1==1)//输入1个负数和0 
		{
			printf("0
");
			continue;
		}
		printf("%lld
",sum1*sum2);
	}
	return 0;
}