杭电-5504GT and sequence

缺乏、安全感 2022-08-18 02:12 43阅读 0赞

# GT and sequence #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)**

Total Submission(s): 2213 Accepted Submission(s): 508

Problem Description

You are given a sequence of  N![Image 1][] 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  2![Image 1][]63![Image 1][]−1![Image 1][].

Input

In the first line there is a number  T![Image 1][] (test numbers).  
  
For each test,in the first line there is a number  N![Image 1][],and in the next line there are  N![Image 1][] numbers.  
  
 1≤T≤1000![Image 1][]  
 1≤N≤62![Image 1][]  
  
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

给一个数列，选择若干个数，求出其最大乘积

#include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    bool cmp(__int64 a,__int64 b)
    {
    	return a<b;
    }
    int main()
    {
    	__int64 a,b,c,i,j,m,n,x[100],y[100],q1,q2,sum;
    	
    	scanf("%I64d",&m);
    	while(m--)
    	{
    		scanf("%I64d",&n);
    		q1=0,q2=0;
    		for(i=0;i<n;i++)
    		{
    			scanf("%I64d",&a);
    			if(a>0)
    			x[q1++]=a;
    			if(a<0)
    			y[q2++]=a;
    		}
    		sort(y,y+q2,cmp);
    		
    		if(!q1&&!q2)
    		{
    			printf("0\n");
    			continue;
    		}
    	    if(!q1&&q2==1&&n>1)
    	    {
    	    	printf("0\n");
    	    	continue;
    		}
    		if(n==1&&q2==1)
    	    {
    	    	printf("%I64d\n",y[0]);
    	    	continue;
    		}
    		sum=1;
    		for(i=0;i<q1;i++)
    			sum*=x[i];
    			
    		if(q2%2==0)
    		{
    			for(i=0;i<q2;i++)
    			sum*=y[i];
    		}
    		else 
    		{
    			for(i=0;i<q2-1;i++)
    			sum*=y[i];
    		}
    		printf("%I64d\n",sum);
    	}
    	return 0;	
     }

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