hdu5504 GT and sequence（BestCoder Round #60 ）

GT and sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 334 Accepted Submission(s): 85

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
   
   
   
   

Source

BestCoder Round #60 

题意：给定一个序列，从中选出任意个数的乘积最大值。

分析：此题太多坑，得慢慢发现。。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
#define ll long long
#define CL(a) memset(a,0,sizeof(a))

int main()
{
    int T;
    ll n,x,sum;
    cin>>T;
    while(T--)
    {
        cin>>n;
        sum = 0;
        int k=0,re=0;
        ll maxx = -INF;
        for(int i=0; i<n; i++)
        {
            cin>>x;
            if(x == 0) {re++; continue;}
            if(x < 0)
            {
                k++;
                maxx = max(maxx, x);
            }
            if(i == re) sum = x;
            else sum *= x;
        }
        //cout<<re<<" "<<k<<endl;
        if(k%2&&n!=1&&!(k==1&&re+k==n)) sum /= maxx;
        if(re>0&&(sum<0||n==1)) sum = 0;
        
        cout<<sum<<endl;
    }
    return 0;
}