hdoj 5504 GT and sequence 【脑子抽了】

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解决一个编程问题，通过选择序列中的元素以获得最大乘积，包括处理正数、负数和零的情况。

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GT and sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 98 Accepted Submission(s): 29

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

题意：给出N个数，让你至少选择其中一个数，使得选出的所有数乘积最大，输出最大乘积。

输出没有带Case，可我却加上了。无脑，找了好几分钟才发现。o(╯□╰)o

枚举所有情况，弱渣写的代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL __int64
using namespace std;
LL a[1010];
bool cmp(LL a, LL b){
    return a > b;
}
int main()
{
    int t, k = 1;
    scanf("%d", &t);
    while(t--)
    {
        int n;
        scanf("%d", &n);
        LL sum = 1; LL b;
        int top = 0;
        bool flag = false;//标记是否全为0
        bool have = false;//标记是否出现正数
        bool zero = false;//标记是否出现0
        for(int i = 0; i < n; i++)
        {
            scanf("%I64d", &b);
            if(b)
                flag = true;
            if(b == 0)
                zero = true;
            if(b > 0)
            {
                have = true;
                sum *= b;
            }
            if(b < 0)
                a[top++] = -b;
        }
        if(!flag)//全为0
        {
            printf("0\n");
            continue;
        }
        sort(a, a+top, cmp);
        if(!have)//没有出现正数
        {
            if(top == 1)
            {
                if(zero)
                    sum = 0;
                else
                    sum = -a[0];
            }
            else
            {
                if(top & 1)
                    top -= 1;
                for(int i = 0; i < top; i++)
                    sum *= a[i];
            }
            printf("%I64d\n", sum);
        }
        else
        {
            if(top & 1)
                top -= 1;
            for(int i = 0; i < top; i++)
                sum *= a[i];
            printf("%I64d\n", sum);
        }
    }
    return 0;
}