A. Array

题目连接：

Description

Vitaly has an array of n distinct integers. Vitaly wants to divide this array into three non-empty sets so as the following conditions hold:

The product of all numbers in the first set is less than zero ( < 0).

The product of all numbers in the second set is greater than zero ( > 0).

The product of all numbers in the third set is equal to zero.

Each number from the initial array must occur in exactly one set.

Help Vitaly. Divide the given array.

Input

The first line of the input contains integer n (3 ≤ n ≤ 100). The second line contains n space-separated distinct integers a1, a2, ..., an (|ai| ≤ 103) — the array elements.

Output

In the first line print integer n1 (n1 > 0) — the number of elements in the first set. Then print n1 numbers — the elements that got to the first set.

In the next line print integer n2 (n2 > 0) — the number of elements in the second set. Then print n2 numbers — the elements that got to the second set.

In the next line print integer n3 (n3 > 0) — the number of elements in the third set. Then print n3 numbers — the elements that got to the third set.

The printed sets must meet the described conditions. It is guaranteed that the solution exists. If there are several solutions, you are allowed to print any of them.

Sample Input

3

-1 2 0

Sample Output

1 -1

1 2

1 0

Hint

题意

给你n个数，然后让你分成三组，要求第一组的乘积是小于0的，第二组是大于0的，第三组是等于0的

让你输出一个方案

题解：

第一组需要奇数个负数，第二组需要偶数个负数

如果负数是偶数，那么扔一个去第三组，那么就可以变成奇数了

奇数 = 奇数+偶数

所以讨论一下就好了

代码

#include
    
     using namespace std;vector
     
      ans[4];int a[400];int main(){    int n;    scanf("%d",&n);    int num = 0;    for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);        if(a[i]<0)num++;    }    if(num==1)    {        for(int i=1;i<=n;i++)        {            if(a[i]<0)ans[0].push_back(a[i]);            else if(a[i]==0)ans[2].push_back(a[i]);            else ans[1].push_back(a[i]);        }    }    else if(num%2==0)    {        int k = 0;        for(int i=1;i<=n;i++)        {            if(a[i]<0&&k==0){ans[0].push_back(a[i]);k++;}            else if(a[i]<0&&k==1){ans[2].push_back(a[i]);k++;}            else if(a[i]==0)ans[2].push_back(a[i]);            else ans[1].push_back(a[i]);        }    }    else    {        int k = 0;        for(int i=1;i<=n;i++)        {            if(a[i]<0&&k==0){ans[0].push_back(a[i]);k++;}            else if(a[i]==0)ans[2].push_back(a[i]);            else ans[1].push_back(a[i]);        }    }    for(int i=0;i<3;i++)    {        printf("%d ",ans[i].size());        for(int j=0;j