coeforces 665D D. Simple Subset(最大团orsb题)

题目链接：

D. Simple Subset

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A tuple of positive integers {x1, x2, ..., xk} is called simple if for all pairs of positive integers (i,  j) (1  ≤ i  <  j ≤ k), xi  +  xj is a prime.

You are given an array a with n positive integers a1,  a2,  ...,  an (not necessary distinct). You want to find a simple subset of the array awith the maximum size.

A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's define a subset of the array a as a tuple that can be obtained from a by removing some (possibly all) elements of it.

Input

 

The first line contains integer n (1 ≤ n ≤ 1000) — the number of integers in the array a.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Output

 

On the first line print integer m — the maximum possible size of simple subset of a.

On the second line print m integers bl — the elements of the simple subset of the array a with the maximum size.

If there is more than one solution you can print any of them. You can print the elements of the subset in any order.

Examples

 

input

2
2 3

output

2
3 2

input

2
2 2

output

1
2

input

3
2 1 1

output

3
1 1 2

input

2
83 14

output

2
14 83

题意：

选一个最大的子序列,满足这个序列里的任意两个数的和是素数;

思路：

可以是一个最大完全数的题,也可以是水题,因为奇数+奇数=偶数,偶数+偶数=偶数,(1除外);所以最多有一个奇数和一个偶数;
我写的分情况讨论的代码真是跟翔一样;

AC代码：

#include <bits/stdc++.h>
using namespace std;
const int N=2e6+;
typedef long long ll;
int n,a[],flag[N];
int get_prime()
{
    for(int i=;i<N;i++)
    {
        if(!flag[i])
        {
            for(int j=*i;j<N;j+=i)
            {
                flag[j]=;
            }
        }
    }
}
queue<int>qu;
void print()
{
    printf("%d\n",qu.size());
    while(!qu.empty())
    {
        printf("%d ",qu.front());
        qu.pop();
    }
}

int main()
{
    get_prime();
    int f=;
    scanf("%d",&n);
    for(int i=;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(a[i]==)
        {
            f++;
            qu.push();
        }
    }
    if(f>)
    {
        for(int i=;i<=n;i++)
        {
          if(a[i]>&&!flag[a[i]+])
          {
              for(int j=i+;j<=n;j++)
              {
                  if(a[j]>&&!flag[a[i]+a[j]]&&!flag[a[j]+])
                  {
                      qu.push(a[i]);
                      qu.push(a[j]);
                      print();
                      return ;
                  }
              }
          }
        }
        for(int i=;i<=n;i++)
        {
            if(a[i]>&&!flag[a[i]+])
            {
                qu.push(a[i]);
                print();
                return ;
            }
        }
        print();
    }
    else if(f==)
    {
        for(int i=;i<=n;i++)
        {
          if(a[i]>&&!flag[a[i]+])
          {
              for(int j=i+;j<=n;j++)
              {
                  if(a[j]>&&!flag[a[i]+a[j]]&&!flag[a[j]+])
                  {
                      qu.push(a[i]);
                      qu.push(a[j]);
                      print();
                      return ;
                  }
              }
          }
        }
        for(int i=;i<=n;i++)
        {
            if(a[i]>)
            {
                for(int j=i+;j<=n;j++)
                {
                    if(a[j]>&&!flag[a[i]+a[j]])
                    {
                        printf("2\n");
                        printf("%d %d",a[i],a[j]);
                        return ;
                    }
                }
            }
        }
        for(int i=;i<=n;i++)
        {
            if(a[i]>&&!flag[a[i]+])
            {
                qu.push(a[i]);
                print();
                return ;
            }
        }
        print();
    }
    else
    {
        for(int i=;i<=n;i++)
        {
            for(int j=i+;j<=n;j++)
            {
                if(!flag[a[i]+a[j]])
                {
                    qu.push(a[i]);
                    qu.push(a[j]);
                    print();
                    return ;
                }
            }
        }
        printf("1\n");
        printf("%d",a[]);

    }

    return ;
}