CodeForces 660A

Description

You are given an array of n elements, you must make it a co-prime array in as few moves as possible.

In each move you can insert any positive integral number you want not greater than 10⁹ in any place in the array.

An array is co-prime if any two adjacent numbers of it are co-prime.

In the number theory, two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.

Input

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

The second line contains n integers a_i (1 ≤ a_i ≤ 10⁹) — the elements of the array a.

Output

Print integer k on the first line — the least number of elements needed to add to the array a to make it co-prime.

The second line should contain n + k integers a_j — the elements of the array a after adding k elements to it. Note that the new array should be co-prime, so any two adjacent values should be co-prime. Also the new array should be got from the original array a by adding k elements to it.

If there are multiple answers you can print any one of them.

我不想说，本来可以一次a掉的  硬是搞了好几次，不需要加数的时候0也要输出，

只要判断相邻是否互质，否的时候在中间插入1，因为1和任何数都互质，

英语是硬伤

#include<iostream>
#include<queue>
#include<cstdio>
#include<cstdio>
#include<cstring>
using namespace std;
int fun(int m,int n)
{
    if(m<n)
    {
        n=m+n;
        m=n-m;
        n=n-m;
    }
    int t;
    while(m%n!=0)
    {
        t=n;
        n=m%n;
        m=t;
    }
    return n;
}
int main()
{
    queue<int>Q;
    int a[1005],b[1005];
    int n,t;
    int i,j;
    while(cin>>n)
    {
        t=1;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        for(i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        b[t]=a[1];
        for(j=2;j<=n;j++)
        {
            if(fun(b[t],a[j])==1)
            {
                b[++t]=a[j];
            }
            else
            {
                b[++t]=1;
                b[++t]=a[j];
            }
        }
        cout<<t-n<<endl;
        cout<<b[1];
        for(i=2;i<=t;i++)
            cout<<" "<<b[i];
        cout<<endl;

    }
    return 0;
}

Sample Input

Input

3
2 7 28

Output

1
2 7 9 28