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!=)
     {
         t=n;
         n=m%n;
         m=t;
     }
     return n;
 }
 int main()
 {
     queue<int>Q;
     int a[],b[];
     int n,t;
     int i,j;
     while(cin>>n)
     {
         t=;
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         for(i=;i<=n;i++)
         {
             cin>>a[i];
         }
         b[t]=a[];
         for(j=;j<=n;j++)
         {
             if(fun(b[t],a[j])==)
             {
                 b[++t]=a[j];
             }
             else
             {
                 b[++t]=;
                 b[++t]=a[j];
             }
         }
         cout<<t-n<<endl;
         cout<<b[];
         for(i=;i<=t;i++)
             cout<<" "<<b[i];
         cout<<endl;

     }
     return ;
 }

Sample Input

Input

3
2 7 28

Output

1
2 7 9 28