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

 1 #include<iostream>
 2 #include<queue>
 3 #include<cstdio>
 4 #include<cstdio>
 5 #include<cstring>
 6 using namespace std;
 7 int fun(int m,int n)
 8 {
 9     if(m<n)
10     {
11         n=m+n;
12         m=n-m;
13         n=n-m;
14     }
15     int t;
16     while(m%n!=0)
17     {
18         t=n;
19         n=m%n;
20         m=t;
21     }
22     return n;
23 }
24 int main()
25 {
26     queue<int>Q;
27     int a[1005],b[1005];
28     int n,t;
29     int i,j;
30     while(cin>>n)
31     {
32         t=1;
33         memset(a,0,sizeof(a));
34         memset(b,0,sizeof(b));
35         for(i=1;i<=n;i++)
36         {
37             cin>>a[i];
38         }
39         b[t]=a[1];
40         for(j=2;j<=n;j++)
41         {
42             if(fun(b[t],a[j])==1)
43             {
44                 b[++t]=a[j];
45             }
46             else
47             {
48                 b[++t]=1;
49                 b[++t]=a[j];
50             }
51         }
52         cout<<t-n<<endl;
53         cout<<b[1];
54         for(i=2;i<=t;i++)
55             cout<<" "<<b[i];
56         cout<<endl;
57 
58     }
59     return 0;
60 }

Sample Input

Input

3
2 7 28

Output

1
2 7 9 28