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 ai (1 ≤ ai ≤ 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 aj — 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
32 7 28
Output
12 7 9 28