CodeForces 660A Co-prime Array

Co-prime Array

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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 addingk elements to it.

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

Sample Input

Input

3
2 7 28

Output

1
2 7 9 28

题解：此题就是让把一个给定序列的数变为互素的一个序列，即任意两个相邻的数互素，如果不互素就在两数之间添加一个数使得这个数既与前一个数互素又与后一个数互素，记录添加的数的最少个数并输出，此外还要输出添加完数之后的序列。

<span style="font-family:SimSun;font-size:18px;">#include<cstdio>#include<cstring>#define N 1000000int a[2010];int gcd(int a,int b){if(a%b==0)return b;return gcd(b,a%b);}int main(){int n,ans;while(scanf("%d",&n)!=EOF){int i=0;ans=0;memset(a,0,sizeof(a));while(n--){scanf("%d",&a[i]);if(i>0&&gcd(a[i],a[i-1])!=1)//判断相邻两个数是否互素{for(int j=2;j<N;j++){if(gcd(a[i-1],j)==1&&gcd(a[i],j)==1){int t=a[i];a[i++]=j;a[i]=t;ans++;break;}}}i++;}printf("%d\n",ans);printf("%d",a[0]);for(int j=1;j<i;j++)printf(" %d",a[j]);printf("\n");}return 0;}</span>