互质——Educational Codeforces Round 11-——A

A. Co-prime Array

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 kelements to it.

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

Example

input

3
2 7 28

output

1
2 7 9 28

题意：给出n个数，在其中插入一些数，使得任意两个相邻的数互质。
题解：1和任意一个数互质，所以当找到两个数不满足条件时，在这之间插入一个1即可。

/*A*/
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn=1000+100;    
int gcd(int a,int b)
{
    return b==0?a:gcd(b,a%b);
}

int main()
{
    int n;
    int a[maxn];
    int insert[maxn];
    while(scanf("%d",&n)!=EOF)
    {
        memset(a,0,sizeof(a));
        memset(insert,0,sizeof(insert));
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        int k=0;
        for(int i=0;i<n-1;i++)
        {
            int t=gcd(a[i],a[i+1]);
            if(t!=1)
            {
                //insert[i]=getinsert(a[i],a[i+1]);
                insert[i]=1;
                k++;
            }    
        }
        printf("%d
%d",k,a[0]);
        if(insert[0])
                printf(" %d",insert[0]);
        for(int i=1;i<n;i++)
        {
            printf(" %d",a[i]);
            if(insert[i])
                printf(" %d",insert[i]);
        }
        printf("
");
    }
    return 0;
}