codeforces 660A A. Co-prime Array(水題)

題目連接：

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 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.

 

Example

input

3
2 7 28

output

1
2 7 9 28

題意：

給一個數組,問插入多少個數字才能使相鄰的兩個數互質;

思路：

暴力找唄,找到的放隊列裏,最後輸出不就行了;

AC代碼：

/*
2014300227    660A - 7    GNU C++11    Accepted    15 ms    2188 KB
*/

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+4;
typedef long long ll;
const double PI=acos(-1.0);
int n,a[3000];
int gcd(int x,int y)
{
if(y==0)return x;
return gcd(y,x%y);
}
queue<int>qu;
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int num=0;
for(int i=1;i<n;i++)
{
qu.push(a[i]);
if(gcd(a[i],a[i+1])>1)
{

qu.push(1);
num++;
}
}
qu.push(a[n]);
printf("%d\n",num);
while(!qu.empty())
{
printf("%d ",qu.front());
qu.pop();
}

return 0;}