Co-prime Array

最新推荐文章于 2018-09-16 20:40:33 发布

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最新推荐文章于 2018-09-16 20:40:33 发布

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给定一个数组，你需要通过最少的操作次数将其变为互质数组。每步操作可以向数组中任意位置插入不超过10^9的正整数。数组互质意味着其任意相邻两个数互质。题目要求输出最小插入次数及其后的互质数组。

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Co-prime Array

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

Submit Status Practice 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 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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int GCD(int a,int b)
{
	if(b==0)
	return a;
	else
	return GCD(b,a%b);
}
int main()
{
	int  n,a[1010],k=0;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	for(int i=1;i<n;i++)
	{
		if(GCD(a[i],a[i+1])!=1)
		k++;
	}
	printf("%d\n",k);
	for(int i=1;i<n;i++)
	{
		printf("%d ",a[i]);
		if(GCD(a[i],a[i+1])!=1)
		printf("1 ");
	}
	printf("%d\n",a[n]);
}