codeforces 660A

最新推荐文章于 2018-12-05 10:56:17 发布

NoDemon

最新推荐文章于 2018-12-05 10:56:17 发布

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本文链接：https://blog.youkuaiyun.com/CookieCTO/article/details/51150808

本文介绍了一种算法，用于将给定数组转换为相邻元素互质的共质数数组。通过插入不超过10^9的正整数，实现数组的最小改动。文章详细解释了输入输出格式及示例。

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 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



#include<iostream>
#include<algorithm>
#include<cstring>
using  namespace std;
int  gcd(int a,int b)
{
    if(a%b==0)
    {
        if(b==1)return 1;
        else return 0;
    }
    else gcd(b,a%b);
}
int main()
{
    int n;
    while(cin>>n)
    {
        int  num[2010]={0};

        int flag=0;
        for(int i=0;i<n;++i)
            cin>>num[2*i];
        for(int i=0;i<2*n-2;i+=2)
        {
            if(num[i]==1||num[i+2]==1)
                continue;
            int a=max(num[i],num[i+2]);
            int b=min(num[i],num[i+2]);
            if(gcd(a,b)==0)
            {
                flag++;
                num[i+1]=1;
            }
        }
        cout<<flag<<endl;
        for(int i=0;i<=2*n-2;i++)
            if(num[i])cout<<num[i]<<" ";

        cout<<endl;

    }
    return 0;
}