Codeforces 894C：Marco and GCD Sequence（構造）

題目鏈接

Marco and GCD Sequence

Time limit:1 seconds Memory limit:256 megabytes

---

Problem Description

In a dream Marco met an elderly man with a pair of black glasses. The man told him the key to immortality and then disappeared with the wind of time.

When he woke up, he only remembered that the key was a sequence of positive integers of some length n, but forgot the exact sequence. Let the elements of the sequence be a1, a2, …, an. He remembered that he calculated gcd(ai, ai + 1, …, aj) for every 1 ≤ i ≤ j ≤ n and put it into a set S. gcd here means the greatest common divisor.

Note that even if a number is put into the set S twice or more, it only appears once in the set.

Now Marco gives you the set S and asks you to help him figure out the initial sequence. If there are many solutions, print any of them. It is also possible that there are no sequences that produce the set S, in this case print -1.

Input

The first line contains a single integer m (1 ≤ m ≤ 1000) — the size of the set S.

The second line contains m integers s1, s2, …, sm (1 ≤ si ≤ 106) — the elements of the set S. It’s guaranteed that the elements of the set are given in strictly increasing order, that means s1 < s2 < … < sm.

Output

If there is no solution, print a single line containing -1.

Otherwise, in the first line print a single integer n denoting the length of the sequence, n should not exceed 4000.

In the second line print n integers a1, a2, …, an (1 ≤ ai ≤ 106) — the sequence.

We can show that if a solution exists, then there is a solution with n not exceeding 4000 and ai not exceeding 106.

If there are multiple solutions, print any of them.

Examples

input
4
2 4 6 12
output
3
4 6 12

input
2
2 3
output
-1

---

題意：

有一個序列a，集合S包含了a中所有gcd(ai, ai + 1, …, aj) 1 ≤ i ≤ j ≤ n，現在給出s，求滿足條件的a，如果不存在則輸出-1

集合：可以理解成set

解題思路：

S中所有數的GCD必須是S中最小的數，否則-1
構造就很簡單了，在輸入的每個數中間都插入最小的這個數，就能保證任意不同兩個數的gcd都是這個最小的數

Code：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <queue>
#include <algorithm>
#include <sstream>

using namespace std;
typedef long long LL;

vector<int> v;

int gcd(int a,int b)
{
    if(a<b)swap(a,b);
    if(a%b==0)return b;
    return gcd(b,a%b);
}

int main()
{
    int n;
    cin>>n;
    int total_gcd;
    for(int i=0;i<n;i++)
    {
        int x;
        cin>>x;
        v.push_back(x);
        if(i==0)
            total_gcd=x;
        else
            total_gcd=gcd(x,total_gcd);
    }
    if(v[0]!=total_gcd)
    {
        cout<<"-1"<<endl;
        return 0;
    }
    cout<<2*n<<endl;
    for(int i=0;i<n;i++)
    {
        cout<<v[i]<<' '<<v[0]<<' ';
    }
    cout<<endl;
    return 0;
}