codeforces894c(构造)

最新推荐文章于 2023-08-25 12:14:01 发布

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最新推荐文章于 2023-08-25 12:14:01 发布

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分类专栏：构造题

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马克梦见一位戴黑眼镜的老人揭示了永生的秘密——一个正整数序列。醒来后，他只记得序列长度，以及通过计算序列中连续子序列的最大公约数得到的集合。现在，他需要你帮助从这个集合重建原始序列。如果存在多个解决方案，输出任意一个；若不存在，则输出-1。

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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 every1 ≤ 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

题意：给定一个集合S，求一个数组a，使得gcd(ai, ai + 1, ..., aj)属于S对所有在范围内的i，j成立。

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int s[1005];
int main()
{
	int m;
	while(~scanf("%d",&m))
	{
		bool flag=true;
		for(int i=0;i<m;i++)
		{
			scanf("%d",&s[i]);
			if(s[i]%s[0]!=0)flag=false;
		}
		/*如果S中某个数不是最小数的倍数，那么无解，
		因为整个数组的gcd必然等于S中最小的数，而这样使得gcd的结果比S[0]小 
		*/ 
		if(!flag)       
		{
			printf("-1\n");
			continue;
		}
		/*
		在S每个间隔中放入S[0]，这样gcd(ai,aj)==ai或S[0] 
		*/
		printf("%d\n",2*m-1);
		for(int i=0;i<m-1;i++)
		{
			printf("%d %d ",s[i],s[0]);
		}
		printf("%d\n",s[m-1]);
	}
}