Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence(构造)

题目原文：

  
   
    C. Marco and GCD Sequence
   
   
    
     time limit per test
    1 second
   
   
    
     memory limit per test
    256 megabytes
   
   
    
     input
    standard input
   
   
    
     output
    standard output
   
  
  
   
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      
        
 Note In the first example 2 = gcd(4, 6), the other elements from the set appear in the sequence, and we can show that there are no values different from 2, 4, 6 and 12 among gcd(ai, ai + 1, …, aj) for every 1 ≤ i ≤ j ≤ n. 

  
一道构造题： 题意：Marco在梦中遇见一个老人，老人告诉了他长生不老的方法，然后Marco想来后却只记得：给你n个所有GCD的集合S，然后让你去构造原来的序列，使得所有原序列的子区间的GCD都在集合S中。如果不存在则输出-1. 思路：首先要使新的序列能构造出来，那么原序列中最小的数肯定是其他数的因子，如果不是，那么肯定就构造不出来，构造的时候只需要把最小的数插入在原序列中间即可，这样可以保证gcd出来的结果不是自己就是最小的数
 #include<iostream>#include<stdio.h>//#define MOD 1000000007using namespace std;int main(){int n;cin >> n;int a[4005];for (int i = 0; i < n; ++i){cin >> a[i];}for (int i = 1; i < n; ++i){if (a[i] % a[0] != 0){printf("-1");return 0;}}printf("%d\n", 2 * n);for (int i = 0; i < n; ++i){printf("%d %d ", a[i], a[0]);}return 0;}