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

来源:互联网 发布:软件数据接口开发合同 编辑:程序博客网 时间:2024/05/19 17:02

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.

这道题刚开始被长度迷惑住了,不超过4000的才算是正确的序列,其实这句话没什么用的,我们取最小的数,然后判断其他数可不可以整除这个数,如果有不能整除的就代表存在更小的gcd,显然与给出的序列是不符的,如果所有都能整除的话,那么我们就可以把这个数加到每两个数字的中间最为一个新序列输出,即可,因为这样就不会有新的公约数出现,满足给出的序列
代码:

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long LL;const int N=50050;int a[N],n;int mn=2333333;int main(){    int i,j,k;    scanf("%d",&n);    for(i=1;i<=n;i++){        scanf("%d",&a[i]);        if(a[i]<mn)mn=a[i];    }    for(i=1;i<=n;i++)        if(a[i]%mn)            break;    if(i<=n)        puts("-1");    else{        printf("%d\n%d",2*n-1,a[1]);        for(i=2;i<=n;i++)            printf(" %d %d",mn,a[i]);    }    return 0;}
原创粉丝点击