codeforces 894C Marco and GCD Sequence

最新推荐文章于 2020-08-16 11:57:26 发布

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本文介绍了一种解决特定数学问题的方法，即如何构造一个数列使得该数列的所有子区间的最大公约数(GCD)恰好构成给定集合。文章详细阐述了通过找到集合内元素的最大公约数，并检查其是否存在于原始集合中的方式来解决此问题。

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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 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.

题意：要求我们构造一个数列满足这里面所有区间的最小公因数都在这里出现过并且不会有没有出现过的首先我把所有gcd做一个gcd求出他们的最大公因数然后我判断这个最大公因数是否在给定的数列中出现如果没有则一定是输出-1 否则则我在原数列中没两个中间加一个他们的最大公因数即可

#include<cstdio>
#define N 1100
inline char gc(){
    static char now[1<<16],*S,*T;
    if (T==S){T=(S=now)+fread(now,1,1<<16,stdin);if (T==S) return EOF;}
    return *S++;
}
inline int read(){
    int x=0;char ch=gc();
    while (ch<'0'||ch>'9') ch=gc();
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=gc();}
    return x;
}
inline int gcd(int x,int y){
    if (y==0) return x;return gcd(y,x%y);
}
int a[N],n;
int main(){
    freopen("cf.in","r",stdin);
    n=read();int gg=0;bool flag=0;
    for (int i=1;i<=n;++i) a[i]=read(),gg=gcd(gg,a[i]);
    for (int i=1;i<=n;++i) if (a[i]==gg) {flag=1;break;}
    if (!flag) {printf("-1");return 0;}printf("%d\n",(n<<1)-1);
    for (int i=1;i<n;++i) printf("%d %d ",a[i],gg);printf("%d",a[n]);
    return 0;
}