codeforces 864C Marco and GCD Sequence
来源:互联网 发布:php直播系统源码下载 编辑:程序博客网 时间:2024/06/06 16:44
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.
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.
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.
42 4 6 12
34 6 12
22 3
-1
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.
#pragma comment(linker,"/STACK:1024000000,1024000000")#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<stack>#include<queue>#include<deque>#include<set>#include<map>#include<cmath>#include<vector>using namespace std;typedef long long ll;typedef unsigned long long ull;typedef pair<int, int> PII;#define pi acos(-1.0)#define eps 1e-10#define pf printf#define sf scanf#define lson rt<<1,l,mi#define rson rt<<1|1,mi+1,r#define it rt,l,r#define root 1,1,n#define e tree[rt]#define _s second#define _f first#define all(x) (x).begin,(x).end#define mem(i,a) memset(i,a,sizeof i)#define for0(i,a) for(int (i)=0;(i)<(a);(i)++)#define for1(i,a) for(int (i)=1;(i)<=(a);(i)++)#define mi ((l+r)>>1)#define sqr(x) ((x)*(x))const int inf=0x3f3f3f3f;const int mod=1e9+7;int a[1010];int main(){ int n; sf("%d",&n); for1(i,n)sf("%d",&a[i]); for1(i,n) if(a[i]%a[1]) { puts("-1"); return 0; } pf("%d\n",2*n-1); for1(i,n-1) pf("%d %d ",a[i],a[1]); pf("%d\n",a[n]); return 0;}
- codeforces 864C Marco and GCD Sequence
- codeforces C. Marco and GCD Sequence
- codeforces-894C Marco and GCD Sequence
- Codeforces 894C C. Marco and GCD Sequence
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence
- Codeforces 894C:Marco and GCD Sequence(构造)
- Marco and GCD Sequence
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence 构造
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence (构造 贪心)
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence(构造,思路)
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence(构造)
- Codeforces Round #447 (Div. 2) C. Marco and GCD Sequence 构造
- CF-Marco and GCD Sequence
- Codeforces894C Marco and GCD Sequence
- Codeforces 798C Mike and gcd problem
- HTML03-表单
- Java递归实现整数各位求和
- HDU
- docker:搭建单机redis主从集群
- C语言之贪心算法(背包问题)
- codeforces 864C Marco and GCD Sequence
- 稀疏矩阵三元组的操作
- Redis开发实例(3)-List
- 二叉树,完全二叉树,满二叉树,平衡二叉树的区别
- jsp生成验证码
- 8张图理解Java
- Linux下解压和压缩文件的方式
- 都是was缓存惹的祸
- c++中vector的用法详解