[ARC122E] Increasing LCMs

Problem Statement

We have a sequence of $N$ positive integers: $A_1,A_2,\cdots,A_N$.
You are to rearrange these integers into another sequence $x_1,x_2,\cdots,x_N$, where $x$ must satisfy the following condition:

• Let us define $y_i=\operatorname{LCM}(x_1,x_2,\cdots,x_i)$, where the function $\operatorname{LCM}$ returns the least common multiple of the given integers. Then, $y$ is strictly increasing. In other words, $y_1<y_2<\cdots<y_N$ holds.

Determine whether it is possible to form a sequence $x$ satisfying the condition, and show one such sequence if it is possible.

Constraints

• $1 \leq N \leq 100$
• $2 \leq A_1 < A_2 \cdots < A_N \leq 10^{18}$
• All values in input are integers.

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Input

Input is given from Standard Input in the following format:

$N$
$A_1$ $A_2$ $\cdots$ $A_N$

Output

If it is possible to form a sequence $x$ satisfying the condition, print your answer in the following format:

Yes
$x_1$ $x_2$ $\cdots$ $x_N$

If it is impossible, print No.

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Sample Input 1

3
3 4 6

Sample Output 1

Yes
3 6 4

For $x=(3,6,4)$, we have:

• $y_1=\operatorname{LCM}(3)=3$
• $y_2=\operatorname{LCM}(3,6)=6$
• $y_3=\operatorname{LCM}(3,6,4)=12$

Here, $y_1<y_2<y_3$ holds.

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Sample Input 2

3
2 3 6

Sample Output 2

No

No permutation of $A$ would satisfy the condition.

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Sample Input 3

10
922513 346046618969 3247317977078471 4638516664311857 18332844097865861 81706734998806133 116282391418772039 134115264093375553 156087536381939527 255595307440611247

Sample Output 3

Yes
922513 346046618969 116282391418772039 81706734998806133 255595307440611247 156087536381939527 134115264093375553 18332844097865861 3247317977078471 4638516664311857

巧妙地，考虑倒着构造整个序列。

想一下如何选出一个可以排在最后的数，当且仅当他存在某一个质因子的次数是严格最大的。

可以先用 Pollard-Pho 分解出来判断。

每次选一个可以放在最后的元素，不会使本来可以放的数变成不能放。

但是真的要 Pollard-Pho 吗？

枚举 \(10^6\) 以内的数进行分解，那么还没分解出来的要不是两个大质数相乘，要不是一个质数。

对于两个大质数的情况，枚举其他的数，取gcd，如果取出来不是 1 我们就分解出来了，否则可以把这个数当成一个数，不影响性质。

#include<bits/stdc++.h>
using namespace std;
const int N=105,M=N*30;
typedef long long LL;
int p[N][M],c,n,vs[N],st[N];
LL a[N],to[M],b[N];
map<LL,LL>v;
multiset<int>s[M];
LL gcd(LL x,LL y)
{
	if(!y)
		return x;
	return gcd(y,x%y);
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",a+i),b[i]=a[i];
		for(int j=2;j<=1000000;j++)
		{
			if(a[i]%j==0)
			{
				if(!v[j])
					to[++c]=j,v[j]=c;
				while(a[i]%j==0)
					a[i]/=j,p[i][v[j]]++;
			}
		}
	}
	for(int i=1;i<=n;i++)
	{
		if(a[i]==1)
			continue;
		int fl=0;
		for(int j=1;j<=n;j++)
		{
			LL d=gcd(a[i],a[j]);
			if(d^a[i]&&d^1)
			{
				if(!v[d])
					v[d]=++c;
				if(!v[a[i]/d])
					to[++c]=a[i]/d,v[a[i]/d]=c;
				++p[i][v[d]],++p[i][v[a[i]/d]];
				fl=1,j=n;
			}
		}
		if(!fl)
		{
			if(!v[a[i]])
				to[++c]=a[i],v[a[i]]=c;
			p[i][v[a[i]]]++;
		}
	}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=c;j++)
			s[j].insert(p[i][j]);
//	for(int i=1;i<=n;i++)
//	{
//		printf("%lld ",b[i]);
//		for(int j=1;j<=c;j++)
//			printf("%lld %d\n",to[j],p[i][j]);
//		puts("");
//
//	}
	for(int i=1;i<=n;i++)
	{
		int pf=0;
		for(int j=1;j<=n;j++)
		{
			if(vs[j])
				continue;
			int fl=0;
			for(int k=1;k<=c;k++)
				if(s[k].size()==1||(*(--s[k].end())==p[j][k]&&(*--s[k].end())^(*(--(--s[k].end())))))
					fl=1,vs[j]=1,st[i]=j,k=c,pf=1;
			if(fl)
			{
				for(int k=1;k<=c;k++)
					s[k].erase(s[k].lower_bound(p[j][k]));
				j=n;
			}
		}
		if(!pf)
			return puts("No"),0;
	}
	puts("Yes");
	for(int i=n;i>=1;i--)
		printf("%lld ",b[st[i]]);
}