D. Divide by three, multiply by two
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Polycarp likes to play with numbers. He takes some integer number xx, writes it down on the board, and then performs with it n−1n−1 operations of the two kinds:
- divide the number xx by 33 (xx must be divisible by 33);
- multiply the number xx by 22.
After each operation, Polycarp writes down the result on the board and replaces xx by the result. So there will be nn numbers on the board after all.
You are given a sequence of length nn — the numbers that Polycarp wrote down. This sequence is given in arbitrary order, i.e. the order of the sequence can mismatch the order of the numbers written on the board.
Your problem is to rearrange (reorder) elements of this sequence in such a way that it can match possible Polycarp's game in the order of the numbers written on the board. I.e. each next number will be exactly two times of the previous number or exactly one third of previous number.
It is guaranteed that the answer exists.
Input
The first line of the input contatins an integer number nn (2≤n≤1002≤n≤100) — the number of the elements in the sequence. The second line of the input contains nn integer numbers a1,a2,…,ana1,a2,…,an (1≤ai≤3⋅10181≤ai≤3⋅1018) — rearranged (reordered) sequence that Polycarp can wrote down on the board.
Output
Print nn integer numbers — rearranged (reordered) input sequence that can be the sequence that Polycarp could write down on the board.
It is guaranteed that the answer exists.
Examples
input
Copy
6 4 8 6 3 12 9
output
Copy
9 3 6 12 4 8
input
Copy
4 42 28 84 126
output
Copy
126 42 84 28
input
Copy
2 1000000000000000000 3000000000000000000
output
Copy
3000000000000000000 1000000000000000000
Note
In the first example the given sequence can be rearranged in the following way: [9,3,6,12,4,8][9,3,6,12,4,8]. It can match possible Polycarp's game which started with x=9x=9.
算法分析：
题意：
n个数叫你排列成一个数是后一个数的三倍或者是后一个数的二分之一。
分析：
dfs思路，但比赛时忘了怎么打印路径了，dfs也几处有错误。
当时不知道为什么不看看课本呢，哎，脑抽。
代码实现：
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
using namespace std;
const double eps = 1e-8;
typedef long long LL;
typedef unsigned long long ULL;
const int INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN=5010;
const int MAXM=100010;
using namespace std;
long long a[105],b[1005];
int cnt=0,n,vis[105];
int flag=0,k=0;
int dfs(int s,int step)
{
if(step==n)
{
flag=1;
for(int i=1;i<=n;i++)
{
cout<<b[i]<<" ";
}
cout<<endl;
return 1;
}
if(flag==1) return 1;
for(int i=1;i<=n;i++)
{
if(vis[i]==0)
{
if(a[s]*2==a[i])
{
k++;
b[k]=a[i];
vis[i]=1;
dfs(i,step+1);
vis[i]=0;
k--;
}
if(a[i]*3==a[s])
{
k++;
b[k]=a[i];
vis[i]=1;
dfs(i,step+1);
vis[i]=0;
k--;
}
}
}
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]);
for(int i=1;i<=n;i++)
{ memset(vis,0,sizeof(vis));
vis[i]=1;
k++;
b[k]=a[i];
dfs(i,1);
vis[i]=0;
k--;
if(flag==1) break;
}
return 0;
}
