題意： 給你一個序列，然後利用這個序列中的元素重新組成一個新的序列，其中每相鄰的兩個元素具有兩種關係的一種，例如x,y
滿足 x/3 = y 或 x*2 = y

思路： 爆搜，對於每一個元素

AC程式碼：

#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,e)      for(int i=s;i<=e;i++)
#define rev(i,s,e)      for(int i=e;i>=s;i--)
#define all(x)          x.begin(),x.end()
#define sz(x)           x.size()
 

#define szz(x)          int(x.size()-1)
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
const int MAXN = 2e5+10;
typedef long long LL;
LL a[105];
LL ans[105];
int n;
int k;
int sum;
map<string,int> m;
void dfs(LL x)
{
    if(sum == n)
        return;
    LL x1 = x/3;
    LL x2 = x*2;
    int pos1 = lower_bound(a+1
 
,a+n+1,x1)-a;
    int pos2 = lower_bound(a+1,a+n+1,x2)-a;
    char tmp1[25];
    char tmp2[25];
    if(x1 == a[pos1])
        sprintf(tmp1,"%I64d",a[pos1]);
    if(x2 == a[pos2])
        sprintf(tmp2,"%I64d",a[pos2]);
    string t1 = tmp1;
    string t2 = tmp2;
    if(m[t1] == 1)
    {
        m[t1] = 0
 
;
        ans[k++] = x1;
        sum++;
        dfs(x1);
    }
    else if(m[t2] == 1)
    {
        m[t2] = 0;
        ans[k++] = x2;
        sum++;
        dfs(x2);
    }
}
void init()
{
    rep(i,1,n)
    {
        char tmp[25];
        sprintf(tmp,"%I64d",a[i]);
        string t = tmp;
        m[t] = 1;
    }
}
int main()
{
    #ifdef LOCAL
    freopen("in.txt","r",stdin);
    #endif // LOCAL
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    cin>>n;
    rep(i,1,n) cin>>a[i];
    sort(a+1,a+n+1);
    int flag = 0;
    rep(i,1,n)
    {
        k = 0;
        sum = 0;
        m.clear();
        init();
        fill(ans,ans+n,0);
        ans[k++] = a[i];
        char tmp[25];
        sprintf(tmp,"%I64d",a[i]);
        string t = tmp;
        m[t] = 0;
        sum++;
        dfs(a[i]);
        if(sum == n)
        {
            flag = 1;
            break;
        }
    }
    if(flag)
    {
        rep(i,0,k-1) cout<<ans[i]<<" ";
    }
    return 0;
}