CF Round 479 (Div. 3)--D. Divide by three, multiply by two(离散化+拓扑排序)

Polycarp likes to play with numbers. He takes some integer number x, writes it down on the board, and then performs with it n−1 operations of the two kinds:

• divide the number x by 3 (x must be divisible by 3);
• multiply the number x by 2.

After each operation, Polycarp writes down the result on the board and replaces x by the result. So there will be n numbers on the board after all.

You are given a sequence of length n — 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 n (2≤n≤100) — the number of the elements in the sequence. The second line of the input contains n integer numbers a1,a2,…,an (1≤ai≤3⋅10^18) — rearranged (reordered) sequence that Polycarp can wrote down on the board.

Output

Print n 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.

input 1

6
4 8 6 3 12 9

output 1

9 3 6 12 4 8 

input 2

2
1000000000000000000 3000000000000000000

output 2

3000000000000000000 1000000000000000000 

题意：其实就是给你n个数，让你求出从一个数出发通过上述两种操作最后经历剩下n-1个数。

解析：因为题目已经保证一定有解，猜了一手每个数必然不同(貌似猜对了😂)，因为答案肯定类似一条链，可以往拓扑排序上想，很正确，因为Ai很大，因此我们需要离散化处理，然后就进行拓扑排序求出答案序列即可。

#include 
using namespace std;
const int N=1e6+5;
typedef long long ll;
ll a[N],d[N],n;//分别记录原数组和入度数组
map mp,id,value;
vector v[N],ans;
void topsort()
{
    queue<int> q;
    for(int i=1;i<=n;i++)
    {
        int w=id[a[i]];//映射值
        if(!d[w]) ans.push_back(w),q.push(w);
    }
    while(q.size())
    {
        int u=q.front();
        q.pop();
        for(int i=0;isize();i++)
        {
            int j=v[u][i];//邻点
            if(--d[j]==0) ans.push_back(j),q.push(j);
        }
    }
}
void solve()
{
    int cnt=0;//用来给每个值映射
    scanf("%lld",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&a[i]);
        mp[a[i]]++;//该数存在
        id[a[i]]=++cnt;//将a[i]映射成cnt
        value[cnt]=a[i];//记录一下该cnt对于哪个a[i]
    }
    for(int i=1;i<=n;i++)
    {
        ll x=a[i],y=x/3,z=x*2;//x可以变成y,z
        ll idx=id[x],idy=id[y],idz=id[z];//x,y,z的映射值
        if(x%3==0&&mp[y]) v[idx].push_back(idy),d[idy]++;
        if(mp[z]) v[idx].push_back(idz),d[idz]++;
    }
    topsort();
    for(int i=0;iprintf("%lld ",value[ans[i]]);//最后再映射回去即可
    printf("\n");
}
int main()
{
    int t=1;
    //scanf("%d",&t);
    while(t--) solve();
    return 0;
}