CF1059C Sequence Transformation 题解

这几天不知道写点什么，状态也不太好，搬个题上来吧

题意：给定一个数n，设一个从1到n的序列，每次删掉一个序列中的数，求按字典序最大化的GCD序列

做法：按2的倍数找，但是如果除2能得到3的这种情况要特殊处理（￥#……%￥……@#￥不知道该怎么描述，看代码吧）

C. Sequence Transformation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's call the following process a transformation of a sequence of length nn.

If the sequence is empty, the process ends. Otherwise, append the greatest common divisor (GCD) of all the elements of the sequence to the result and remove one arbitrary element from the sequence. Thus, when the process ends, we have a sequence of nn integers: the greatest common divisors of all the elements in the sequence before each deletion.

You are given an integer sequence 1,2,…,n1,2,…,n. Find the lexicographically maximum result of its transformation.

A sequence a1,a2,…,ana1,a2,…,an is lexicographically larger than a sequence b1,b2,…,bnb1,b2,…,bn, if there is an index ii such that aj=bjaj=bj for all j<ij<i, and ai>biai>bi.

Input

The first and only line of input contains one integer nn (1≤n≤1061≤n≤106).

Output

Output nn integers  — the lexicographically maximum result of the transformation.

Examples

input

Copy

3

output

Copy

1 1 3 

input

Copy

2

output

Copy

1 2 

input

Copy

1

output

Copy

1 

Note

In the first sample the answer may be achieved this way:

• Append GCD(1,2,3)=1(1,2,3)=1, remove 22.
• Append GCD(1,3)=1(1,3)=1, remove 11.
• Append GCD(3)=3(3)=3, remove 33.

We get the sequence [1,1,3][1,1,3] as the result.

 #include <iostream>
 using namespace std;

 int p[];

 int _pow(int a, int b)
 {
     int ans = ;
     int temp = a;
     while (b)
     {
         if (b & )
             ans *= temp;
         temp *= temp;
         b >>= ;
     }
     return ans;
 }

 int main()
 {
     ios::sync_with_stdio(false);
     cout.tie();
     for (int i = ; i <= ; i++)
     {
         p[i] = _pow(, i);
     }

     int n;
     cin >> n;
     if (n == )
     {
         cout << "1 1 3" << endl;
         return ;
     }

     int step = ;
     int flag = ;
     int n1 = n;
     while (n1)
     {
         if (n1 ==  && !flag)
         {
             int temp = ;
             while (temp <= n)
             {
                 temp *= ;
             }
             flag = temp / ;
             break;
         }
         n1 /= ;
     }
     while (n)
     {
         int num = n - n / ;
         n -= num;

         for (int i = ; i < num; i++)
         {
             if (flag && num == )
                 cout << flag;
             else
                 cout << p[step];
             if (i != num -  || n)
                 cout << " ";
         }
         step++;
     }

     cout << endl;
     return ;
 }