let 如果 targe seq input 給定 wing using common

這幾天不知道寫點什麽，狀態也不太好，搬個題上來吧

題意：給定一個數n，設一個從1到n的序列，每次刪掉一個序列中的數，求按字典序最大化的GCD序列

做法：按2的倍數找，但是如果除2能得到3的這種情況要特殊處理（￥#……%￥……@#￥不知道該怎麽描述，看代碼吧）

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 $\mathrm{nn\; integers:\; the\; greatest\; common\; divisors\; of\; all\; the\; elements\; in\; the\; sequence\; before\; each\; deletion.}$

You are given an integer sequence $\mathrm{1,2,\dots ,n1,2,\dots ,n.\; Find\; the\; lexicographically\; maximum\; result\; of\; its\; transformation.}$

A sequence ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an\; is\; lexicographically\; larger\; than\; a\; sequence\; b1,b2,\dots ,bnb1,b2,\dots ,bn,\; if\; there\; is\; an\; index\; ii\; such\; that\; aj=bjaj=bj\; for\; all}}^{}$

j<ij<i, and ${\mathrm{ai>biai>bi.}}^{}$

Input

The first and only line of input contains one integer $\mathrm{nn\; (1\le n\le 1061\le n\le 106).}$

Output

Output $\mathrm{nn\; integers\; —\; the\; lexicographically\; maximum\; result\; of\; the\; transformation.}$

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 $[1,1,3][1,1,3]\; as\; the\; result.$

 1 #include <iostream>
 2 using namespace std;
 3 
 4 int p[25];
 5 
 6 int _pow(int a, int b)
 7 {
 8     int ans = 1;
 9     int temp = a;
10     while (b)
11     {
12         if (b & 1)
13             ans *= temp;
14         temp *= temp;
15         b >>= 1;
16     }
17     return ans;
18 }
19 
20 int main()
21 {
22     ios::sync_with_stdio(false);
23     cout.tie(0);
24     for (int i = 0; i <= 20; i++)
25     {
26         p[i] = _pow(2, i);
27     }
28 
29     int n;
30     cin >> n;
31     if (n == 3)
32     {
33         cout << "1 1 3" << endl;
34         return 0;
35     }
36 
37     int step = 0;
38     int flag = 0;
39     int n1 = n;
40     while (n1)
41     {
42         if (n1 == 3 && !flag)
43         {
44             int temp = 6;
45             while (temp <= n)
46             {
47                 temp *= 2;
48             }
49             flag = temp / 2;
50             break;
51         }
52         n1 /= 2;
53     }
54     while (n)
55     {
56         int num = n - n / 2;
57         n -= num;
58 
59         for (int i = 0; i < num; i++)
60         {
61             if (flag && num == 1)
62                 cout << flag;
63             else
64                 cout << p[step];
65             if (i != num - 1 || n)
66                 cout << " ";
67         }
68         step++;
69     }
70 
71     cout << endl;
72     return 0;
73 }

CF1059C Sequence Transformation 題解