CodeForces 1059C

Posted 2021-01-11 daybreaking

Description

Let‘s call the following process a transformation of a sequence of length $\mathrm{nn\; .}$

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 ${\mathrm{a1,a2,\dots ,ana1,a2,\dots ,an\; is\; lexicographically\; larger\; than\; a\; sequence\; b1,b2,\dots ,bnb1,b2,\dots ,bn\; ,\; if\; there\; is\; an\; index\; ii\; such\; that\; aj=bjaj=bj\; for\; all\; j<ij<i\; ,\; and\; ai>biai>bi\; .}}^{}$

Input

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

Output

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

Sample Input

 

Input

3

Output

1 1 3 

Input

2

Output

1 2 

Input

1

Output

1 

Sample Output

 

Hint

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

 

 

尽可能的让大的gcd值尽快出现。

有一条规则可以推出来，两个连续的数的gcd是1，所以第一步是将原数列变成奇数数列或偶数数列，又因为对于长度n大于3时，偶数数列肯定要先出现大的gcd，所以第一步将原数列转成偶数数列。

之后有趣的事情就出现了，可以发现，可以将形成的数列，奇数位上的数看“奇数数列”，偶数位上的数看成“偶数数列”，又重复第一步的过程。

在以上整个程中n都是大于3的，对于小于3的直接按“偶奇奇”的顺序删。

 

 1 #include<cstdio>
 2 #include<cstdlib>
 3 #include<cstring>
 4 #include<string>
 5 #include<cmath>
 6 #include<algorithm>
 7 #include<queue>
 8 #include<stack>
 9 #include<deque>
10 #include<map>
11 #include<iostream>
12 using namespace std;
13 typedef long long  LL;
14 const double pi=acos(-1.0);
15 const double e=exp(1);
16 const int N = 100009;
17 
18 int con[1000009];
19 
20 int gcd(int a,int b)
21 {
22     int c;
23     while(b)
24     {
25         c=b;
26         b=a%b;
27         a=c;
28     }
29     return a;
30 }
31 
32 int main()
33 {
34     int i,p,j,n;
35     int cnt=0;
36     scanf("%d",&n);
37     for(i=1;i<=n;i++)
38         con[i]=i;
39     p=n;
40     while(p>0)
41     {
42         if(p==3)
43         {
44             printf("%d %d %d
",gcd(gcd(con[1],con[2]),con[3]),gcd(con[2],con[3]),con[3]);
45             break;
46         }
47         else if(p>=2)
48         {
49             cnt=0;
50             int k=gcd(con[1],con[2]);
51             for(i=1;i<=p;i+=2)
52             {
53                 printf("%d ",k);
54                 if(i+1<=p)
55                     con[++cnt]=con[i+1];
56             }
57             p=cnt;
58         }
59         else if(p==1)
60         {
61             printf("%d
",con[p]);
62             break;
63         }
64 
65     }
66     return 0;
67 }

View Code