CF 286B(Shifting-deque)

B. Shifting

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

John Doe has found the beautiful permutation formula.

Let's take permutation p = p₁, p₂, ..., p_n. Let's define transformation f of this permutation:

 k (k > 1) 是每段长度, r 是最大满足 rk ≤ n 的整数  把 r 段和尾剩余部分（如果有）左移.

求序列 f(f( ... f(p = [1, 2, ..., n], 2) ... , n - 1), n) .

Input

一行 n (2 ≤ n ≤ 10⁶).

Output

Print n distinct space-separated integers from 1 to n — a beautiful permutation of size n.

Sample test(s)

input

2

output

2 1 

input

3

output

1 3 2 

input

4

output

4 2 3 1 

Note

A note to the third test sample:

• f([1, 2, 3, 4], 2) = [2, 1, 4, 3]
• f([2, 1, 4, 3], 3) = [1, 4, 2, 3]
• f([1, 4, 2, 3], 4) = [4, 2, 3, 1]

这题我是用WJMZBMR的神模拟过的。

先普及一下deque< >

deque deq;   //建立双端队列

deq.push_back(x) 

deq.push_front(x) 

deq.pop_back(x) 

deq.pop_back(x) 

然后可以模拟了，每次把每段的最后搬上来。

#include
#include
#include
#include
#include
using namespace std;
#define MAXN (1000000+10)
deque deq;
int n;
int main()
{
	cin>>n;
	for (int i=1;i<=n;i++) deq.push_back(i);
	for (int i=2;i<=n;i++)
	{
		int l=(n-1)/i*i;deq.push_back(deq[l]);
		while (l-i>=0)		
		{
			deq[l]=deq[l-i];
			l-=i;
		}
		deq.pop_front();		
	}	
	for (int i=0;i