Codeforces Round #176 (Div. 2) D. Shifting(模拟,STLdeque应用)

题目链接

D. 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, ..., pn.
Let's define transformation f of this permutation: 



where k (k > 1) is
an integer, the transformation parameter, r is such maximum integer that rk ≤ n.
If rk = n, then elements prk + 1, prk + 2and
so on are omitted. In other words, the described transformation of permutation p cyclically shifts to the left each consecutive block
of length k and the last block with the length equal to the remainder after dividing n by k. 

John Doe thinks that permutation f(f( ... f(p = [1, 2, ..., n], 2) ... , n - 1), n) is
beautiful. Unfortunately, he cannot quickly find the beautiful permutation he's interested in. That's why he asked you to help him.

Your task is to find a beautiful permutation for the given n. For clarifications, see the notes to the third sample.

Input

A single line contains integer n (2 ≤ n ≤ 106).

Output

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

Examples

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]

题意：

定义排列p = p1, p2, ..., pn。

定义函数f(p, k)用来转换p排列，k是转换参数, k > 1。

假设排列p的长度为n，首先把p按长度k进行分块，若最后一块长度不足k，则其长度为n%k，然后分别对每一块循环左移一位，得到一个新的排列。

给定n(2 <= n <= 1e6)。请输出 f(f(...f(p = [1,2,...,n],2)...,n-1,),n)的结果。

对于样例3:

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

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

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

题解：

对f(p,k)循环右移一位之后可以发现得到的序列相当于对原序列中下标mod k=1位置构成的子序列循环右移一位，

因此只需要对这些位置循环右移一位再将整个序列循环左移一位（也就是把第一个数字挪到最后）即可得到f(p,k)，
可以用一个双端队列维护，第k次操作需要操作n/k+O(1)的元素，总复杂度是O(nlogn)。

quailty代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
#include<deque>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const int inf=0x3fffffff;
const ll mod=1000000007;
const int maxn=20+10;
deque<int> dq;
int main()
{
int n;
scanf("%d",&n);
rep(i,1,n+1) dq.push_back(i);
rep(i,2,n+1)
{
for(int j=(n-1)/i;j>0;j--)
swap(dq[(j-1)*i],dq[j*i]);
dq.push_back(dq.front());
dq.pop_front();
}
rep(i,0,n) printf("%d ",dq[i]);
return 0;
}