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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.

2

2 1

3

1 3 2

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]

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)。

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        #include
        
         #include
         
          #include
          
           using namespace std;#define rep(i,a,n) for (int i=a;i
           
            =a;i--)#define pb push_back#define fi first#define se secondtypedef vector
            
              VI;typedef long long ll;typedef pair
             
               PII;const int inf=0x3fffffff;const ll mod=1000000007;const int maxn=20+10;deque
              
                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;}
              
             
            
           
          
         
        
       
      
     
    
   

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