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

最新推荐文章于 2022-12-16 16:19:21 发布

___fouzhe

最新推荐文章于 2022-12-16 16:19:21 发布

阅读量473

点赞数 1

分类专栏： STL 模拟

本文链接：https://blog.youkuaiyun.com/fouzhe/article/details/56486259

版权

模拟同时被 2 个专栏收录

4 篇文章

订阅专栏

STL

1 篇文章

订阅专栏

John Doe has found the beautiful permutation formula.

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

$\text{[math]}$

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;
}