CF 286B(Shifting-deque)

Posted 2022-10-25

B. Shifting

time limit per test

memory limit per test

input

output

John Doe has found the beautiful permutation formula.

p = p₁, p₂, ..., p_n. Lets define transformation f

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.

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

Input

n (2 ≤ n ≤ 10⁶).

Output

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<int> 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<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
#define MAXN (1000000+10)
deque<int> 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<deq.size();i++) cout<<deq[i]<< ;
  return 0;