[CF1748F] Circular Xor Reversal

题目描述

You have an array $ a_0, a_1, \ldots, a_{n-1} $ of length $ n $ . Initially, $ a_i = 2^i $ for all $ 0 \le i \lt n $ . Note that array $ a $ is zero-indexed.

You want to reverse this array (that is, make $ a_i $ equal to $ 2^{n-1-i} $ for all $ 0 \le i \lt n $ ). To do this, you can perform the following operation no more than $ 250,000 $ times:

Select an integer $ i $ ( $ 0 \le i \lt n $ ) and replace $ a_i $ by $ a_i \oplus a_{(i+1)\bmod n} $ .

Here, $ \oplus $ denotes the bitwise XOR operation.

Your task is to find any sequence of operations that will result in the array $ a $ being reversed. It can be shown that under the given constraints, a solution always exists.

输入格式

The first line contains a single integer $ n $ ( $ 2 \le n \le 400 $ ) — the length of the array $ a $ .

输出格式

On the first line print one integer $ k $ ( $ 0 \le k \le 250,000 $ ) — the number of operations performed.

On the second line print $ k $ integers $ i_1,i_2,\ldots,i_k $ ( $ 0 \le i_j \lt n $ ). Here, $ i_j $ should be the integer selected on the $ j $ -th operation.

Note that you don't need to minimize the number of operations.

样例 #1

样例输入 #1

样例输出 #1

3

1 0 1

样例 #2

样例输入 #2

样例输出 #2

9

1 0 1 0 2 1 0 1 0

提示

In the notes, the elements on which the operations are performed are colored red.

In the first test case, array $ a $ will change in the following way:

$ [1,\color{red}{2}] \rightarrow [\color{red}{1},3] \rightarrow [2,\color{red}{3}] \rightarrow [2,1] $ .

In the second test case, array $ a $ will change in the following way:

$ [1,\color{red}{2},4] \rightarrow [\color{red}{1},6,4] \rightarrow [7,\color{red}{6},4] \rightarrow [\color{red}{7},2,4] \rightarrow [5,2,\color{red}{4}] \rightarrow [5,\color{red}{2},1] \rightarrow [\color{red}{5},3,1] \rightarrow [6,\color{red}{3},1] \rightarrow [\color{red}{6},2,1] \rightarrow [4,2,1] $

交换两个数，有一个经典的通过异或达到的方式。相信各位卡常高手应该都知道。

x^=y

y^=x

x^=y

简写完后就是 x^=y^=x^=y

拆成二进制后，发现每个二进制位都交换了，所以两个数可以成功交换。

所以，要实现交换，只要想办法使一个数 $a_x^=a_y$ 就行了。

经过一波尝试，有这样一种方法：

先异或一波

然后把要异或上那个数移过来

然后上面两种操作全部进行逆操作，将一切都变回去，就成功让一个数异或上另一个数，且其他数不变。

但是写完后跑了 400 时，出来的次数是 400000。大概多了一倍吧。

仔细观察上面的过程，会发现让一个数异或上另一个数用了 4 轮此操作，但是有两次操作是废的。那么如果我们把第一次操作完后反复使用。由于我们是要 $a_x\oplus =a_y$，$a_{x+1}\oplus=a_{y-1}$，所以可以比如说这样。

这样子就可以把常数除以2.

#include<cstdio>

const int N=405,M=3e5+5;

int n,m,st[M],k;

void run(int l,int r)

{

	if(l>=r||l+1==r)

		return;

	for(int i=r-2;i>=l;i--)

		st[++m]=i%n;

	for(int i=l+1;i<r;i++)

		st[++m]=i%n;

	run(l+1,r-1);

}

void todo(int l,int r)

{

	for(int i=l;i<r;i++)

		st[++m]=i%n;

}

int main()

{

	scanf("%d",&n);

	k=n-2>>1;

	todo(0,n-1);

	run(0,n-1);

	todo(n-1-k,n+k);

	run(n-1-k,n+k);

	todo(0,n-1);

	run(0,n-1);

	printf("%d\n",m);

	for(int i=1;i<=m;i++)

		printf("%d ",st[i]);

	return 0;

}