Score : $600$ points

问题描述

存在长度为 $N$ 的序列 $A=(A_0,A_1,\ldots,A_{N-1})$ 和 $B=(B_0,B_1,\ldots,B_{N-1})$。

Takahashi 可以对序列 $A$ 进行任意次数（包括零次）的以下操作：

• 对序列 $A$ 应用左循环移位。换句话说，将 $A$ 替换为由 $A'_i=A_{(i+1)\% N}$ 定义的新序列，其中 $x\% N$ 表示 $x$ 除以 $N$ 后的余数。

Takahashi 的目标是最大化 $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$，其中 $x|y$ 表示 $x$ 和 $y$ 的按位逻辑和（按位 OR）。

求最大可能的 $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$。

什么是按位逻辑和（bitwise OR）？逻辑和（或 OR 操作）是一种针对两个一比特整数（$0$ 或 $1$）的操作，定义如下表格所示。按位逻辑和（bitwise OR） 是一种逐比特应用逻辑和的操作。

$x$	$y$	$x|y$
$0$	$0$	$0$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$1$

逻辑和在至少一个比特 $x$ 和 $y$ 为 $1$ 时产生 $1$。相反，只有当它们两者都为 $0$ 时才产生 $0$。

示例

0110 | 0101 = 0111```

 `以上为通义千问 qwen-max 翻译，仅供参考。`
## Problem Statement

There are length-$N$ sequences $A=(A_0,A_1,\ldots,A_{N-1})$ and $B=(B_0,B_1,\ldots,B_{N-1})$.  
Takahashi may perform the following operation on $A$ any number of times (possibly zero):

-   apply a left cyclic shift to the sequence $A$. In other words, replace $A$ with $A'$ defined by $A'_i=A_{(i+1)\% N}$, where $x\% N$ denotes the remainder when $x$ is divided by $N$.

Takahashi's objective is to maximize $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$, where $x|y$ denotes the bitwise logical sum (bitwise OR) of $x$ and $y$.

Find the maximum possible $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$.

What is the bitwise logical sum (bitwise OR)? The **logical sum** (or the OR operation) is an operation on two one-bit integers ($0$ or $1$) defined by the table below.  
The **bitwise logical sum (bitwise OR)** is an operation of applying the logical sum bitwise.<table><thead><tr align="center"><td align="center">$x$</td><td align="center">$y$</td><td align="center">$x|y$</td></tr></thead><tbody><tr align="center"><td align="center">$0$</td><td align="center">$0$</td><td align="center">$0$</td></tr><tr align="center"><td align="center">$0$</td><td align="center">$1$</td><td align="center">$1$</td></tr><tr align="center"><td align="center">$1$</td><td align="center">$0$</td><td align="center">$1$</td></tr><tr align="center"><td align="center">$1$</td><td align="center">$1$</td><td align="center">$1$</td></tr></tbody></table>

The logical sum yields $1$ if at least one of the bits $x$ and $y$ is $1$. Conversely, it yields $0$ only if both of them are $0$.

##### Example

0110 | 0101 = 0111```

Constraints

• $2 \leq N \leq 5\times 10^5$
• $0\leq A_i,B_i \leq 31$
• All values in the input are integers.

Input

The input is given from Standard Input in the following format:

$N$

$A_0$ $A_1$ $\ldots$ $A_{N-1}$

$B_0$ $B_1$ $\ldots$ $B_{N-1}$

Output

Print the maximum possible $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$.

Sample Input 1

3
0 1 3
0 2 3

Sample Output 1

8

If Takahashi does not perform the operation, $A$ remains $(0,1,3)$, and we have $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)=(0|0)+(1|2)+(3|3)=0+3+3=6$;
if he performs the operation once, making $A=(1,3,0)$, we have $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)=(1|0)+(3|2)+(0|3)=1+3+3=7$; and
if he performs the operation twice, making $A=(3,0,1)$, we have $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)=(3|0)+(0|2)+(1|3)=3+2+3=8$.
If he performs the operation three or more times, $A$ becomes one of the sequences above, so the maximum possible $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)$ is $8$, which should be printed.

Sample Input 2

5
1 6 1 4 3
0 6 4 0 1

Sample Output 2

23

The value is maximized if he performs the operation three times, making $A=(4,3,1,6,1)$,
where $\displaystyle\sum_{i=0}^{N-1} (A_i|B_i)=(4|0)+(3|6)+(1|4)+(6|0)+(1|1)=4+7+5+6+1=23$.

update @ 2024/3/10 12:09:58