A - Interview（c）

最新推荐文章于 2025-12-23 09:11:29 发布

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本文介绍了一个面试算法题，题目要求从两个数组中找到能使区间或运算结果相加最大的组合。通过暴力枚举的方法，文章提供了一段C语言代码实现，该算法遍历所有可能的区间，计算并比较或运算结果，最终输出最大值。

A - Interview

Blake is a CEO of a large company called “Blake Technologies”. He loves his company very much and he thinks that his company should be the best. That is why every candidate needs to pass through the interview that consists of the following problem.We define function f(x, l, r) as a bitwise OR of integers xl, xl + 1, …, xr, where xi is the i-th element of the array x. You are given two arrays a and b of length n. You need to determine the maximum value of sum f(a, l, r) + f(b, l, r) among all possible 1 ≤ l ≤ r ≤ n.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the length of the arrays.The second line contains n integers ai (0 ≤ ai ≤ 109).The third line contains n integers bi (0 ≤ bi ≤ 109).

Output

Print a single integer — the maximum value of sum f(a, l, r) + f(b, l, r) among all possible 1 ≤ l ≤ r ≤ n.

Examples

Input

5
1 2 4 3 2
2 3 3 12 1

Output

Input

10
13 2 7 11 8 4 9 8 5 1
5 7 18 9 2 3 0 11 8 6

Output

Note

Bitwise OR of two non-negative integers a and b is the number c = a OR b, such that each of its digits in binary notation is 1 if and only if at least one of a or b have 1 in the corresponding position in binary notation.In the first sample, one of the optimal answers is l = 2 and r = 4, because f(a, 2, 4) + f(b, 2, 4) = (2 OR 4 OR 3) + (3 OR 3 OR 12) = 7 + 15 = 22. Other ways to get maximum value is to choose l = 1 and r = 4, l = 1 and r = 5, l = 2 and r = 4, l = 2 and r = 5, l = 3 and r = 4, or l = 3 and r = 5.In the second sample, the maximum value is obtained for l = 1 and r = 9.

题目意思就是让我们在每串数字里面找出或运算值最大的区间相加得出结果

我学习参考的大佬链接暴力枚举

下面是我的代码。

#include <stdio.h>
#include <stdlib.h>
int main()
{
 	 int a[1005],b[1005],i,j,x,y,max=0,n;
	 scanf("%d",&n);
 	 for(i=1;i<=n;i++)
 		 scanf("%d",&a[i]);
	 for(i=1;i<=n;i++)
 		 scanf("%d",&b[i]);
	 if(n==1)
		  max=a[1]+b[1];
	 else
	 {
  		for(i=1;i<=n;i++)
  		{
   		x=a[i];
  		 y=b[i];
  		 for(j=i+1;j<=n;j++)
  		 {
   			 x=(x|a[j]);
   			 y=(y|b[j]);
    			if(x+y>max)
   			 max=x+y;
  		 }
  	 }
	 }
	 printf("%d",max);
	 return 0;
	}