Codeforces Round #344 (Div. 2) A. Interview 水题

A. Interview

题目连接:

http://www.codeforces.com/contest/631/problem/A

Description

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.

Sample Input

5
1 2 4 3 2
2 3 3 12 1

Sample Output

22

Hint

题意

给你2*n的矩阵

然后定义一个函数f(a,l,r)表示a数组在l到r的或值

然后让你找到一对l,r,使得f(a,l,r)+f(b,l,r)最大

题解:

由于是或嘛,所以就把所有数全部或起来

代码

#include<bits/stdc++.h>
using namespace std;

long long ans = 0;
int main()
{
    int n;
    scanf("%d",&n);
    long long tmp = 0;
    for(int i=1;i<=n;i++)
    {
        long long x;
        scanf("%lld",&x);
        tmp|=x;
    }
    ans+=tmp;
    tmp = 0;
    for(int i=1;i<=n;i++)
    {
        long long x;
        scanf("%lld",&x);
        tmp|=x;
    }
    ans+=tmp;
    cout<<ans<<endl;
}
    原文作者:qscqesze
    原文地址: https://www.cnblogs.com/qscqesze/p/5243147.html
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