Interview（大水题）

C - Interview
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status

Practice

CodeForces 631A
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
Input
5
1 2 4 3 2
2 3 3 12 1
Output
22
Input
10
13 2 7 11 8 4 9 8 5 1
5 7 18 9 2 3 0 11 8 6
Output
46
Hint
Bitwise OR of two non-negative integers a and b is the number c = aORb, 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.

这题是codeforces上的一道题，读完题目一片迷茫，怎么下手，排序？将数组中的每个数化为二进制数，每一个二进制数循环进行与运算？望着题目发呆，后来感觉真的一点思路也没有，果断百度。百度完我的内心是崩溃的。。。原来按位与运算在直接用运算符 | 就可以了。（这个知识点真是没想到），然后更无语，根本不需要什么排序啊，循环啊，直接从第一个加到最后一个就是最大的。。。
因为两个数相与不可能变小。。。
代码：

#include <iostream>
using namespace std;
int main()
{
    int n, res1=0, res2=0, a;
    cin>>n;
    for(int i=0; i<n; i++)
    {
        cin>>a;
        res1=res1|a;
    }
    for(int i=0; i<n; i++)
    {
        cin>>a;
        res2=res2|a;
    }
    //cout<<res1<<endl;
     //cout<<res2<<endl;
    cout<<res1+res2<<endl;
    return 0;
}