CodeForces - 631A Interview （思想）水

CodeForces - 631A

Interview

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

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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 x_l, x_l + 1, ..., x_r, where x_i 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 sumf(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 a_i (0 ≤ a_i ≤ 10⁹).

The third line contains n integers b_i (0 ≤ b_i ≤ 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

51 2 4 3 22 3 3 12 1

Output

22

Input

1013 2 7 11 8 4 9 8 5 15 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 4OR 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.

Source

Codeforces Round #344 (Div. 2)

//题意：定义方程f(a,2,4)=a[2]|a[3]|a[4]。

先输入一个n，表示数组的长度，再输入a[],b[]。让求f(a,l,r)+f(b,l,r)。l表示数组的左端点，r表示数组的右端点。

//思路：

因为是或运算，所以不管几个数或运算（|），值只会变大，所以直接求出所有a[i]的取或（|）的值再加上所有b[i]的取或（|）的值即可。

#include<stdio.h>#include<string.h>#include<algorithm>#define INF 0x3f3f3f3f#define ll long long#define N 1010#define M 1000000007#define PI acos(-1.0)using namespace std;int a[N];int b[N];int main(){int i,j,k;int n;while(scanf("%d",&n)!=EOF){int aa=0,bb=0;for(i=0;i<n;i++){scanf("%d",&a[i]);aa=aa|a[i];}for(i=0;i<n;i++){scanf("%d",&b[i]);bb=bb|b[i];}printf("%d\n",aa+bb);}return 0;}