原题：
You are given two arrays a and b, consisting of n integers each.

Let’s define a function f(a,b) as follows:

let’s define an array c of size n, where ci=ai⊕bi (⊕ denotes bitwise XOR);
the value of the function is c1&c2&⋯&cn (i.e. bitwise AND of the entire array c).
Find the maximum value of the function f(a,b) if you can reorder the array b in an arbitrary way (leaving the initial order is also an option).
Input
The first line contains one integer t (1≤t≤10^4) — the number of test cases.

The first line of each test case contains one integer n (1≤n≤10^5) — the size of arrays a and b.

The second line contains n integers a1,a2,…,an (0≤ai<2^30).

The third line contains n integers b1,b2,…,bn (0≤bi<2^30).

The sum of n over all test cases does not exceed 10^5.

Output
For each test case print one integer — the maximum value of the function f(a,b) if you can reorder the array b in an arbitrary way.

input
3
5
1 0 0 3 3
2 3 2 1 0
3
1 1 1
0 0 3
8
0 1 2 3 4 5 6 7
7 6 5 4 3 2 1 0
output
2
0
7

中文：
给你两个序列a和b，让你计算函数f(a, b)的最大值。函数f(a, b)的计算方法为：
设置ci = ai ⊕ bi，该函数返回的结果是c1 & c2 & … cn，求c1 & c2 & … cn的最大值。

代码：

#include
using namespace std;
const int maxn = 1e5 + 1;
const int N = 30;
int t, n;
int a[maxn], b[maxn];
int ta[maxn], rb[maxn];bool judge(int val) {for (int i = 1; i <=n ;i++) {ta[i] = val & a[i];rb[i] = val & (~b[i]);}sort(ta + 1, ta + 1 + n);sort(rb + 1, rb + 1 + n);for (int i = 1; i <= n; i++) {if (ta[i] != rb[i]) {return false;}}return true;
}int main() {ios::sync_with_stdio(false);cin >> t;while (t--) {cin >> n;for (int i = 1; i <=n ;i++) {cin >> a[i];}for (int i = 1; i <=n ;i++) {cin >> b[i];}int ans = 0;for (int i = 30; i >= 0; i--) {if (judge(ans | (1 << i))) {ans |= (1 << i);}}cout << ans << endl;}return 0;
}

思路：
设c1 & c2 & … cn = ans，将ans转换成二进制考虑，如果ans二进制的某一位为1，需要让c1到cn对应的位都为1，由于ci = ai xor bi，那么就要求ai和bi对应位相反。要想c1 & c2 & … cn的值最大，按照贪心的策略，就进行要让ans的最高二进制位为1。

可以枚举枚举ans结果的每一位，判断时候可以设置为1，判断方法即如judge函数中设置，排序后判断是满足即可。