给定两个长度为n的数组a，b，定义f(a, b): 设ci = ai ^ bi, f(a, b) = c1 & c2 & ... & cn。可以随意对b数组排序，求f(a, b)的最大值

题目

思路:

#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define fi first
#define se second
#define lson p << 1
#define rson p << 1 | 1
const int maxn = 1e6 + 5, inf = 1e18, maxm = 4e4 + 5;
const int mod = 1e9 + 7;
// const int mod = 998244353;
const int N = 1e6;
// int a[505][5005];
// bool vis[505][505];
// char s[505][505];
int a[maxn], b[maxn];
bool vis[maxn];
string s;
int n, m;

struct Node{
    // int val, id;
    // bool operator<(const Node &u)const{
    //     return val < u.val;
    // }
    // int x, y;
    int l, r, j;
}c[maxn];

bool check(int x){
    vector<int> c, d;
    for(int i = 1; i <= n; i++){
        c.pb(a[i] & x);//&x, 因为只关心答案x为1的位
        d.pb(~b[i] & x);//先取反再与
    }
    sort(c.begin(), c.end());
    sort(d.begin(), d.end());
    return c == d;//如果相等，说明每个ai都能跟bj匹配，例如，ai = 01, bj = 10
}

void solve(){
    int res = 0;
    int q, k;
    cin >> n;
    for(int i = 1; i <= n; i++){
        cin >> a[i];
    }
    for(int i = 1; i <= n; i++){
        cin >> b[i];
    }
    for(int j = 29; j >= 0; j--){
        if(check(res | (1LL << j))){
            res |= (1LL << j);
        }
    }
    cout << res << '\n';
}
    
signed main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    
    int T = 1;
    cin >> T;
    while (T--)
    {
        solve();
    }
    return 0;
}