The XOR Largest Pair

思路

树板子题,直利用异或的性质,在树上遍历就行了。

代码

/*
  Author : lifehappy
*/
#include <bits/stdc++.h>

using namespace std;

const int N = 5e6 + 10;

int trie[N][2], a[N], n, tot;

void add(int x) {
    int rt = 0;
    for(int i = 30; i >= 0; i--) {
        int now = x >> i & 1;
        if(!trie[rt][now]) trie[rt][now] = ++tot;
        rt = trie[rt][now];
    }
}

int query(int x) {
    int rt = 0, ans = 0;
    for(int i = 30; i >= 0; i--) {
        int now = x >> i & 1;
        if(trie[rt][now ^ 1]) {
            ans |= 1 << i;
            rt = trie[rt][now ^ 1];
        }
        else {
            rt = trie[rt][now];
        }
    }
    return ans;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n;
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
        add(a[i]);
    }
    int ans = 0;
    for(int i = 1; i <= n; i++) {
        ans = max(ans, query(a[i]));
    }
    cout << ans << "\n";
    return 0;
}