codeforces 842 D. Vitya and Strange Lesson

给定$n$个数字，分别为$a_1, a_2, \dots , a_n$

有若干次操作，每次给定一个$x$，并将所有的$a_i$变为$a_i \oplus x$，之后询问$\mathop{mex}\{a_1,a_2, \dots , a_n\}$

01trie维护数字，然后全部异或相当于在二进制下为1的位置进行子树交换，打上标记就没了……

#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, m;

int cnt[N], ch[N][2], tot = 1, tag, a[N];

void ins(int val) {
    int x = 1;
    for(int i = 20 ; ~ i ; -- i) {
        ++ cnt[x];
        int c = (val >> i) & 1;
        if(!ch[x][c]) ch[x][c] = ++ tot;
        x = ch[x][c];
    }
    ++ cnt[x];
}

int query() {
    int res = 0, x = 1, l = 0, r = (1 << 21) - 1;
    for(int i = 20 ; ~ i ; -- i) {
        int bit = (tag >> i) & 1;
        int lc = ch[x][bit], rc = ch[x][!bit];
        int mid = (l + r) >> 1;
        if(cnt[lc] == mid - l + 1) x = rc, l = mid + 1, res |= 1 << i;
        else x = lc, r = mid;
    }
    return res;
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1 ; i <= n ; ++ i) scanf("%d", &a[i]);
    sort(a + 1, a + 1 + n);
    n = unique(a + 1, a + 1 + n) - a - 1;
    for(int i = 1 ; i <= n ; ++ i) ins(a[i]);
    for(int i = 1, x ; i <= m ; ++ i) {
        scanf("%d", &x);
        tag ^= x;
        printf("%d\n", query());
    }
}