不再根据son[p][u]来走，而是根据cnt[son[p][u]]。cnt记录经过此节点的数的数量

原题链接：最大异或和

#include<iostream>
using namespace std;

const int N = 1e5 * 31 + 10, M = 1e5 + 10;
int son[N][2], idx, cnt[N], sum[M];
int n, m;

void insert(int x, int op){
    int p = 0;
    for (int i = 30; i >= 0; i -- ){
        int u = x >> i & 1;
        if (!son[p][u]) son[p][u] = ++ idx;
        p = son[p][u];
        cnt[p] += op;
    }
}

int query(int x){
    int res = 0, p = 0;
    for (int i = 30; i >= 0; i -- ){
        int u = x >> i & 1;
        if (cnt[son[p][!u]]) p = son[p][!u], res = res * 2 + 1;
        else p = son[p][u], res = res * 2;
    }
    return res;
}

int main(){

    cin >> n >> m;
    for (int i = 1; i <= n; i ++ ){
        cin >> sum[i];
        sum[i] ^= sum[i - 1];
    }

    insert(sum[0], 1);
    int res = -1;
    for (int i = 1; i <= n; i ++ ){
        if (i > m) insert(sum[i - m - 1], -1);
        res = max(res, query(sum[i]));
        insert(sum[i], 1);
    }

    cout << res << endl;

    return 0;
}