滴滴xor(nlogn)的还有一个O（N）的

思路是xor有性质a ^ b ^b =a;
所以统计前缀和是否出现过
 #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
map <int, bool> mp;
int n, t, q, ans;
int main() {
    ans = 0;
    scanf("%d", &n);
    mp[0] = true;
    while (n--) {
        scanf("%d", &t);
        q ^= t;
        if (mp.count(q)) {
            q = 0;
            mp.clear();
            mp[0] = true;
            ans ++;
        } else {
            mp[q] = true;
        }
    }
    printf("%d\n", ans);
    return 0;
} 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
bool mp[100001];
queue <int> que;
int n, t, q, ans;
int main() {
    ans = 0;
    scanf("%d", &n);
    memset(mp, 0, sizeof mp);
    mp[0] = true;
    while (n--) {
        scanf("%d", &t);
        q ^= t;
        if (mp[q]) {
            q = 0;
            while (!que.empty()) {
                mp[que.front()] = false;
                que.pop();
            }
            ans ++;
        } else {
            que.push(q);
            mp[q] = true;
        }
    }
    printf("%d\n", ans);
    return 0;
}