题意
Sol
设(sum[i])表示(1 - i)的异或和
首先把每个询问的(x oplus sum[n])就变成了询问前缀最大值
可持久化Trie树维护前缀xor，建树的时候维护一下每个节点被遍历了多少次
注意设置好偏移量，不然询问区间为([1, 1])的时候可能挂掉
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 6e5 + 10;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-')f =- 1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M, a[MAXN], sum[MAXN], cnt[MAXN * 25], ch[MAXN * 25][2], tot, root[MAXN];
void insert(int i, int x) {
x = (sum[i] = sum[i - 1] ^ x);
int p = root[i - 1], now = (root[i] = ++tot);
for(int i = 23; i >= 0; i--) {
bool nxt = x >> i & 1;
ch[now][nxt ^ 1] = ch[p][nxt ^ 1];
now = ch[now][nxt] = ++tot; p = ch[p][nxt];
cnt[now] = cnt[p] + 1;
}
}
int Query(int l, int r, int x) {
r = root[r], l = root[l];
int ans = 0;
for(int i = 23; i >= 0; i--) {
int nxt = x >> i & 1;
if(cnt[ch[r][nxt ^ 1]] - cnt[ch[l][nxt ^ 1]] > 0) ans += 1 << i, r = ch[r][nxt ^ 1], l = ch[l][nxt ^ 1];
else r = ch[r][nxt], l = ch[l][nxt];
}
return ans;
}
int main() {
N = read(); M = read();
for(int i = 1; i <= N; i++) a[i] = read(), insert(i, a[i]);
char ss[4];
for(int i = 1; i <= M; i++) {
scanf("%s", ss + 1);
if(ss[1] == 'A')
N++, a[N] = read(), insert(N, a[N]);
else {
int l = read() - 1, r = read() - 1, val = read();
printf("%d
", Query(l - 1, r, val ^ sum[N]));
}
}
}
/*
5 5
2 6 4 3 6
A 1
Q 3 5 4
A 4
Q 5 7 0
Q 3 6 6
*/