題意

題目鏈接

Sol

設$sum[i]$表示$1 - i$的異或和

首先把每個詢問的$x \oplus sum[n]$就變成了詢問前綴最大值

可持久化Trie樹維護前綴xor，建樹的時候維護一下每個節點被遍歷了多少次

註意設置好偏移量，不然詢問區間為$[1, 1]$的時候可能掛掉

#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\n", 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 
*/
 

BZOJ3261: 最大異或和(可持久化trie樹)