luoguP4735 最大异或和

https://www.luogu.org/problemnew/show/P4735

令 s 数组为 a 数组的异或前缀，则题目要求的式子可变为 s[p - 1] ^ s[n] ^ x，s[n] ^ x不变，则可以高效的从高位到低位贪心，凑出一个和 s[n] ^ x 异或最大的值，查询一个区间内的 s 值可用可持久化 trie 实现，方法与主席树类似，这里不再赘述

#include <bits/stdc++.h>
using namespace std;

const int N = 600000 + 10; 

int trie[N * 26][2], siz[N * 26], tot = 0;
int s[N], root[N], n, m;

template <typename T>
inline void read(T &f) {
    f = 0; T fu = 1; char c = getchar();
    while(c < '0' || c > '9') {if(c == '-') fu = -1; c = getchar();}
    while(c >= '0' && c <= '9') {f = (f << 3) + (f << 1) + (c & 15); c = getchar();}
    f *= fu;
}

void ins(int &u, int pre, int d, int x) {
    u = ++tot; siz[u] = siz[pre] + 1; if(d == -1) return;
    trie[u][0] = trie[pre][0]; trie[u][1] = trie[pre][1];
    int now = (x & (1 << d)) >> d; ins(trie[u][now], trie[pre][now], d - 1, x);
}

int find(int l, int r, int d, int x) {
    if(d == -1) return 0;
    int now = ((x & (1 << d)) >> d) ^ 1;
    if(siz[trie[r][now]] - siz[trie[l][now]]) return (1 << d) * now + find(trie[l][now], trie[r][now], d - 1, x);
    return (1 << d) * (now ^ 1) + find(trie[l][now ^ 1], trie[r][now ^ 1], d - 1, x);
}

int main() {
    scanf("%d %d", &n, &m);
    for(int i = 1; i <= n; i++) {
        int a; read(a);
        s[i] = s[i - 1] ^ a;
        ins(root[i], root[i - 1], 23, s[i - 1]);
    }
    for(int i = 1; i <= m; i++) {
        char c = getchar();
        while(c != 'A' && c != 'Q') c = getchar();
        if(c == 'A') {
            int a; read(a); n++;
            s[n] = s[n - 1] ^ a;
            ins(root[n], root[n - 1], 23, s[n - 1]);
        } else {
            int l, r, x;
            read(l); read(r); read(x);
            int Ans = s[n] ^ x;
            // 在 l 和 r 之间找和 Ans ^ 最大的
            printf("%d
", find(root[l - 1], root[r], 23, Ans) ^ Ans);
        }
    }
    return 0;
}