Luogu4735 最大異或和

題目藍鏈

Description

給你一個序列，你需要支持以下兩個操作：

1. A x: 在序列尾部添加一個整數\(x\)，序列的長度增加\(1\)
2. Q l r x: 詢問操作，你需要找到一個位置\(p \in [l, r]\)，使得：\(x \bigoplus a_p \bigoplus a_{p + 1} \bigoplus \ldots \bigoplus a_n\)最大，輸出最大值是多少

Solution

首先我們需要打一個可持久化的\(trie\)樹來維護\(a_i\)的前綴和，這樣我們就可以快速在一段區間對應的\(trie\)樹上查詢，查詢的時候我們只需要貪心的找就可以了

另外，我們需要把詢問的式子轉化一下
\[ x \bigoplus a_p \bigoplus a_{p + 1} \bigoplus \ldots \bigoplus a_n = (x \bigoplus all) \bigoplus a_1 \bigoplus a_2 \bigoplus \ldots \bigoplus a_{p - 1} \]
由于對于當前的詢問，\(x \bigoplus all\)為定值，所以我們只需要在\(trie\)樹上找到一個\(a_1 \bigoplus a_2 \bigoplus \ldots \bigoplus a_{p - 1}\)，使其與\(x \bigoplus all?\)的異或和最大即可

Code

#include <bits/stdc++.h>using namespace std;#define fst first
#define snd second
#define mp make_pair
#define squ(x) ((LL)(x) * (x))
#define debug(...) fprintf(stderr, __VA_ARGS__)typedef long long LL;
typedef pair<int, int> pii;template<typename T> inline bool chkmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
template<typename T> inline bool chkmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }inline int read() {int sum = 0, fg = 1; char c = getchar();for (; !isdigit(c); c = getchar()) if (c == '-') fg = -1;for (; isdigit(c); c = getchar()) sum = (sum << 3) + (sum << 1) + (c ^ 0x30);return fg * sum;
}const int maxn = 3e5 + 10;
const int inf = (1 << 24) - 1;int n, m, a[maxn << 1], rt[maxn << 1];namespace ST {struct node {int ls, rs, v;}A[maxn << 6];
#define ls(x) A[x].ls
#define rs(x) A[x].rsint cnt;void change(int &nrt, int rt, int l, int r, int x) {A[nrt = ++cnt] = A[rt], ++A[cnt].v;if (l == r) return;int mid = (l + r) >> 1;if (x <= mid) change(ls(nrt), ls(rt), l, mid, x);else change(rs(nrt), rs(rt), mid + 1, r, x);}int query(int x, int y, int l, int r, int v) {if (l == r) return l;int mid = (l + r) >> 1, len = r - mid;if (v <= mid) {if (A[ls(y)].v - A[ls(x)].v) return query(ls(x), ls(y), l, mid, v);return query(rs(x), rs(y), mid + 1, r, v + len);} else {if (A[rs(y)].v - A[rs(x)].v) return query(rs(x), rs(y), mid + 1, r, v);return query(ls(x), ls(y), l, mid, v - len);}return 0;}
}int main() {
#ifdef xunzhenfreopen("xor.in", "r", stdin);freopen("xor.out", "w", stdout);
#endifn = read(), m = read();for (int i = 1; i <= n; i++) {a[i] = a[i - 1] ^ read();ST::change(rt[i], rt[i - 1], 0, inf, a[i - 1]);}for (int i = 1; i <= m; i++) {static char s[10];scanf("%s", s);if (s[0] == 'A') {++n, a[n] = a[n - 1] ^ read();ST::change(rt[n], rt[n - 1], 0, inf, a[n - 1]);} else {int l = read(), r = read(), x = read() ^ a[n];printf("%d\n", ST::query(rt[l - 1], rt[r], 0, inf, inf ^ x) ^ x);}}return 0;
}