题意：

已知\(n\)个数字，进行以下操作：

• \(1.\)区间\([L,R]\) 按位与\(x\)
• \(2.\)区间\([L,R]\) 按位或\(x\)
• \(3.\)区间\([L,R]\) 询问最大值

思路：

吉司机线段树。
我们按位考虑，维护区间或\(\_or\)和区间与\(\_and\)，那么得到区间非公有的\(1\)为\((\_or \oplus \_and)\)，那么如果对所有的非公有的\(1\)影响都一样就不会对最大值有影响，那么就直接打标机，否则继续往下更新。即
\[ [(\_or[rt] \oplus \_and[rt]) \& x] == 0 || [(\_or[rt] \oplus \_and[rt]) \& x] == (\_or[rt] \oplus \_and[rt]) \]
时就直接打标记。

代码：

#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<sstream>
#include<iostream>
#include<algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 2e5 + 5;
const int MAXM = 3e6;
const ll MOD = 998244353;
const ull seed = 131;
const int INF = 0x3f3f3f3f;#define lson (rt << 1)
#define rson (rt << 1 | 1)
inline bool read(int &num){char in;bool IsN=false;in = getchar();if(in == EOF) return false;while(in != '-' && (in < '0' || in > '9')) in = getchar();if(in == '-'){ IsN = true; num = 0;}else num = in - '0';while(in = getchar(),in >= '0' && in <= '9'){num *= 10, num += in-'0';}if(IsN) num = -num;return true;
}int a[maxn], all = (1 << 21) - 1;
int _or[maxn << 2], _and[maxn << 2], Max[maxn << 2];
int lazya[maxn << 2], lazyo[maxn << 2];
inline void pushup(int rt){_or[rt] = _or[lson] | _or[rson];_and[rt] = _and[lson] & _and[rson];Max[rt] = max(Max[lson], Max[rson]);
}
inline void pushdown(int rt, int l, int r){int m = (l + r) >> 1;if(lazya[rt] != all){Max[lson] &= lazya[rt];Max[rson] &= lazya[rt];_or[lson] &= lazya[rt];_or[rson] &= lazya[rt];_and[lson] &= lazya[rt];_and[rson] &= lazya[rt];lazya[lson] &= lazya[rt];lazya[rson] &= lazya[rt];lazyo[lson] &= lazya[rt];lazyo[rson] &= lazya[rt];lazya[rt] = all;}if(lazyo[rt] != 0){Max[lson] |= lazyo[rt];Max[rson] |= lazyo[rt];_or[lson] |= lazyo[rt];_or[rson] |= lazyo[rt];_and[lson] |= lazyo[rt];_and[rson] |= lazyo[rt];lazya[lson] |= lazyo[rt];lazya[rson] |= lazyo[rt];lazyo[lson] |= lazyo[rt];lazyo[rson] |= lazyo[rt];lazyo[rt] = 0;}
}
void build(int l, int r, int rt){lazya[rt] = all, lazyo[rt] = 0;if(l == r){_and[rt] = _or[rt] = Max[rt] = a[l];return;}int m = (l + r) >> 1;build(l, m, lson);build(m + 1, r, rson);pushup(rt);
}
void update(int L, int R, int l, int r, int op, int x, int rt){if(L <= l && R >= r){if(op == 1){    //&if(((_or[rt] ^ _and[rt]) & x) == 0 || ((_or[rt] ^ _and[rt]) & x) == (_or[rt] ^ _and[rt])){   Max[rt] &= x;_or[rt] &= x;_and[rt] &= x;lazya[rt] &= x;lazyo[rt] &= x;return;}}else{   //|if(((_or[rt] ^ _and[rt]) & x) == 0 || ((_or[rt] ^ _and[rt]) & x) == (_or[rt] ^ _and[rt])){    Max[rt] |= x;_or[rt] |= x;_and[rt] |= x;lazya[rt] |= x;lazyo[rt] |= x;return;}}}int m = (l + r) >> 1;pushdown(rt, l, r);if(L <= m)update(L, R, l, m, op, x, lson);if(R > m)update(L, R, m + 1, r, op, x, rson);pushup(rt);
}
int query(int L, int R, int l, int r, int rt){if(L <= l && R >= r){return Max[rt];}pushdown(rt, l, r);int m = (l + r) >> 1, MAX = -INF;if(L <= m)MAX = max(MAX, query(L, R, l, m, lson));if(R > m)MAX = max(MAX, query(L, R, m + 1, r, rson));return MAX;}
int main(){int n, m;read(n), read(m);for(int i = 1; i <= n; i++) read(a[i]);build(1, n, 1);while(m--){int op, l, r, x;read(op), read(l), read(r);if(op < 3) read(x);if(op < 3){update(l, r, 1, n, op, x, 1);}else{printf("%d\n", query(l, r, 1, n, 1));}}return 0;
}