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
\[ [(\_or[rt] \oplus \_and[rt]) \& x] == 0 || [(\_or[rt] \oplus \_and[rt]) \& x] == (\_or[rt] \oplus \_and[rt]) \]
时就直接打标机。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;
}