E - Many Operations (bitwise consideration + dp thought to record the result after the operation

E - Many Operations
思路：
虽然第 $i$ The initial value of the operation is th $i-1$ the result after the operation,But consider bitwise,Each bit can only be at the beginning of the operation $0$ 或者 $1$,And the sequence of operations is the same every time
可以想到用 $dp$ The idea is to record the result of each bit in sequence
$f[i][j][k]$ 表示第 $j$ 位初始值为 $k$ ,进行 $i$ 次操作后的值
code：

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
#define ull unsigned long long
#define ld long double
#define all(x) x.begin(), x.end()
#define mem(x, d) memset(x, d, sizeof(x))
#define eps 1e-6
using namespace std;
const int maxn = 2e5 + 9;
const int mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
ll n, m;
bool f[maxn][30][2];
// f[i][j][k] 表示第j位初始值为k,进行 i 次操作后的值
 
void work()
{
    
	cin >> n >> m;
	
	vector <int>t(n + 1), a(n + 1);
	for(int i = 1; i <= n; ++i){
    
		cin >> t[i] >> a[i];		
	}
	
	for(int i = 0; i < 30; ++i){
    
		f[0][i][0] = 0;
		f[0][i][1] = 1;
	}
	for(int i = 1; i <= n; ++i){
    
		for(int j = 0; j < 30; ++j){
    
			int x = (a[i] >> j & 1);
			for(int k = 0; k < 2; ++k){
    
				if(t[i] == 1){
    
					f[i][j][k] = f[i-1][j][k] & x;  
				}
				else if(t[i] == 2){
    
					f[i][j][k] = f[i-1][j][k] | x;
				}
				else {
    
					f[i][j][k] = f[i-1][j][k] ^ x;
				}
			}
		}
	}
	for(int i = 1; i <= n; ++i){
    
		int ans = 0;
		for(int j = 0; j < 30; ++j){
    
			int k = (m >> j & 1);
			if(f[i][j][k]) ans += 1 << j;
		}
		cout << ans << endl;
		m = ans;
	}
}

int main()
{
    
	ios::sync_with_stdio(0);
// int TT;cin>>TT;while(TT--)
	work();
	return 0;
}