Many Operations

Problem Statement

We have a variable \(X\) and \(N\) kinds of operations that change the value of \(X\). Operation \(i\) is represented as a pair of integers \((T_i,A_i)\), and is the following operation:

• if \(T_i=1\), it replaces the value of \(X\) with \(X\) and \(A_i\);
• if \(T_i=2\), it replaces the value of \(X\) with \(X\) or \(A_i\);
• if \(T_i=3\),, it replaces the value of X with X xor \(A_i\).

Initialize \(X\) with the value of \(C\) and execute the following procedures in order:

• Perform Operation 1, and then print the resulting value of \(X\).
• Next, perform Operation 1,2 in this order, and then print the value of \(X\).
• next, perform Operation 1,2,3 in this order, and then print the value of X.
  ⋮
• Next, perform Operation 1,2,…,N in this order, and then print the value of X.

Constraints

• \(1≤N≤2×10^5\)
• \(1≤T_i≤3\)
• \(0≤A_i<2^{30}\)
• \(0≤C<2^{30}\)
• All values in input are integers.

Input

Input is given from Standard Input in the following format:
\(N\) \(C\)
\(T_1\) \(A_1\)
\(T_2\) \(A_2\)
⋮
\(T_N\) \(A_N\)

Output

Print \(N\) lines, as specified in the Problem Statement.

Sample Input 1

3 10
3 3
2 5
1 12

Sample Output 1

9
15
12

The initial value of \(X\) is \(10\).

• Operation \(1\) changes \(X\) to \(9\).
• Next, Operation \(1\) changes \(X\) to \(10\), and then Operation \(2\) changes it to \(15\).
• Next, Operation \(1\) changes X to \(12\), and then Operation \(2\) changes it to \(13\), and then Operation \(3\) changes it to \(12\).

Sample Input 2

9 12
1 1
2 2
3 3
1 4
2 5
3 6
1 7
2 8
3 9

Sample Output 2

0
2
1
0
5
3
3
11
2

为了方便，称执行操作1,2\(\cdots\)为第i轮操作

位运算问题，考虑按位计算。把C拆成每一位，求出他的运算后的答案。这是就只用考虑and,xor,or 0/1

• xor 0,and 1,or 0可以看作不变，不操作

• xor 1其实就是取反，可以打上取反标记。

• and 0=直接赋值为0，取反标记清空，答案直接为0，不管前面有哪些多少操作。

• or 1=直接赋值为1，同上

但是有很多轮操作，所以我们可以推出第i轮结束时赋值和取反操作情况。如果是没有赋值，那么第i轮结束后这一位答案就是上一轮结束答案^取反操作，否则就直接赋值。注意取反后赋值操作跟着取反。

每一位都这样操作，相加，就求出了答案。

#include<cstdio>
const int N=2e5+5;
int n,c,ans[N],rt,tx,tt,t[N],a[N];
int main()
{
	scanf("%d%d",&n,&c);
	for(int i=1;i<=n;i++)
		scanf("%d%d",t+i,a+i);
	for(int i=30;~i;i--)
	{
		tx=0,rt=-1,tt=c>>i&1;//rt:赋值,tx:取反
		for(int j=1;j<=n;j++)
		{
			if(t[j]==1)
				if(!(a[j]&1<<i))
					rt=tx=0;
			if(t[j]==2)
				if(a[j]&1<<i)
					rt=1,tx=0;
			if(t[j]==3)
				if(a[j]&1<<i)
					tx^=1;
			if(rt==-1)
				tt=tt^tx;
			else
				tt=rt^tx;
			ans[j]+=tt<<i;
		}
	}
	for(int i=1;i<=n;i++)
		printf("%d\n",ans[i]);
}

标签：Operations,12,Many,取反,value,Operation,changes,赋值
来源： https://www.cnblogs.com/mekoszc/p/16655713.html