[ABC246Ex] 01? Queries

Problem Statement

You are given a string $S$ of length $N$ consisting of 0, 1, and ?.

You are also given $Q$ queries $(x_1, c_1), (x_2, c_2), \ldots, (x_Q, c_Q)$.
For each $i = 1, 2, \ldots, Q$, $x_i$ is an integer satisfying $1 \leq x_i \leq N$ and $c_i$ is one of the characters 0 , 1, and ?.

For $i = 1, 2, \ldots, Q$ in this order, do the following process for the query $(x_i, c_i)$.

1. First, change the $x_i$-th character from the beginning of $S$ to $c_i$.
2. Then, print the number of non-empty strings, modulo $998244353$, that can be obtained as a (not necessarily contiguous) subsequence of $S$ after replacing each occurrence of ? in $S$ with 0 or 1 independently.

Constraints

• $1 \leq N, Q \leq 10^5$
• $N$ and $Q$ are integers.
• $S$ is a string of length $N$ consisting of 0, 1, and ?.
• $1 \leq x_i \leq N$
• $c_i$ is one of the characters 0 , 1, and ?.

---

Input

Input is given from Standard Input in the following format:

$N$ $Q$
$S$
$x_1$ $c_1$
$x_2$ $c_2$
$\vdots$
$x_Q$ $c_Q$

Output

Print $Q$ lines. For each $i = 1, 2, \ldots, Q$, the $i$-th line should contain the answer to the $i$-th query $(x_i, c_i)$ (that is, the number of strings modulo $998244353$ at the step 2. in the statement).

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Sample Input 1

3 3
100
2 1
2 ?
3 ?

Sample Output 1

5
7
10

• The $1$-st query starts by changing $S$ to 110. Five strings can be obtained as a subsequence of $S = $ 110: 0, 1, 10, 11, 110. Thus, the $1$-st query should be answered by $5$.

• The $2$-nd query starts by changing $S$ to 1?0. Two strings can be obtained by the ? in $S = $ 1?0: 100 and 110. Seven strings can be obtained as a subsequence of one of these strings: 0, 1, 00, 10, 11, 100, 110. Thus, the $2$-nd query should be answered by $7$.

• The $3$-rd query starts by changing $S$ to 1??. Four strings can be obtained by the ?'s in $S = $ 1??: 100, 101, 110, 111. Ten strings can be obtained as a subsequence of one of these strings: 0, 1, 00, 01, 10, 11, 100, 101, 110, 111. Thus, the $3$-rd query should be answered by $10$.

---

Sample Input 2

40 10
011?0??001??10?0??0?0?1?11?1?00?11??0?01
5 0
2 ?
30 ?
7 1
11 1
3 1
25 1
40 0
12 1
18 1

Sample Output 2

746884092
532460539
299568633
541985786
217532539
217532539
217532539
573323772
483176957
236273405

Be sure to print the count modulo $998244353$.

如果这个问题不是动态的，那要怎么做？想到dp做法。
定义 \(dp_{i,0/1}\) 为在前 \(i\) 个字符的所有子序列中，如果再加上 \(0/1\) 这个字符后，就不是前 \(i\) 个字符的子序列了的子序列个数。

那么如果遇到一个 \(1\)，那么就相当于给所有加上 \(1\) 不属于前 \(i\) 个数的子序列加上了一个 \(1\)，\(dp_{i,1}=dp_{i-1,1}\)，然后新生成的这些子序列肯定再加上 \(0\) 后不属于前面的子序列，\(dp_{i,0}=dp_{i-1,1}+dp_{i-1,0}\).

如果遇到一个 \(0\) ，同理。遇到一个问好，\(dp_{i,1}=dp_{i,0}=dp_{i-1,1}+dp_{i-1,0}\)。

另开一个变量统计答案就可以了。

然后就要开始动态 dp，设 \((dp_0,dp_1,ans)\)为一个向量
遇到一个\(1\)，向量乘上 \(\begin{Bmatrix}1&0&0\\1&1&1\\0&0&1 \end{Bmatrix}\)

遇到一个\(0\)，向量乘上 \(\begin{Bmatrix}1&1&1\\0&1&0\\0&0&1 \end{Bmatrix}\)

遇到一个\(?\)，向量乘上 \(\begin{Bmatrix}1&1&1\\1&1&1\\0&0&1 \end{Bmatrix}\)

剩下的就是用线段树维护矩阵乘法，单点修改，区间查询就可以了。

#include<bits/stdc++.h>
const int N=1e5+5,P=998244353;
int n,q,x;
char c;
struct matrix{
	int a[4][4];
}t[3],tr[N<<2],p,dw;
matrix cheng(matrix a,matrix b)
{
	matrix c;
	memset(c.a,0,sizeof(c.a));
	for(int i=1;i<=3;i++)
		for(int j=1;j<=3;j++)
			for(int k=1;k<=3;k++)
				c.a[i][j]+=1LL*a.a[i][k]*b.a[k][j]%P,c.a[i][j]%=P;
	return c;
}
int turn(char c)
{
	if(c<='1')
		return c-'0';
	return 2;
}
void copy(matrix&a,matrix b)
{
	for(int i=1;i<=3;i++)
		for(int j=1;j<=3;j++)
			a.a[i][j]=b.a[i][j];
}
void build(int o,int l,int r)
{
//	printf("%d %d %d\n",o,l,r);
	if(l>r)	
		return;
	if(l==r)
	{
		scanf("  %c",&c);
		copy(tr[o],t[turn(c)]);
		return;
	}
	int md=l+r>>1;
	build(o<<1,l,md);
	build(o<<1|1,md+1,r);
	copy(tr[o],cheng(tr[o<<1],tr[o<<1|1]));
}
void update(int o,int l,int r,int x,int y)
{
	if(l==r)
	{
		copy(tr[o],t[y]);
		return;
	}
	int md=l+r>>1;
	if(md>=x)
		update(o<<1,l,md,x,y);
	else
		update(o<<1|1,md+1,r,x,y);
	copy(tr[o],cheng(tr[o<<1],tr[o<<1|1]));
}
int main()
{
	scanf("%d%d",&n,&q);
	p.a[1][1]=p.a[1][2]=1;
	dw.a[1][1]=dw.a[2][2]=dw.a[3][3]=1;
	for(int i=0;i<(N<<2);i++)
		copy(tr[i],dw);
	t[0].a[1][1]=t[0].a[2][1]=t[0].a[2][2]=t[0].a[2][3]=t[0].a[3][3]=1;
	t[1].a[1][1]=t[1].a[1][2]=t[1].a[1][3]=t[1].a[2][2]=t[1].a[3][3]=1;
	t[2].a[1][1]=t[2].a[1][2]=t[2].a[1][3]=t[2].a[2][1]=t[2].a[2][2]=t[2].a[2][3]=t[2].a[3][3]=1;
	build(1,1,n);
	while(q--)
	{
		scanf("%d %c",&x,&c);
		update(1,1,n,x,turn(c));
		printf("%d\n",cheng(p,tr[1]).a[1][3]);
	}
}