[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 $(x_i, c_i)$.

First, change the $x_i$-th character from the beginning of $S$ to $c_i$.
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).

Sample Input 1

Sample Output 1

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

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]);

	}

}