CF1051D Bicolorings

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原题链接：http://codeforces.com/contest/1051/problem/D

Bicolorings

You are given a grid, consisting of $2$ rows and $n$ columns. Each cell of this grid should be colored either black or white.

Two cells are considered neighbours if they have a common border and share the same color. Two cells $A$ and $B$ belong to the same component if they are neighbours, or if there is a neighbour of $A$ that belongs to the same component with $B$.

Let’s call some bicoloring beautiful if it has exactly $k$ components.

Count the number of beautiful bicolorings. The number can be big enough, so print the answer modulo $998244353$.

Input

The only line contains two integers $n$ and $k (1≤n≤1000, 1≤k≤2n)$ — the number of columns in a grid and the number of components required.

Output

Print a single integer — the number of beautiful bicolorings modulo $998244353$.

Examples

input

3 4

output

12

input

4 1

output

2

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input

1 2

output

2

Note

One of possible bicolorings in sample $1$:

题解

一眼状压 $dp$题， $dp[i][j][k]$表示前 $i$列，联通块数量为 $j$，该列的 $mask$为 $k$时的方案数，日常转移没有什么问题.

代码

#include<bits/stdc++.h>
using namespace std;
const int M=1005,mod=998244353;
long long dp[M][M<<1][4];
int n,m,i,j,k;
void in(){scanf("%d%d",&n,&m);}
void ac()
{
	dp[1][1][0]=dp[1][1][3]=dp[1][2][1]=dp[1][2][2]=1;
	for(i=1;i<n;++i)for(j=1;j<=(i<<1);++j)
	{
		for(k=0;k<4;++k)(dp[i+1][j][k]+=dp[i][j][k])%=mod;
		for(k=1;k<4;++k)(dp[i+1][j+1][k]+=dp[i][j][0])%=mod;
		for(k=1;k<3;++k)(dp[i+1][j+1][k]+=dp[i][j][3])%=mod,(dp[i+1][j][3]+=dp[i][j][k])%=mod,(dp[i+1][j][0]+=dp[i][j][k])%=mod;
		(dp[i+1][j+2][1]+=dp[i][j][2])%=mod,(dp[i+1][j+2][2]+=dp[i][j][1])%=mod,(dp[i+1][j+1][0]+=dp[i][j][3]);
	}
	printf("%lld",(dp[n][m][0]+dp[n][m][1]+dp[n][m][2]+dp[n][m][3])%mod);
}
int main(){in();ac();}