You are given a grid, consisting of 22 rows and nn 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 AA and BB belong to the same component if they are neighbours, or if there is a neighbour of AA that belongs to the same component with BB.
Let's call some bicoloring beautiful if it has exactly kk components.
Count the number of beautiful bicolorings. The number can be big enough, so print the answer modulo 998244353998244353.
Input
The only line contains two integers nn and kk (1≤n≤10001≤n≤1000, 1≤k≤2n1≤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 998244353998244353.
Examples
Input
3 4
Output
12
Input
4 1
Output
2
Input
1 2
Output
2
题意：
有2行，n列，求k个连通块的方法。
自己想实在是想不出来啊，看了看题解才勉强明白。
设一个三维dp数组。
dp[i][j][k]表示前i列有j个联通块，k表示i列地板的颜色，k=0表示11,k=1表示10,k=2表示01,k=3表示00。
然后有下面的递推关系：
if(m==0)
dp[i][j][0]=(dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j-1][3])%mod;
if(m==1)
dp[i][j][1]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j-2][2]+dp[i-1][j-1][3])%mod;
if(m==2)
dp[i][j][2]=(dp[i-1][j-1][0]+dp[i-1][j-2][1]+dp[i-1][j][2]+dp[i-1][j-1][3])%mod;
if(m==3)
dp[i][j][3]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j][3])%mod;
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
const long long int mod=998244353;
const int maxn=1005;
int n,k;
ll dp[maxn][maxn<<1][4];
void init ()
{
memset (dp,0,sizeof(dp));
dp[1][1][0]=1;
dp[1][2][1]=1;
dp[1][2][2]=1;
dp[1][1][3]=1;
}
int main()
{
scanf("%d%d",&n,&k);
init();
for (int i=2;i<=n;i++)
{
for (int j=1;j<=k;j++)
{
for (int m=0;m<4;m++)
{
if(m==0)
{
dp[i][j][0]=(dp[i-1][j][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j-1][3])%mod;
}
else if(m==1)
{
dp[i][j][1]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j-2][2]+dp[i-1][j-1][3])%mod;
}
else if(m==2)
{
dp[i][j][2]=(dp[i-1][j-1][0]+dp[i-1][j-2][1]+dp[i-1][j][2]+dp[i-1][j-1][3])%mod;
}
else
{
dp[i][j][3]=(dp[i-1][j-1][0]+dp[i-1][j][1]+dp[i-1][j][2]+dp[i-1][j][3])%mod;
}
}
}
}
printf("%lld\n",(dp[n][k][0]+dp[n][k][1]+dp[n][k][2]+dp[n][k][3])%mod);
return 0;
}