CodeForces

odeforces 1051 D. Bicolorings (DP)

 

D. Bicolorings

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a grid, consisting of $\mathrm{22\; rows\; and\; nn\; columns.\; Each\; cell\; of\; this\; grid\; should\; be\; colored\; either\; black\; or\; white.}$

Two cells are considered neighbours if they have a common border and share the same color. Two cells $\mathrm{AA\; and\; BB\; belong\; to\; the\; same\; component\; if\; they\; are\; neighbours,\; or\; if\; there\; is\; a\; neighbour\; of\; AA\; that\; belongs\; to\; the\; same\; component\; with\; BB.}$

Let's call some bicoloring beautiful if it has exactly $\mathrm{kk\; components.}$

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 $\mathrm{nn\; and\; kk\; (1\le n\le 10001\le n\le 1000,\; 1\le k\le 2n1\le k\le 2n)\; —\; the\; number\; of\; columns\; in\; a\; grid\; and\; the\; number\; of\; components\; required.}$

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

Note

One of possible bicolorings in sample $\mathrm{11:}$

题意：就是求2行n列 k个联通块的方案数这个题注意前面列的染色情况，就可以知道当前列染色后的方案数，i表示列，k表示联通块，后面一维表示当前位置的染色情况，这个情况总共四种，这样转移方程就很好想，只和前一列的联通块和染色情况有关

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,k;

const int mod = 998244353;
ll dp[1005][2005][4];
int main()
{
    scanf("%d%d",&n,&k);
    dp[1][1][0] = 1;
    dp[1][2][1] = 1;
    dp[1][2][2] = 1;
    dp[1][1][3] = 1;
    for(int i=2;i<=n;i++)
    {
        for(int j=1;j<=(i<<1);j++)
        {
            dp[i][j][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;
            dp[i][j][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;
            dp[i][j][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;
            dp[i][j][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;
        }
    }
    printf("%lld
",(dp[n][k][0]+dp[n][k][1]+dp[n][k][2]+dp[n][k][3])%mod);
}