CodeForces 1051D Bicolorings (DP)

最新推荐文章于 2020-06-19 10:56:05 发布

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#动态规划

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本文探讨了一个2xn矩阵的涂色问题，目标是计算有k个联通块的涂色方案数量。通过动态规划方法，记录每列的涂色状态，实现了对复杂度的有效控制。

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

Input

4 1

Output

Input

1 2

Output

给一个2*n的矩阵，每个格子可以涂成黑或白色，问有k个联通块的涂色方式有几种。

很容易想到dp，只需要记录最后一列是黑黑、黑白、白黑、还是白白即可。

用0,1,2,3分别表示，则dp[i][j][k]表示在2*i的矩阵中，有j个联通块且,最后一行涂成k的方式。

dp即可

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define ll long long
#define ull unsigend long long
const int M=1e3+5;
const int N=998244353;
ll dp[M][M*2][4],ans;
void init()
{
    ans=0;
    memset(dp,0,sizeof(dp));
}
void solve(int n,int m)
{
    dp[1][1][0]=1;
    dp[1][1][3]=1;
    dp[1][2][1]=1;
    dp[1][2][2]=1;
    for(int i=2; i<=n; i++)
    {
        for(int j=1; j<=i*2; j++)
        {
            dp[i][j][0]+=dp[i-1][j][0];
            dp[i][j][0]+=dp[i-1][j][1];
            dp[i][j][0]+=dp[i-1][j][2];
            if(j>1) dp[i][j][0]+=dp[i-1][j-1][3];

            dp[i][j][1]+=dp[i-1][j][1];
            if(j>1)   dp[i][j][1]+=dp[i-1][j-1][0];
            if(j>2) dp[i][j][1]+=dp[i-1][j-2][2];
            if(j>1) dp[i][j][1]+=dp[i-1][j-1][3];

            dp[i][j][2]+=dp[i-1][j][2];
            if(j>1)   dp[i][j][1]+=dp[i-1][j-1][0];
            if(j>2) dp[i][j][2]+=dp[i-1][j-2][1];
            if(j>1) dp[i][j][2]+=dp[i-1][j-1][3];

            dp[i][j][3]+=dp[i-1][j][3];
            dp[i][j][3]+=dp[i-1][j][1];
            dp[i][j][3]+=dp[i-1][j][2];
            if(j>1) dp[i][j][3]+=dp[i-1][j-1][0];
            for(int k=0; k<3; k++) dp[i][j][k]%=N;
        }
    }
    for(int i=0; i<4; i++) ans+=dp[n][m][i];
    ans%=N;
    printf("%lld\n",ans);
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        solve(n,m);
    }
    return 0;
}