codeforce 1051D Bicolorings(DP)

最新推荐文章于 2022-05-26 19:50:53 发布

Amovement

最新推荐文章于 2022-05-26 19:50:53 发布

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本文链接：https://blog.youkuaiyun.com/Amovement/article/details/82833981

本文探讨了在一个由两行组成的黑白网格中，寻找特定数量连通块的方案数问题。通过动态规划方法，详细阐述了解决方案的构建过程，包括状态转移方程和模运算的实现。

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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 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

Copy

3 4

output

Copy

input

Copy

4 1

output

Copy

input

Copy

1 2

output

Copy

Note

One of possible bicolorings in sample 11:

一、原题地址

原题地址

二、大致题意

一个两行的黑白格，有n列。询问n列存在k块连通块的方案数。

三、大致思路

设dp[ i ][ j ][ k ]

i 表示当前转移到第 i 列，j 表示当前存在 j 个连通块，k 表示当前更新的这列的方案。

k=0 表示摆两个白块，k=1表示上面一个白下面黑，k=2表示上面黑下面白，k=3表示两个黑块。

四、代码

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<map>
#include<set>
#include<string>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<functional>
using namespace std;
#define inf 0x3f3f3f3f
typedef long long LL;
int gcd(int a, int b) { return a == 0 ? b : gcd(b % a, a); }


LL n, m;
LL dp[1005][2005][5];
const LL MOD = 998244353;
int main()
{
	scanf("%lld %lld", &n, &m);
	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;

	for (int i = 2; i <= n; i++)
	{
		for (int j = 1; j <= (i << 1); j++)
		{
			dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1] + dp[i - 1][j][2] + dp[i - 1][max(j - 1, 0)][3];
			dp[i][j][0] %= MOD;

			dp[i][j][1] = dp[i - 1][max(j - 1, 0)][0] + dp[i - 1][j][1] + dp[i - 1][max(j - 2, 0)][2] + dp[i - 1][max(j - 1, 0)][3];
			dp[i][j][1] %= MOD;

			dp[i][j][2] = dp[i - 1][max(j - 1, 0)][0] + dp[i - 1][max(j - 2, 0)][1] + dp[i - 1][j][2] + dp[i - 1][max(j - 1, 0)][3];
			dp[i][j][2] %= MOD;

			dp[i][j][3] = dp[i - 1][max(j - 1, 0)][0] + dp[i - 1][j][1] + dp[i - 1][j][2] + dp[i - 1][j][3];
			dp[i][j][3] %= MOD;
		}
	}

	LL ans = 0;
	for (int i = 0; i < 4; i++)
	{
		ans += dp[n][m][i];
		ans %= MOD;
	}
	cout << ans << endl;
	getchar();
	getchar();

}