Educational Codeforces Round 49 (Rated for Div. 2)-E. Inverse Coloring-基础dp

Educational Codeforces Round 49 (Rated for Div. 2)-E. Inverse Coloring-基础dp

Description

You are given a square board, consisting of n rows and n

columns. Each tile in it should be colored either white or black.

Let's call some coloring beautiful if each pair of adjacent rows are either the same or
different in every position. The same condition should be held for the columns as well.

Let's call some coloring suitable if it is beautiful and there is no rectangle of the
single color, consisting of at least k tiles.

Your task is to count the number of suitable colorings of the board of the given size.

Since the answer can be very large, print it modulo 998244353.

Input

A single line contains two integers n and k (1≤n≤500, 1≤k≤n^2) — the number of rows and 
columns of the board and the maximum number of tiles inside the rectangle of the single
color, respectively.

Output

Print a single integer — the number of suitable colorings of the board of the given size
modulo 998244353.

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Examples

Input

1 1

Output

0

Input

2 3

Output

6

Input

49 1808

Output

359087121

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Problem Description

给你n*n的矩阵格子，矩阵能只能填入黑色或白色，并且任意两行或两列的颜色要么相同，要么相反。求使得矩阵
内连续的相同颜色形成的矩形的面积小于k的填法次数。

Solution

基础dp
矩形面积就是长*宽，也就是连续黑色或白色方块的个数。
所以我们通过dp算出所有可能情况，枚举相乘，保证面积小于k，求和相加即可得到答案。

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Code

/*
 * @Author: Simon 
 * @Date: 2018-08-22 19:42:38 
 * @Last Modified by: Simon
 * @Last Modified time: 2018-08-22 21:31:02
 */
#include<bits/stdc++.h>
using namespace std;
typedef int Int;
#define int long long
#define INF 0x3f3f3f3f
#define maxn 505
#define mod 998244353
int dp[maxn][maxn];//前i个长度，最长连续不超过j的方案数
int ans[maxn];
Int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n,k;
    cin>>n>>k;
    for(int i=1;i<=n;i++) dp[0][i]=1;
    for(int i=1;i<=n;i++)//枚举所有情况
    {
        for(int j=1;j<=n;j++)
        {
            for(int k=1;k<=min(i,j);k++)
            {
                dp[i][j]=(dp[i][j]+dp[i-k][j])%mod;
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        ans[i]=(dp[n][i]-dp[n][i-1]+mod)%mod;//前n个长度，连续个数为i的方案数。
    }
    int cnt=0;
    for(int i=1;i<=n;i++)//枚举长和宽相乘
    {
        for(int j=1;j<=n;j++)
        {
            if(i*j>=k) continue;//保证面积大小符合条件
            cnt=(cnt+ans[i]*ans[j]%mod)%mod;//求和最终方案数
        }
    }
    cout<<(cnt*2)%mod<<endl;//乘2是因为，要加上整张图颜色全部取反的方案数
    cin.get(),cin.get();
    return 0;
}