cf1027E Inverse Coloring dp

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本文探讨了在一个n*n的棋盘上，如何计算满足特定条件的染色方案数量。这些条件包括每对相邻行和列要么完全相同要么在每个位置都不同，且不存在至少包含k个相同颜色单元格的单色矩形。由于答案可能非常大，最终结果需对998244353取模。

Description

You are given a square board, consisting of n n n rows and n n 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 k 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 998244353 998244353 .

Solution

首先要知道，如果存在一行和一列01组成的数列x[]y[]，令a[i,j]=x[i]^y[i]即可得到满足条件1和2 的矩形
对于第三个条件考虑dp，设f[i,j,k]表示满足长度恰好为i，最大连续为j，末尾连续为k的01数组数量，我们统计一下乘起来就可以了

Code

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define rep(i,st,ed) for (register int i=st,_=ed;i<=_;++i)
#define fill(x,t) memset(x,t,sizeof(x))

typedef long long LL;
const int MOD=998244353;
const int N=505;

LL f[2][N][N],g[N][N];

void upd(LL &x,LL v) {
	x=x+v; x%=MOD;
}

int main(void) {
	freopen("data.in","r",stdin);
	int n,m; scanf("%d%d",&n,&m);
	f[0][0][0]=1;
	rep(i,0,n) {
		int x=i&1;
		fill(f[!x],0);
		rep(j,0,i) {
			rep(k,0,std:: min(i,j)) {
				upd(f[!x][std:: max(j,k+1)][k+1],f[x][j][k]);
				upd(f[!x][std:: max(j,1)][1],f[x][j][k]);
				upd(g[i][j],f[x][j][k]);
			}
		}
	}
	LL ans=0;
	rep(i,1,n) rep(j,1,n) {
		if (i*j<m) upd(ans,1LL*g[n][i]*g[n][j]%MOD);
		else break;
	}
	ans=499122177LL*ans%MOD;
	printf("%lld\n", ans);
	return 0;
}