2020ICPC济南 A Matrix Equation（高斯消元、自由元数量）

链接

题目描述

We call a matrix “01 Square” if and only if it’s a $N\times N$ matrix and its elements are all $0$ or $1$.

For two 01 Squares $X$,$Y$, we define two operators $X\times Y$ and $X\odot Y$. The value of them are also 01 Square matrices and calculated below(we use $Z$ to abbreviate $X\times Y$ and $D$ to abbreviate $X\odot Y$):

$Z_{i,j}=({\sum }_{k=1}^N{X}_{i,k}{Y}_{k,j})mod2$

$D_{i,j}=X_{i,j}{Y}_{i,j}$​

Now MianKing has two 01 Squares $A,B$, he wants to solve the matrix equation below:

$A\times C=B\odot C$

You need to help MainKing solve this problem by calculating how many 01 Squares $C$ satisfy this equation.

The answer may be very large, so you only need to output the answer module $998244353$.

输入描述:

The first line has one integer $N$

Then there are $N$ lines and each line has $N$ integers, the j-th integer of the i-th line denotes $A_{i,j}$

Then there are $N$ lines and each line has $N$ integers, the j-th integer of the i-th line denotes $B_{i,j}$

$1\le N\le 200$, $A_{i,j},B_{i,j}\in \{0,1\}$

输出描述:

Output the answer module $998244353$.

输入

2
0 1
1 1
1 0
0 1

输出

2

输入

3
1 0 0
0 1 0
0 0 1
1 1 1
1 1 1
1 1 1

输出

512

输入

4
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 1
1 0 1 1
0 1 1 1
1 0 0 1
1 1 1 0

输出

8

思路

设矩阵 $A$ 、$B$ 分别如下：

$$A=\left[\begin{array}{ccc}1& 0& 1\\ 1& 0& 0\\ 0& 1& 1\end{array}\right],B=\left[\begin{array}{ccc}1& 0& 1\\ 1& 0& 0\\ 0& 1& 1\end{array}\right]$$

设矩阵 $C$ 为
$$C=\left[\begin{array}{ccc}a& d& g\\ b& e& h\\ c& f& i\end{array}\right]$$

由题意有：$A\times C=B\odot C$

首先解 $C$ 第一列的三个变量： $a,b,c$。有如下关系：

$$\{\begin{array}{r}a+c\equiv 0\\ a+b\equiv 0\\ b+c\equiv 0\end{array}$$
相当于得到如下增广矩阵：
$$\left[\begin{array}{cccc}1& 0& 1& 0\\ 1& 1& 0& 0\\ 0& 1& 1& 0\end{array}\right]$$

用高斯消元法解得：
$$\left[\begin{array}{cccc}1& 0& 1& 0\\ 0& 1& 1& 0\\ 0& 0& 0& 0\end{array}\right]$$

这一列有一个自由元。用同样的方法，计算出 $n$ 列自由元数量之和 $cnt$，那么答案就是 $2^{cnt}$。

注意

1. 高斯消元过程中始终要进行模 $2$ 计算。
2. 如果使用 $\%$ 运算符，可能会被卡常，使用异或运算来优化。

#include
#define ll long long
using namespace std;
const int N=210,mod=998244353;
int x[N][N],y[N][N],a[N][N],ans,n;

ll qpow(ll a,ll b){
	ll ret=1;
	while(b){
		if(b&1LL) ret=ret*a%mod;
		a=a*a%mod; b>>=1LL;
	}
	return ret;
}

void gauss(){
	int ri=1,add=0;
	for(int i=1;i<=n;i++){
		if(add) ri++; add=1;
		int mx=ri;
		for(int j=ri+1;j<=n;j++)
			if(a[j][i]>a[mx][i]) mx=j;
		if(a[mx][i]==0){ ans++,add=0; continue; }
		for(int j=i;j<=n+1;j++) swap(a[ri][j],a[mx][j]);
		for(int k=1;k<=n;k++) if(ri!=k&&a[k][i]!=0)
			for(int j=n+1;j>=i;j--) a[k][j]^=a[ri][j];
	}
}

signed main(){
	cin>>n;
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) cin>>x[i][j];
	for(int i=1;i<=n;i++) for(int j=1;j<=n;j++) cin>>y[i][j];
	for(int i=1;i<=n;i++){
		memcpy(a,x,sizeof(x));
		for(int j=1;j<=n;j++)
			if(y[j][i]==1) a[j][j]^=1;
		gauss();
	}
	cout<<qpow(2,ans)<<"\n";
}