2020ICPC济南站 A . Matrix Equation (高斯消元)

​​linkkk​​ 题意：

思路：
$A×C=B⊙C$
$\sum_{k=1}^{n}A_{i,k}*C_{k,j}=B_{i,j}*C_{i,j}$
对于$1<=j<=n$
$(A_{1,1}C_{1,j}+A_{1,2}C_{2,j}+……+A_{1,n}C_{n,j})\%2=B_{1,j}C_{1,j}$
$01$的加法和取余$2$相当于异或。
式子变成了
$(A_{1,1}C_{1,j} ⊕ A_{1,2}C_{2,j}⊕…⊕A_{1,n}C_{n,j})=B_{1,j}C_{1,j}$
相当于解一个异或线性方程组，对于$1<=j<=n$中，是相互独立的。最后$2^{总自由元个数}$就是答案。
如果$A_{1,1}=B_{1,j}$，则$a_{1,1}=0$，这样右侧的结果位都是$0$；
$A_{1,1}!=B_{1,j}$，则$a_{1,1}=1$

代码：

// Problem: Matrix Equation
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/10662/A
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
typedef pair<string,string>PSS;
#define I_int ll
inline ll read(){ll x = 0, f = 1;char ch = getchar();while(ch < '0' || ch > '9'){if(ch == '-')f = -1;ch = getchar();}while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0';ch = getchar();}return x * f;}
 
inline void write(ll x){if (x < 0) x = ~x + 1, putchar('-');if (x > 9) write(x / 10);putchar(x % 10 + '0');}
 
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)

ll ksm(ll a, ll b,ll mod){ll res = 1;while(b){if(b&1)res=res*a%mod;a=a*a%mod;b>>=1;}return res;}
 
const int maxn=2010;
const double pi = acos(-1);
ll n,m=31;
ll a[210][210],g[210];
int A[210][210],B[210][210];
const int mod=998244353;
int Gauss(int n,int m)
{
    int r, c;
    for(r = 0, c = 0; c < n; c++)
    {
        int t = -1;
        for(int i = r; i < m; i++)
        {
            if(a[i][c])
            {
                t = i;
                break;
            }
        }
        if(t==-1) continue;

        for(int i = c; i < n; i++)///交换
        {
            swap(a[t][i], a[r][i]);
        }

        for(int i = r + 1; i < m; i++)
        {
            if(a[i][c])
                for(int j = c; j <n; j++)
                {
                    a[i][j] =  a[i][j] ^ a[r][j];
                }
        }

        r++;
    }
    return n-r;
}
void solve()
{
    n=read;
    for(int i=0;i<n;i++) 
        for(int j=0;j<n;j++)
            A[i][j]=read;
    for(int i=0;i<n;i++) 
        for(int j=0;j<n;j++)
        B[i][j]=read;
    ll cnt=0;
    for(int j=0;j<n;j++){
        
        for(int i=0;i<n;i++){
            for(int k=0;k<n;k++){
                a[i][k]=A[i][k];
            }
            a[i][n]=0;
            if(A[i][i]==B[i][j]) a[i][i]=0;
            else a[i][i]=1;
        }
        ll tot=Gauss(n,n);
        if(tot<0){
            puts("0");
            return ;
        }
        else{
            cnt+=tot;
            
        }
    }
    cout<<ksm(2,cnt,mod)<<endl;
    
}

int main()
{
    int _;_=1;
    while(_--) solve();
    return 0;
}