【线性代数】

【线性代数】

【矩阵封装】

矩阵加法/乘法

//矩阵加乘封装：1-based
//如不用取模：mod请用超大质数100000000000000003
const int MAX_SIZE=200; // 最大矩阵大小
struct Matrix{
    int rows,cols;
    i64 M[MAX_SIZE+5][MAX_SIZE+5];

    Matrix(int r=0,int c=0):rows(r),cols(c){
        clear();
    }

    void clear(){
        memset(M,0,sizeof(M));
    }
    
    //重置为单位矩阵
    void reset(){
        clear();
        for(int i=1;i<=min(rows,cols);i++){
            M[i][i]=1;
        }
    }
    
    //矩阵乘法: nxm mxp -> nxp
    Matrix friend operator*(const Matrix &A,const Matrix &B){
        if (A.cols!=B.rows){
            return Matrix(0,0); //不合法返回空矩阵
        }
        
        Matrix Ans(A.rows,B.cols);
        Ans.clear();
        
        for(int i=1;i<=A.rows;i++){
            for(int j=1;j<=B.cols;j++){
                for(int k=1;k<=A.cols;k++){
                    Ans.M[i][j]=(Ans.M[i][j]+A.M[i][k]*B.M[k][j])%mod;
                }
            }
        }
        return Ans;
    }
    
    //矩阵加法
    Matrix friend operator+(const Matrix &A, const Matrix &B){
        if(A.rows!=B.rows || A.cols!=B.cols){
            return Matrix(0, 0); //不合法返回空矩阵
        }
        
        Matrix Ans(A.rows,A.cols);
        Ans.clear();
        
        for(int i=1;i<=A.rows;i++){
            for(int j=1;j<=A.cols;j++){
                Ans.M[i][j]=(A.M[i][j]+B.M[i][j])%mod;
            }
        }
        return Ans;
    }
};

示例

int n,m,p;
void solve(){
    cin>>n>>m;
    Matrix s1(n,m);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            cin>>s1.M[i][j];
        }
    }
    cin>>p;
    Matrix s2(m,p);
    for(int i=1;i<=m;i++){
        for(int j=1;j<=p;j++){
            cin>>s2.M[i][j];
        }
    }
    Matrix ans=s1*s2;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=p;j++){
            cout<<ans.M[i][j]<<" ";
        }
        cout<<endl;
    }
}

矩阵求逆

//矩阵求逆:仅适用于方阵
bool ok=1;
Matrix getinv(Matrix a){
    // 检查是否为方阵
    if(a.rows!=a.cols){
        puts("-1");
        ok=0;
        return Matrix(0,0);
    }
    
    int n=a.rows;
    int m=2*n;
    
    for(int i=1;i<=n;i++){
        a.M[i][i+n]=1;
    }
    
    for(int i=1;i<=n;i++){
        int pos=i;
        for(int j=i+1;j<=n;j++){
            if(abs(a.M[j][i])>abs(a.M[pos][i])){
                pos=j;
            }
        }
        
        if(i!=pos){
            swap(a.M[i],a.M[pos]);
        }
        
        if(!a.M[i][i]){
            puts("-1");
            ok=0;
            return Matrix(0,0);
        }
        
        i64 inv=qmi(a.M[i][i],mod-2LL,mod);
        
        for(int j=1;j<=n;j++){
            if(j!=i){
                i64 mul=a.M[j][i]*inv%mod;
                for(int k=i;k<=m;k++){
                    a.M[j][k]=((a.M[j][k]-a.M[i][k]*mul)%mod+mod)%mod;
                }
            }
        }
        
        for(int j=1;j<=m;j++){
            a.M[i][j]=a.M[i][j]*inv%mod;
        }
    }
    
    Matrix res(n,n);
    res.clear();
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            res.M[i][j]=a.M[i][n+j];
        }
    }
    
    return res;
}

矩阵快速幂

不需要使用Matrix封装

const int N = 110, mod = 1e9 + 7;
i64 n, k, a[N][N], b[N][N], t[N][N];
void matrixQp(i64 y){
    while (y){
        if (y & 1){
            memset(t, 0, sizeof t);
            for (int i = 1; i <= n; i ++ )
                for (int j = 1; j <= n; j ++ )
                    for (int k = 1; k <= n; k ++ )
                        t[i][j] = ( t[i][j] + (a[i][k] * b[k][j]) % mod ) % mod;
            memcpy(b, t, sizeof t);
        }
        y >>= 1;
        memset(t, 0, sizeof t);
        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= n; j ++ )
                for (int k = 1; k <= n; k ++ )
                    t[i][j] = ( t[i][j] + (a[i][k] * a[k][j]) % mod ) % mod;
        memcpy(a, t, sizeof t);
    }
}
int main(){
    cin >> n >> k;
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ ){
            cin >> b[i][j];
            a[i][j] = b[i][j];
        }
    matrixQp(k - 1);//注意这里是k-1
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
            cout << b[i][j] << " \n"[j == n];
    return 0;
}