矩阵快速幂计算hdu1575

矩阵快速幂计算和整数快速幂计算相同。在计算A^7时,7的二进制为111，从而A^7=A^(1+2+4)=A*A^2*A^4.而A^2可以由A*A得到，A^4可以由A^2*A^2得到。计算两个n阶方阵的乘积复杂度为O(n^3)。k的二进制大约有logk位,总的复杂度为O(n^3*logk).

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<queue>
#include<iomanip>
#include<vector>
#include<string.h>
#include<algorithm>
using namespace std;
const int MOD = 9973;
class Matrix{
    int n;
    int M[12][12];
    Matrix Matrix::operator*(Matrix& X);  //矩阵乘法
public:
    void input(int i, int j, int m){ M[i][j] = m; }
    Matrix(int n);  //初始化
    int trSum();   //计算矩阵迹
    Matrix quickPower(int k);//矩阵快速幂
};
Matrix::Matrix(int n){
    this->n = n;
    int i, j;
    for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++)
        M[i][j] = (i == j ? 1 : 0);
}

Matrix Matrix::operator*(Matrix& X){
    Matrix res(n);
    int i, j, k, ans;
    for (i = 1; i <= n; i++)
    for (j = 1; j <= n; j++){
        ans = 0;
        for (k = 1; k <= n; k++)
            ans += M[i][k] * X.M[k][j];
        res.M[i][j] = ans%MOD;
    }
    res.n = n;
    return res;
}

int Matrix::trSum(){
    int res = 0, i;
    for (i = 1; i <= n; i++)
        res += M[i][i];
    return res%MOD;
}

Matrix Matrix::quickPower(int k){
    Matrix Y(n);
    Matrix X = *this;
    while (k){
        if (k & 1)
            Y = Y*X;
        k >>= 1;
        X = X*X;
    }
    return Y;
}

int main(){
    int i, j, T, k, n,ans;
    scanf("%d", &T);
    while (T--){
        scanf("%d%d", &n,&k);
        Matrix X(n);
        for (i = 1; i <= n;i++)
        for (j = 1; j <= n; j++){
            scanf("%d", &ans);
            X.input(i, j, ans);
        }
        X = X.quickPower(k);
        printf("%ld\n",X.trSum());
    }
    return 0;
}