矩阵相关操作和矩阵快速幂

矩阵相关操作和矩阵快速幂

• 矩阵基本运算以及快速幂模板
• POJ3070-Fibonacci题解
• Hdu1757-A Simple Math Problem
• Codefroces-185A- Plant

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矩阵基本运算以及快速幂模板

先看一下矩阵的乘法规则:

直接给出一个模板题，直接包含了基本的乘法和求幂，求幂的详细解释，可以看这篇乘法快速幂。

题目链接

题目

注意:

• 矩阵的乘法必须满足第一个矩阵的列 = 第二个矩阵的行；
• 矩阵的求幂必须满足矩阵是一个方阵；

import java.io.BufferedInputStream;
import java.util.Scanner;

public class Main {

    private static class Matrix{
        public int row; 
        public int col; 
        public int[][] m;

        public Matrix(int row, int col) {
            this.row = row;
            this.col = col;
            m = new int[row][col];
        }
    }

    // 两个矩阵相加  -->  a,b必须为 同型矩阵
    public static Matrix add(Matrix a,Matrix b){
        Matrix c = new Matrix(a.row,a.col);
        for(int i = 0; i < a.row; i++){
            for(int j = 0; j < a.col; j++){
                c.m[i][j] = a.m[i][j] + b.m[i][j]; // sub 减法换成-
            }
        }
        return c;
    }

    // 必须满足a.col = b.row  才能相乘
    public static Matrix mul(Matrix a,Matrix b){
        Matrix c = new Matrix(a.row,b.col); //注意这里

        for(int i = 0; i < a.row; i++){
            for(int j = 0; j < b.col; j++){
                for(int k = 0; k < a.col; k++)
                    c.m[i][j] = c.m[i][j] + a.m[i][k]*b.m[k][j];
            }
        }
        return c;
    }

    // 必须为 方阵才能 求幂
    public static Matrix pow(Matrix a,int k){ // 矩阵 a 的 k次幂
        Matrix res = new Matrix(a.row,a.col);  //求幂必须满足 a.row = a.col(也就是方阵)
        for(int i = 0; i < a.row; i++)
            res.m[i][i] = 1;

        // 真正的快速幂
        while(k > 0){
            if( (k&1) != 0)
                res = mul(res,a);
            a = mul(a,a);
            k >>= 1;
        }
        return res;
    }

    public static void main(String[] args) {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));

        int n = cin.nextInt();
        int k = cin.nextInt();

        Matrix a = new Matrix(n,n);
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                a.m[i][j] = cin.nextInt();
            }
        }

        Matrix res = pow(a,k);
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if( j == n- 1){
                    System.out.println(res.m[i][j]);
                }else {
                    System.out.print(res.m[i][j] + " ");
                }
            }
        }
    }
}

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POJ3070-Fibonacci题解

题目链接

题目

解析

关键在于推导出递推式，也就是左边是一个A矩阵，B一般是一个列向量；

类似的规律:

import java.io.BufferedInputStream;
import java.util.Scanner;

public class Main {

    private static class Matrix{
        public int row;
        public int col;
        public int[][] m;

        public Matrix(int row, int col) {
            this.row = row;
            this.col = col;
            m = new int[row][col];
        }
    }

    private static final int mod = 10000;


    public static Matrix mul(Matrix a,Matrix b){
        Matrix c = new Matrix(a.row,b.col); //注意这里

        for(int i = 0; i < a.row; i++){
            for(int j = 0; j < b.col; j++){
                for(int k = 0; k < a.col; k++)
                    c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j]) % mod;
            }
        }
        return c;
    }

    public static Matrix pow(Matrix a,int k){
        Matrix res = new Matrix(a.row,a.col); // 方阵
        for(int i = 0; i < a.row; i++)
            res.m[i][i] = 1;
        while(k > 0){
            if( (k&1) != 0)
                res = mul(res,a);
            a = mul(a,a);
            k >>= 1;
        }
        return res;
    }

    public static void main(String[] args) {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));

        while(cin.hasNext()){
            int n = cin.nextInt();
            if( n == -1)break;
            if(n == 0){
                System.out.println(0);
                continue;
            }

            Matrix a = new Matrix(2,2);
            a.m[0][0] = a.m[0][1] = a.m[1][0] = 1;
            a.m[1][1] = 0;

            Matrix res = pow(a,n-1);

            System.out.println(res.m[0][0] % mod);
        }
    }

}

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Hdu1757-A Simple Math Problem

题目链接

题目

解析

继续递推:

import java.io.BufferedInputStream;
import java.util.Scanner;

public class Main {

    private static class Matrix{
        public int row;
        public int col;
        public int[][] m;

        public Matrix(int row, int col) {
            this.row = row;
            this.col = col;
            m = new int[row][col];
        }
    }

    public static Matrix mul(Matrix a,Matrix b,int mod){
        Matrix c = new Matrix(a.row,b.col); //注意这里

        for(int i = 0; i < a.row; i++){
            for(int j = 0; j < b.col; j++){
                for(int k = 0; k < a.col; k++)
                    c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j]) % mod;
            }
        }
        return c;
    }

    public static Matrix pow(Matrix a,int k,int mod){
        Matrix res = new Matrix(a.row,a.col); // 方阵
        for(int i = 0; i < a.row; i++)
            res.m[i][i] = 1;
        while(k > 0){
            if( (k&1) != 0)
                res = mul(res,a,mod);
            a = mul(a,a,mod);
            k >>= 1;
        }
        return res;
    }


    public static void main(String[] args) {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));

        while(cin.hasNext()){
            int k = cin.nextInt();
            int mod = cin.nextInt();

            if(k < 10){
                System.out.println(k);
                continue;
            }

            Matrix a = new Matrix(10,10);

            // init
            for(int i = 0; i < 10; i++)
                a.m[0][i] = cin.nextInt();
            for(int i = 1; i < 10; i++)
                a.m[i][i-1] = 1;

            // computer  matrix ^ (k-9)
            Matrix res = pow(a,k-9,mod);

            int sum = 0;
            for(int i = 0; i < 10; i++)
                sum += (res.m[0][i] * (9 - i)) % mod;

            System.out.println(sum % mod); // also should mod
        }
    }
}

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Codefroces-185A- Plant

题目链接

题目

解析

import java.io.BufferedInputStream;
import java.util.Scanner;

public class Main {

    private static class Matrix{
        public int row;
        public int col;
        public long[][] m;

        public Matrix(int row, int col) {
            this.row = row;
            this.col = col;
            m = new long[row][col];
        }
    }

    private static final int mod = 1000000007;

    public static Matrix mul(Matrix a,Matrix b){
        Matrix c = new Matrix(a.row,b.col); //注意这里

        for(int i = 0; i < a.row; i++){
            for(int j = 0; j < b.col; j++){
                for(int k = 0; k < a.col; k++)
                    c.m[i][j] = (c.m[i][j] + a.m[i][k]*b.m[k][j]) % mod;
            }
        }
        return c;
    }

    public static Matrix pow(Matrix a,long k){
        Matrix res = new Matrix(a.row,a.col); // 方阵
        for(int i = 0; i < a.row; i++)
            res.m[i][i] = 1;
        while(k > 0){
            if( (k&1) != 0)
                res = mul(res,a);
            a = mul(a,a);
            k >>= 1;
        }
        return res;
    }

    public static void main(String[] args) {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));

        long n = cin.nextLong();

        Matrix a = new Matrix(2,2);
        a.m[0][0] = a.m[1][1] = 3;
        a.m[0][1] = a.m[1][0] = 1;

        Matrix res = pow(a,n);

        System.out.println(res.m[0][0] % mod);
    }
}