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由于偷懒不想用泛型，所以直接用了整型来写了一份
①首先你得有一个矩阵的class matrix
②matrix为了方便用下标进行运算，
matrix的结构如图：(我知道我的字丑。。。)
matrix.h代码如下：(个人并不喜欢把代码全写在一块，对于阅读者是相当巨大的负担，其实自己受不了（逃）)
我要评论 1 #pragma once
2 #include<vector>
3 using namespace std;
4 class matrix
5 {
6 public:
7 vector<vector<int>> nums;
8 int x0, y0;
9
10 int size;
11 matrix();
12 ~matrix();
13 matrix(int size, int x0, int y0);
14 matrix(matrix input, int x0, int y0, int size);
15 matrix(vector<vector<int>>input);
16 void display();
17 static void matrixmultiinit(matrix &a, matrix &b);
18 static matrix matrixmulti(matrix a, matrix b);
19 static matrix matrixadd(matrix a, matrix b);
20 static matrix matrixsub(matrix a, matrix b);
21 };
matrix.cpp对类的实现
一，构造，析构函数
1 matrix::matrix()
2 {
3 }
4
5
6 matrix::~matrix()
7 {
8 }
9
10 matrix::matrix(int size, int x0, int y0)
11 {
12 vector<vector<int>> temp(size, *new vector<int>(size));
13 for (int i = 0; i < size; i++)
14 for (int j = 0; j < size; j++)
15 temp[i][j] = 0;
16 this->nums = temp;
17 this->x0 = x0;
18 this->y0 = y0;
19 this->size = size;
20
21 }
22
23 matrix::matrix(matrix input, int x0, int y0, int size)
24 {
25 this->nums = input.nums;
26 this->x0 = x0;
27 this->y0 = y0;
28 this->size = size;
29 }
30
31 matrix::matrix(vector<vector<int>> input)
32 {
33 this->nums = input;
34 this->x0 = 0;
35 this->y0 = 0;
36 this->size = input.size();
37 }
二，a+b，a-b实现
1 matrix matrix::matrixadd(matrix a, matrix b)
2 {
3 matrix result(a.size, 0, 0);
4 for (int i = 0; i < result.nums.size(); i++)
5 for (int j = 0; j < result.nums.size(); j++)
6 result.nums[i][j] = a.nums[a.x0 + i][a.y0 + j] + b.nums[b.x0 + i][b.y0 + j];
7 return result;
8 }
9
10 matrix matrix::matrixsub(matrix a, matrix b)
11 {
12
13 matrix result(a.size, 0, 0);
14 for (int i = 0; i < result.nums.size(); i++)
15 for (int j = 0; j < result.nums.size(); j++)
16 result.nums[i][j] = a.nums[a.x0 + i][a.y0 + j] - b.nums[b.x0 + i][b.y0 + j];
17 return result;
18 }
三，a*b的实现
1 matrix matrix::matrixmulti(matrix a, matrix b)
2 {
3 int n = a.size;
4 int halfsize =n / 2;
5 matrix result(n, 0, 0);
6 if (n == 1)
7 result.nums[0][0] = a.nums[a.x0][a.y0] * b.nums[b.x0][b.y0];
8 else
9 {
10 matrix temps[10];
11 for (int i = 0; i < 10; i++)
12 {
13 temps[i] = *new matrix(halfsize, 0, 0);
14 }
15
16 //00-- a.x0,a.y0,halfsize
17 //01-- a.x0,a.y0+halfsize,halfsize
18 //10-- a.x0+halfsize,a.y0,halfsize
19 //11-- a.x0+halfsize,a.y0+halfsize,halfsize
20
21 //00-- b.x0,b.y0,halfsize
22 //01-- b.x0,b.y0+halfsize,halfsize
23 //10-- b.x0+halfsize,b.y0,halfsize
24 //11-- b.x0+halfsize,b.y0+halfsize,halfsize
25 temps[0] = temps[0].matrixsub(*new matrix(b, b.x0, b.y0 + halfsize, halfsize), *new matrix(b, b.x0 + halfsize, b.y0 + halfsize, halfsize));//01-11 b
26 temps[1] = temps[1].matrixadd(*new matrix(a, a.x0, a.y0, halfsize), *new matrix(a, a.x0, a.y0 + halfsize, halfsize));//00+01 a
27 temps[2] = temps[2].matrixadd(*new matrix(a, a.x0 + halfsize, a.y0, halfsize), *new matrix(a, a.x0 + halfsize, a.y0 + halfsize, halfsize));//10+11 a
28 temps[3] = temps[3].matrixsub(*new matrix(b, b.x0 + halfsize, b.y0, halfsize), *new matrix(b, b.x0, b.y0, halfsize));//10-00 b
29 temps[4] = temps[4].matrixadd(*new matrix(a, a.x0, a.y0, halfsize), *new matrix(a, a.x0 + halfsize, a.y0 + halfsize, halfsize));//00+11 a
30 temps[5] = temps[5].matrixadd(*new matrix(b, b.x0, b.y0, halfsize), *new matrix(b, b.x0 + halfsize, b.y0 + halfsize, halfsize));//00+11 b
31 temps[6] = temps[6].matrixsub(*new matrix(a, a.x0, a.y0 + halfsize, halfsize), *new matrix(a, a.x0 + halfsize, a.y0 + halfsize, halfsize));//01-11 a
32 temps[7] = temps[7].matrixadd(*new matrix(b, b.x0 + halfsize, b.y0, halfsize), *new matrix(b, b.x0 + halfsize, b.y0 + halfsize, halfsize));//10+11 b
33 temps[8] = temps[8].matrixsub(*new matrix(a, a.x0, a.y0, halfsize), *new matrix(a, a.x0 + halfsize, a.y0, halfsize));//00-10 a
34 temps[9] = temps[9].matrixadd(*new matrix(b, b.x0, b.y0, halfsize), *new matrix(b, b.x0, b.y0 + halfsize, halfsize));//00+01 b
35
36
37 matrix tempp[7];
38 for (int i = 0; i < 7; i++)
39 {
40 tempp[i] = *new matrix(n / 2, 0, 0);
41 }
42 tempp[0] = tempp[0].matrixmulti(*new matrix(a, a.x0, a.y0, halfsize), *new matrix(temps[0], 0, 0, halfsize));
43 tempp[1] = tempp[1].matrixmulti(*new matrix(temps[1], 0, 0,halfsize), *new matrix(b, b.x0 + halfsize, b.y0 + halfsize, halfsize));
44 tempp[2] = tempp[2].matrixmulti(*new matrix(temps[2], 0, 0, halfsize), *new matrix(b, b.x0, b.y0, halfsize));
45 tempp[3] = tempp[3].matrixmulti(*new matrix(a, a.x0 + halfsize, a.y0 + halfsize, halfsize), *new matrix(temps[3], 0, 0, halfsize));
46 tempp[4] = tempp[4].matrixmulti(*new matrix(temps[4], 0, 0, halfsize), *new matrix(temps[5], 0, 0, halfsize));
47 tempp[5] = tempp[5].matrixmulti(*new matrix(temps[6], 0, 0, halfsize), *new matrix(temps[7], 0, 0, halfsize));
48 tempp[6] = tempp[6].matrixmulti(*new matrix(temps[8], 0, 0, halfsize), *new matrix(temps[9], 0, 0, halfsize));
49
50
51
52 matrix result00 = result00.matrixadd(tempp[4], tempp[3]);
53 result00 = result00.matrixsub(result00, tempp[1]);
54 result00 = result00.matrixadd(result00, tempp[5]);
55 matrix result01 = result01.matrixadd(tempp[0], tempp[1]);
56 matrix result10 = result10.matrixadd(tempp[2], tempp[3]);
57 matrix result11 = result11.matrixadd(tempp[4], tempp[0]);
58 result11 = result11.matrixsub(result11, tempp[2]);
59 result11 = result11.matrixsub(result11, tempp[6]);
60
61 if (n == 3) {
62 for(int i=0;i<n/2+1;i++)
63 for (int j = 0; j < n / 2 + 1; j++) {
64
65 result.nums[i][j]= result00.nums[i][j];
66 result.nums[i][j+n/2+1] = result01.nums[i][j];
67 result.nums[i+n/2+1][j] = result10.nums[i][j];
68 result.nums[i+n/2+1][j+n/2+1] = result11.nums[i][j];
69 }
70 }
71
72 for(int i=0;i<n/2;i++)
73 for (int j = 0; j < n / 2; j++) {
74
75 result.nums[i][j]= result00.nums[i][j];
76 result.nums[i][j+n/2] = result01.nums[i][j];
77 result.nums[i+n/2][j] = result10.nums[i][j];
78 result.nums[i+n/2][j+n/2] = result11.nums[i][j];
79 }
80
81 }
82 return result;
83 }
四，防止size%2！=0的处理函数（即矩阵的行列数为奇数）
1 void matrix::matrixmultiinit(matrix &a, matrix &b) {
2
3 if (a.nums.size() % 2 != 0)
4 {
5 for (int i = 0; i < a.nums.size(); i++)
6 a.nums[i].push_back(0);
7 for (int i = 0; i < b.nums.size(); i++)
8 b.nums[i].push_back(0);
9 a.nums.push_back(*new vector<int>(a.nums[0].size(), 0));
10 b.nums.push_back(*new vector<int>(b.nums[0].size(), 0));
11 a.size++;
12 b.size++;
13 }
14 }
五，输出函数（这个读者随意）
1 void matrix::display()
2 {
3 for (int i = 0; i < this->nums.size(); i++) {
4
5 cout << "||";
6 for (int j = 0; j < this->nums[i].size(); j++) {
7 cout << this->nums[i][j] << " ";
8 }
9 cout << "||" << endl;
10
11 }
12 }
六，测试函数
1 #include <iostream>
2 #include"matrix.h"
3 int main()
4 {
5 vector<vector<int>> input = {
6 {1,2,3},
7 {1,2,3},
8 {1,1,1},
9 };
10 matrix test0 = * new matrix(input);
11 matrix test1 = * new matrix(input);
12 matrix test2;
13 test2.matrixmultiinit(test0, test1);
14 test2= test2.matrixmulti(test0, test1);
15 test2.display();
16
17 }
本人比较愚笨，耗时一天半才完成，不知道是不是天气热的原因，人太燥了，沉不下心来思考bug。
a*b有一点需要注意的是分块的逻辑应该怎么表示，一开始我用了两个顶点来表示一个矩阵的分块，如图：
然后halfsize还得一个一个的算，然后自己敲错的几率还会加大，并且还不一定表示正确每个分块，然后就逼疯自己了。
感觉这两天被这一堆bug都弄自闭了。。。。
幸好还是撑过来了，这算是我啃算法导论的第一个坎吧。
幸好第二天在csdn里面看到了别人怎么分块的。看到了一个变量halfsize，于是才开始改自己matrix的构造。
虽然一开始不愿意，但是改完之后，竟然一次过了！woc！
给我了一个教训，以后能“少”一个变量尽量“少”一个变量。
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