vc矩阵计算（转置，点乘，逆矩阵）

vc计算矩阵的转置，矩阵的点乘，矩阵的逆矩阵，参考网上的例子

矩阵点乘的例子：

矩阵逆矩阵计算例子：

#include "stdafx.h"
#include <math.h>
//#include<complex.h>
#include <iostream>  
#include <complex> 
#include <cstdlib> // 包含rand()和srand()  
#include <ctime>   // 包含time()  
using namespace std;
///计算 两个矩阵的点乘
void dotMatrix(float *a, float *b, float c[][50],int rowsa, int colsa,int colsb, int n) {///rowsa 前面矩阵的行数，colsa前面矩阵的列数,colsb后面矩阵的列数,n最大维数for (int i = 0; i < rowsa; i++){for (int j = 0; j < colsb; j++){int sum = 0;for (int k = 0; k < colsa; k++){sum += a[i*n+k] * b[k*n+j];}c[i][j] = sum;}}
}
// 函数定义：计算矩阵的转置  
void transposeMatrix(float *source, float *dest, int rows, int cols,int n) {///rows 原始矩阵的行数，cols原始矩阵的列数// 对于每一个元素，将其放到dest数组的相应位置（行变列，列变行） ,n最大维数for (int i = 0; i < rows; i++) {for (int j = 0; j < cols; j++) {float *p = (float *)&source[i*n + j];float *dp = (float *)&dest[j*n + i];**&dp = *p; // 注意这里的行列索引交换  }}
}
计算矩阵的逆矩阵
float** Matrix_inver(float** src, int row, int col,int nn)//原始矩阵的行与列,nn最大的维数
{//step 1//判断指针是否为空if (src == NULL)exit(-1);int i, j, k, n,principal;float** res, ** res2, tmp;//res为增广矩阵，res为输出的逆矩阵float Max;//判断矩阵维数//row = (double)_msize(src) / (double)sizeof(double*);//col = (double)_msize(*src) / (double)sizeof(double);if (row != col)exit(-1);//step 2res = (float**)malloc(sizeof(float*) * row);res2 = (float**)malloc(sizeof(float*) * row);n = 2 * row;for (i = 0; i < row; i++){res[i] = (float*)malloc(sizeof(float) * n);res2[i] = (float*)malloc(sizeof(float) * col);memset(res[i], 0, sizeof(res[0][0]) * n);//初始化memset(res2[i], 0, sizeof(res2[0][0]) * col);}//step 3//进行数据拷贝for (i = 0; i < row; i++){//此处源代码中的n已改为col，当然方阵的row和col相等，这里用col为了整体逻辑更加清晰memcpy(res[i], &src[i*nn], sizeof(res[0][0]) * col);}//将增广矩阵右侧变为单位阵for (i = 0; i < row; i++){for (j = col; j < n; j++){if (i == (j - row))res[i][j] = 1.0;}}for (j = 0; j < col; j++){//step 4//整理增广矩阵，选主元principal = j;Max = fabs(res[principal][j]); // 用绝对值比较// 默认第一行的数最大// 主元只选主对角线下方for (i = j; i < row; i++){if (fabs(res[i][j]) > Max){principal = i;Max = fabs(res[i][j]);}}if (j != principal){for (k = 0; k < n; k++){tmp = res[principal][k];res[principal][k] = res[j][k];res[j][k] = tmp;}}//step 5//将每列其他元素化0for (i = 0; i < row; i++){if (i == j || res[i][j] == 0)continue;float b = res[i][j] / res[j][j];for (k = 0; k < n; k++){res[i][k] += b * res[j][k] * (-1);}}//阶梯处化成1float a = 1.0 / res[j][j];for (i = 0; i < n; i++){res[j][i] *= a;}}//step 6//将逆矩阵部分拷贝到res2中for (i = 0; i < row; i++){memcpy(res2[i], res[i] + row, sizeof(res[0][0]) * row);}//必须释放res内存！for (i = 0; i < row; i++){free(res[i]);}free(res);return res2;
}int _tmain(int argc, _TCHAR* argv[])
{float q[50] = { 0 };///保存视电阻率值float r[10] = { 0 };float h[10] = { 0 };r[0] = 100;r[1] = 1000;r[2] = 500;r[3] = 3000;
//	r[4] = 400;h[0] = 500;h[1] = 1000;h[2] = 2000;
//	h[3] = 2000;for (int i = 0; i < 50; i++)///计算模型的正演值，为模拟退火计算做准备{//q[i]=calcR(r, h, 3, f[i]);q[i] = calcR(r, h, 4, f[i]);}// 假设我们有一个3x3的矩阵  float matrix[50][50] = {{ 1, 2,3 },{ 4, 5,6},{ 7, 8 ,9}};// 用于存储转置后的矩阵的数组  float transposed[50][50] = { 0 };// 调用函数计算转置  transposeMatrix((float *)&matrix[0], (float *)&transposed[0], 3, 2,50);//矩阵点乘计算float a[50][50] = { { 1, 2, 4 }, { 2, 0, 3 } };float b[50][50] = { { 1, 2 }, { 3, 2 }, { 0, 5 } };/  7,26/  2,19dotMatrix((float *)&a, (float *)&b, transposed, 2, 3, 2, 50);///下面计算点乘的结果printf("%f,%f,\n%f,%f\n", transposed[0][0], transposed[0][1], transposed[1][0], transposed[1][1]);下面计算矩阵aa的逆矩阵float aa[50][50] = {{1,0,0},{1,1,0},{1,1,1}};float **ret = Matrix_inver((float **)aa, 3, 3, 50);printf("下面计算矩阵aa的逆矩阵\n");printf("%f,%f,%f,\n%f,%f,%f,\n%f,%f,%f\n", ret[0][0], ret[0][1], ret[0][2], ret[1][0], ret[1][1], ret[1][2], ret[2][0], ret[2][1], ret[2][2]);下面计算矩阵aa与它的逆矩阵bb的点乘float bb[50][50] = { 0 };bb[0][0] = ret[0][0];bb[0][1] = ret[0][1];bb[0][2] = ret[0][2];bb[1][0] = ret[1][0];bb[1][1] = ret[1][1];bb[1][2] = ret[1][2];bb[2][0] = ret[2][0];bb[2][1] = ret[2][1];bb[2][2] = ret[2][2];dotMatrix((float *)&aa, (float *)&bb, transposed, 3, 3, 3, 50);///下面计算点乘的结果printf("下面计算矩阵aa与它的逆矩阵bb的点乘结果\n");printf("%f,%f,%f,\n%f,%f,%f,\n%f,%f,%f", transposed[0][0], transposed[0][1], transposed[0][2], transposed[1][0], transposed[1][1], transposed[1][2], transposed[2][0], transposed[2][1], transposed[2][2]);return 0;
}

最终显示结果