代码搜索：矩阵分析
找到约 10,000 项符合「矩阵分析」的源代码
代码结果 10,000
www.eeworm.com/read/311990/13620418
vbp 矩阵相乘.vbp
Type=Exe
Reference=*\G{00020430-0000-0000-C000-000000000046}#2.0#0#C:\WINDOWS\system32\stdole2.tlb#OLE Automation
Form=矩阵想乘.frm
Object={5E9E78A0-531B-11CF-91F6-C2863C385E30}#1.0#0; MSFLXGRD.OCX
St
www.eeworm.com/read/308771/13692889
txt 导纳矩阵.txt
(0,-17.3611) (0,0) (0,0) (0,17.3611) (0,0) (0,0) (0,0) (0,0) (0,0)
(0,0) (0,-16) (0,0) (0,0) (0,0) (0,0) (0,16) (0,0) (0,0)
(0,0) (0,0) (0,-17.0648) (0,0) (0,0) (0,0) (0,0) (0,0) (0,17.0648)
(0,
www.eeworm.com/read/308157/13707088
c 逆矩阵.c
#define N 5 /*[注]：修改6为你所要的矩阵阶数*/
#include "stdio.h"
#include "conio.h"
/*js()函数用于计算行列式,通过递归算法实现*/
int js(s,n)
int s[][N],n;
{int z,j,k,r,total=0;
int b[N][N];/*b[N][N]用于存放，在矩阵s[
www.eeworm.com/read/308157/13707120
c 乘法矩阵.c
#include
void main()
{
int i,j;
int big[8][8];
for (i = 0;i < 8;i++) /*循环嵌套*/
for (j = 0;j < 8;j++)
big[i][j] = i * j; /* 乘法表 */
big[2][6] = 748;
www.eeworm.com/read/308157/13707214
c 矩阵转换.c
void trans(int *p,int n)
{
int i,j,temp;
int *pi,*pj;
for(i=0;i
www.eeworm.com/read/304950/13782492
cpp 矩阵计算.cpp
#include
#include
template void inverse(T1 *mat1,T2*mat2,int a,int b);
template void multi(T1 *mat1,T2 *mat2,T2 *re,int a,in
www.eeworm.com/read/136148/13871442
c 矩阵转换.c
void trans(int *p,int n)
{
int i,j,temp;
int *pi,*pj;
for(i=0;i
www.eeworm.com/read/136101/13872414
c 矩阵操作.c
用C写的，做成DLL使用很方便。
double * MatrixOpp(double A[],int m,int n) /*矩阵求逆*/
{
int i,j,x,y,k;
double *SP=NULL,*AB=NULL,*B=NULL,X,*C;
SP=(double *)malloc(m*n*sizeof(double));
www.eeworm.com/read/151133/5685884
c 逆矩阵.c
#define N 5 /*[注]：修改6为你所要的矩阵阶数*/
#include "stdio.h"
#include "conio.h"
/*js()函数用于计算行列式,通过递归算法实现*/
int js(s,n)
int s[][N],n;
{int z,j,k,r,total=0;
int b[N][N];/*b[N][N]用于存放，在矩阵s[
www.eeworm.com/read/151133/5685914
c 乘法矩阵.c
#include
void main()
{
int i,j;
int big[8][8];
for (i = 0;i < 8;i++) /*循环嵌套*/
for (j = 0;j < 8;j++)
big[i][j] = i * j; /* 乘法表 */
big[2][6] = 748;