代码搜索：方波逆变
找到约 4,865 项符合「方波逆变」的源代码
代码结果 4,865
www.eeworm.com/read/481249/6646410
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/477233/6740346
m 逆系统.m
%-----------------------------------------------------------------
% exa020805_residues.m, for example 2.8.5
% to test residuez.m and to obtain the residues of H(z).
%------------------------------
www.eeworm.com/read/405283/11466937
c 逆阵.c
#include "stdio.h"
float z[4][4],*y=z; /*定义一个全局二维数组用来存放N-1阶余子式,因为A的伴随矩阵除以|A|时会产生小数，因此定义成float而非int*/
int js(int *p,int n) /*计算行列式的函数*/
{int k=0,i,s2=0,s1=0,j,s,t;
printf("\n"
www.eeworm.com/read/405283/11466938
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/346772/11723688
m 逆系统.m
%-----------------------------------------------------------------
% exa020805_residues.m, for example 2.8.5
% to test residuez.m and to obtain the residues of H(z).
%------------------------------
www.eeworm.com/read/256774/11972541
doc 逆置.doc
www.eeworm.com/read/132104/14110299
bas 求逆.bas
Attribute VB_Name = "Module1"
'Public mat(5, 5) As Double
Public Function Dinvert(mat() As Double, ByVal n As Integer)
Dim i As Integer, j As Integer, k As Integer
Dim sum As Double
'
www.eeworm.com/read/233105/14169137
c 逆阵.c
#include "stdio.h"
float z[4][4],*y=z; /*定义一个全局二维数组用来存放N-1阶余子式,因为A的伴随矩阵除以|A|时会产生小数，因此定义成float而非int*/
int js(int *p,int n) /*计算行列式的函数*/
{int k=0,i,s2=0,s1=0,j,s,t;
printf("\n"
www.eeworm.com/read/233105/14169140
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/227639/14418663
c 逆阵.c
#include "stdio.h"
float z[4][4],*y=z; /*定义一个全局二维数组用来存放N-1阶余子式,因为A的伴随矩阵除以|A|时会产生小数，因此定义成float而非int*/
int js(int *p,int n) /*计算行列式的函数*/
{int k=0,i,s2=0,s1=0,j,s,t;
printf("\n"