代码搜索：LED Matrix
找到约 10,000 项符合「LED Matrix」的源代码
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www.eeworm.com/read/435895/7781732
c copy av matrix.c
//程序的调试可以用"串口调试助手V2.1"辅助完成2008.2.10.20.24
#include "p18cxxx.h"
//#include "p18f6585.h"
#include
#pragma config OSC = XT, OSCS = OFF
#pragma config PWRT = ON
#pragma config WDTPS =3
www.eeworm.com/read/435895/7781791
o copy av matrix.o
www.eeworm.com/read/435895/7781796
o av2_matrix.o
www.eeworm.com/read/435895/7781807
c av3_matrix.c
//程序的调试可以用"串口调试助手V2.1"辅助完成2008.2.10.20.24
#include "p18cxxx.h"
#include
#pragma config OSC = XT, OSCS = OFF
#pragma config PWRT = OFF
#pragma config WDTPS =32768
#pragma config CCP2
www.eeworm.com/read/435895/7781838
c av2_matrix.c
//程序的调试可以用"串口调试助手V2.1"辅助完成2008.2.10.20.24
#include "p18cxxx.h"
#include
#pragma config OSC = XT, OSCS = OFF
#pragma config PWRT = OFF
#pragma config WDTPS =32768
#pragma config CCP2
www.eeworm.com/read/434916/7800420
txt spiral matrix1.txt
#include
#include
#include
/*
1 8 7
2 9 6
3 4 5
输出如图所示的矩正,第一个是3*3的,后面的4*4的,
算法就是随便输入一个N,就可以输出一个N*N的矩正???
1 12 11 10
2 13 16 9
3 14 15 8
4
www.eeworm.com/read/299955/7818816
m form_ref_matrix.m
function p=form_ref_matrix(P)
if P==1
a=[-1 1];
b=a;
elseif P==2
bits_in=[1 0 1 1 0 1 0 0];
full_len=length(bits_in);
% Angle [pi/4 3*pi/4 -3*pi/4 -pi/4] corresponds to
www.eeworm.com/read/299242/7870532
m leader_follower_matrix.m
function [xy_matrix,w_class,w_class_num,class_number]=leader_follower_matrix(xy_matrix,i,w_class,w_class_num,constant)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
www.eeworm.com/read/198889/7904613
c matrix2d.c
/*
EZW编解码器的数据操作
*/
#include "matrix2d.h"
#include
#include
/*
* Allocate memory for a two-dimensional RxC matrix.
* Returns NULL on failure.
*/
matrix_2d *mat
www.eeworm.com/read/198889/7904649
h matrix2d.h
//EZW编解码数据结构的定义：矩阵结构
#ifndef __MATRIX_2D_H__
#define __MATRIX_2D_H__
typedef int element_type;
#define min_element_type -32768
#define max_element_type 32767
typedef struct __matrix_2d