代码搜索：Digit
找到约 4,689 项符合「Digit」的源代码
代码结果 4,689
www.eeworm.com/read/208175/15251641
c fg439_rtc.c
#include
char* LCD = LCDMEM;
const char digit[10] =
{
0xB7, // "0" LCD segments a+b+c+d+e+f
0x12, // "1"
0x8F, // "2"
0x1F, // "3"
0x3A, // "4"
0x3D,
www.eeworm.com/read/14719/403658
m testpnn.m
%程序是正确的,但泛化能力差
clc;
clear;
t1=cputime;
num=10;
disp('正在计算数字识别PNN模型...')
for i=1:num
fname = sprintf('..\\digit\\%da.wav',i-1);
x = wavread(fname);
[x1 x2] = vad(x);
m = mfcc(x);
m = m(
www.eeworm.com/read/14719/403663
asv testpnn.asv
%程序是正确的,但泛化能力差
clc;
clear;
t1=cputime;
num=10;
disp('正在计算数字识别PNN模型...')
for i=1:num
fname = sprintf('..\\digit\\%da.wav',i-1);
x = wavread(fname);
[x1 x2] = vad(x);
m = mfcc(x);
m = m(
www.eeworm.com/read/17194/722731
51
//DS18B20温度检测及其液晶显示
#include //包含单片机寄存器的头文件
#include //包含_nop_()函数定义的头文件
unsigned char code digit[10]={"0123456789"}; //定义字符数组显示数字
unsigned char code Str[]={"Test by DS
www.eeworm.com/read/25885/951926
51
//DS18B20温度检测及其液晶显示
#include //包含单片机寄存器的头文件
#include //包含_nop_()函数定义的头文件
unsigned char code digit[10]={"0123456789"}; //定义字符数组显示数字
unsigned char code Str[]={"Test by DS
www.eeworm.com/read/27624/986580
c ex90.c
//DS18B20温度检测及其液晶显示
#include //包含单片机寄存器的头文件
#include //包含_nop_()函数定义的头文件
unsigned char code digit[10]={"0123456789"}; //定义字符数组显示数字
unsigned char code Str[]={"Test by DS
www.eeworm.com/read/287283/4024616
c 8-3.c
#include
#include
#include
int card [4][13]; //僇乕僪傪堦搙堷偄偨偐偳偆偐婰榐偡傞
char total[13] = { 11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10 };
char *digit[13] = { "A", "2", "3"
www.eeworm.com/read/135243/13948825
asv testpnn.asv
%程序是正确的,但泛化能力差
clc;
clear;
t1=cputime;
num=10;
disp('正在计算数字识别PNN模型...')
for i=1:num
fname = sprintf('..\\digit\\%da.wav',i-1);
x = wavread(fname);
[x1 x2] = vad(x);
m = mfcc(x);
m = m(
www.eeworm.com/read/433217/8539112
h ds18b20.h
#ifndef DS18B20_1_H
#define DS18B20_1_H
#define uchar unsigned char
#define uint unsigned int
sbit DQ1=P2^0;
bit tempflag,DS18B20F;//温度正负标志位
//unsigned char Digit[5]={0,0,0,0,0};
uchar do
www.eeworm.com/read/272894/10937492
dpr ex.dpr
{
X is Answer.
Set Cn that Cn*2^n=Xn
C(n+1)*2^(n+1)
=X(n+1)
=Xn+10^n*k, Here k is 1 or 2, namely the first digit of X(n+1)
=Cn*2^n+10^n*k
=>
2*C(n+1)=Cn+5^n*k
if Cn is Odd then k=