代码搜索：Num
找到约 10,000 项符合「Num」的源代码
代码结果 10,000
www.eeworm.com/read/296909/8072780
m simple_nand2_gate_design.m
% Two-input NAND gate sizing (a simple, hard-coded example).
% (a figure is generated if the tradeoff flag is turned on)
%
% This is an example taken directly from the paper:
%
% Digital circuit opt
www.eeworm.com/read/296824/8075664
m example.m
%bpnet举例，因为BP网络的权值初始化都是随即生成，所以每次运行的状态可能不一样。
%如果初始化的权值有利于训练，那么可能很快能结束训练，反之则反之
clear all
clc
figure
randn('state',sum(100*clock))
num1=5; %隐节点数
num2=10000; %最大迭代次数
a1=0.02; %期望误差
a2=
www.eeworm.com/read/296805/8076384
c avg.c
/* avg.c: Averages 2 integers */
#include
int main()
{
int num1, num2;
float sum;
puts("Enter the 1st number:");
scanf("%d",&num1);
puts("Enter the 2nd nu
www.eeworm.com/read/396837/8087852
c gkim.c
#include
#define uint unsigned int
#define uchar unsigned char
sbit we1=P2^4;
sbit we2=P2^5;
sbit we3=P2^6;
sbit we4=P2^7;
sbit s1=P3^2;
sbit s2=P3^3;
sbit s3=P3^4;
sbit s4=P3^5;
u
www.eeworm.com/read/296569/8092103
cpp hash查找.cpp
/*
扫描一个C源程序,用Hash表存储该程序中出现的关键字,并统计该程序中的关键字出现频度.
用线性探测法解决Hash冲突.
Hash(key)=[(key的第一个字母序号)*100+(key的最后一个字母序号)] MOD 41
*/
#include
#include
#include
#define MAXS
www.eeworm.com/read/296569/8092106
c a.c
/*
扫描一个C源程序,用Hash表存储该程序中出现的关键字,并统计该程序中的关键字出现频度.
用线性探测法解决Hash冲突.
Hash(key)=[(key的第一个字母序号)*100+(key的最后一个字母序号)] MOD 41
*/
#include
#include
#include
#define MAXS
www.eeworm.com/read/196428/8092417
c lt525.c
# include "stdio.h"
void main( )
{ float sum (float *p,int n);
float num[10],total;
int i;
for (i=0;i
www.eeworm.com/read/196428/8092712
c lt425.c
# include "stdio.h"
void main( )
{ char string[30]; //得到字符数组string中的单词个数
int i,num=0,word=0;
char a;
gets(string);
for (i=0;(a=string[i])!='\0';i++)
if (a==' ')
word=0;
el
www.eeworm.com/read/196358/8097956
h amr_config.h
#define MAX_NUM_METERS 100 // maximum of meters in the system
www.eeworm.com/read/196355/8098075
h amr_config.h
#define MAX_NUM_METERS 100 // maximum of meters in the system