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/456568/7345256
c sub_t.c
//*****************************************************************************
// Filename: sub_t.c
// Version: 0.01
// Description: test for sub routine with scaling option
//**************
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cpp leftover.cpp
// leftover.cpp -- overloading the left() function
#include
unsigned long left(unsigned long num, unsigned ct);
char * left(const char * str, int n = 1);
int main()
{
using nam
www.eeworm.com/read/456533/7346106
cpp leftover.cpp
// leftover.cpp -- overloading the left() function
#include
unsigned long left(unsigned long num, unsigned ct);
char * left(const char * str, int n = 1);
int main()
{
using nam
www.eeworm.com/read/456482/7348589
m expert.m
%Expert PID Controller
clear all;
close all;
ts=0.001;
sys=tf(5.235e005,[1,87.35,1.047e004,0]);
dsys=c2d(sys,ts,'z');
[num,den]=tfdata(dsys,'v');
u_1=0.0;u_2=0.0;u_3=0.0;
y_1=0;y_2=0;y_3=0
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cpp qpl.cpp
#include
#include
using namespace std;
void swap(int &a, int &b)
{
int temp;
temp = a;
a = b;
b = temp;
}
/*
根据当前的排列p，计算下一个排列。
原则是从1234-->4321，若p已经
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c dynamic.c
# include
# include
# define NUM 10
int main()
{
char *str[NUM]; /* 定义一个字符性的指针数组 */
int t;
/* 为数组中的每个指针分配内存 */
for(t=0; t
www.eeworm.com/read/456367/7350658
cpp leftover.cpp
// leftover.cpp -- overloading the left() function
#include
unsigned long left(unsigned long num, unsigned ct);
char * left(const char * str, int n = 1);
int main()
{
using nam
www.eeworm.com/read/456367/7350942
cpp leftover.cpp
// leftover.cpp -- overloading the left() function
#include
unsigned long left(unsigned long num, unsigned ct);
char * left(const char * str, int n = 1);
int main()
{
using nam
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m ip_07_09.m
% MATLAB script for Illustrative Problem 9, Chapter 7
echo on
num=[0.01 1];
den=[1 1.01 1];
[a,b,c,d]=tf2ss(num,den);
dt=0.01;
u=ones(1,2000);
x=zeros(2,2001);
for i=1:2000
x(:,i+1)=x(:,i)+dt.*a*x(:
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java exercise4_10.java
// Exercise4_10.java: Display prime numbers
public class Exercise4_10 {
public static void main(String[] args) {
int count = 0;
for (int i = 2; count < 1000; i++) {
// Display e