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/463459/7180447
bat 批量改文件夹名.bat
@echo off
for /f %%i in ('dir /ad /b /on') do set last=%%i
for /f %%i in ('dir /ad /b /o-n') do set fist=%%i
set /a fist=1%fist:~-4%-10000
set /a last=1%last:~-4%-10000
set num=1
setlocal enable
www.eeworm.com/read/463457/7180508
bat 九九乘法表.bat
@echo off
set num=0
for /l %%i in (1,1,9) do (
for /l %%j in (1,1,%%i) do call :multiply %%i %%j
)
pause>nul
goto :eof
:multiply
set /a num+=1
set /a var=%1*%2
set var=%2×%1=%var%
set
www.eeworm.com/read/462985/7191063
cpp mofang.cpp
//该程序用来计算并输出奇数阶的魔方阵
#include
#include
void main()
{
const int num = 7;
int A[num][num] = {0};
int row=0,column=num/2;
// cin >> num; //num is a odd num
www.eeworm.com/read/462672/7198344
cpp josephus.cpp
#include
void main()
{
const int num=10;
int interval;
int a[num];
for(int i1=0;i1
www.eeworm.com/read/462436/7201749
cpp genetic.cpp
/*遗传算法中：
个体编码方案：整数编码
交配方法：常规交配法
变异方法：打乱变异（逆序交换）
新种群构成方法：交配及变异后产生的所有子代
算法结束条件：繁衍2000代
*/
#include
#include
#include
#include
#include
www.eeworm.com/read/462318/7202832
m ip_07_11.m
% MATLAB script for Illustrative Problem 11, 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)
www.eeworm.com/read/462075/7209572
c standard_func.c
//定义数据类型
#define uchar unsigned char
#define uint unsigned int
//延时子程序 time 为多少ms
void delay_us(uchar time)
{
while(time--);
}
//延时子程序 time 为多少ms
void delay_ms(uchar time)
{
ucha
www.eeworm.com/read/462075/7209596
c standard_func.c
//定义数据类型
#define uchar unsigned char
#define uint unsigned int
//延时子程序 time 为多少ms
void delay_us(uchar time)
{
while(time--);
}
//延时子程序 time 为多少ms
void delay_ms(uchar time)
{
ucha
www.eeworm.com/read/211241/7216176
h com.h
//*----------------------------------------------------------------------------
//* ATMEL Microcontroller Software Support - ROUSSET -
//*----------------------------------------------------
www.eeworm.com/read/461489/7226478
cpp lpt.cpp
#include
#include
const int N = 50;
using namespace std;
int main()
{
int n,m; // n为任务数，m为机器数
scanf("%d%d",&n,&m);
int job[N];
int i,j,k;
for(i = 0; i <