代码搜索：预置数
找到约 10,000 项符合「预置数」的源代码
代码结果 10,000
www.eeworm.com/read/349925/10781215
bat 取得硬盘数.bat
@echo off
cd.>script.txt
>>script.txt echo list disk
for /f %%i in ('diskpart /s script.txt^|find /c ^"联机^"') do Set HardDrivers=%%i
del script.txt /q
echo 您的计算机上硬盘安装数量为：%HardDrivers%
pause
www.eeworm.com/read/349332/10835108
txt 何意数排序.txt
using System;
public class Test
{
public static void Main()
{
//不知道怎么搞的,两个不能同进用,想不明白,可能是前一种方法已经把值给改变了
int x;
int temp;
Console.WriteLine ("你想排几个数的序:");
x=int.Parse (Console.ReadLi
www.eeworm.com/read/272424/10957722
java 猜数游戏.java
import java.awt.*;
import java.awt.event.*;
public class GiveFigure extends Thread
{
private int maxFigure=99;
private int minFigure=0;
private int figure;
private int guess;
public
www.eeworm.com/read/463456/7180552
bat 取得硬盘数.bat
@echo off
cd.>script.txt
>>script.txt echo list disk
for /f %%i in ('diskpart /s script.txt^|find /c ^"联机^"') do Set HardDrivers=%%i
del script.txt /q
echo 您的计算机上硬盘安装数量为：%HardDrivers%
pause
www.eeworm.com/read/458654/7292158
e 判断回文数.e
www.eeworm.com/read/457549/7322955
c 阿姆斯特朗数.c
#include
main()
{
int i,t,k,a[3];
printf("There are following Armstrong number smaller than 1000: \n");
for(i=152;i=10;t++)
{
a[t]=(i%k)/(k/1
www.eeworm.com/read/450798/7476667
c 阿姆斯特朗数.c
#include
main()
{
int i,t,k,a[3];
printf("There are following Armstrong number smaller than 1000: \n");
for(i=152;i=10;t++)
{
a[t]=(i%k)/(k/1
www.eeworm.com/read/449694/7497953
c 阿姆斯特朗数.c
#include
main()
{
int i,t,k,a[3];
printf("There are following Armstrong number smaller than 1000: \n");
for(i=152;i=10;t++)
{
a[t]=(i%k)/(k/1
www.eeworm.com/read/446914/7562933
doc 猜数游戏.doc
www.eeworm.com/read/440225/7691791
bat 取得硬盘数.bat
@echo off
cd.>script.txt
>>script.txt echo list disk
for /f %%i in ('diskpart /s script.txt^|find /c ^"联机^"') do Set HardDrivers=%%i
del script.txt /q
echo 您的计算机上硬盘安装数量为：%HardDrivers%
pause