-
Win-in-the-clouds
赢在云端,一本描述云计算的书籍,希望对云计算入门的同学有所帮助。(Win in the clouds, a description of cloud computing books, we hope the cloud started to help students.)
- 2013-08-06 17:12:56下载
- 积分:1
-
Node.js-f46c
深入浅出node.js便于初学者和老手学习node
用Javascript写后台(Node.js easy for beginners to learn node and veteran
Write the background with Javascript)
- 2016-02-25 00:43:16下载
- 积分:1
-
299255
说明: 基础练习,这是Java入门课程的基础练习。包括了许多的知识点(Basic exercises, which are the basic exercises of the Java introductory course. It includes many knowledge points.)
- 2018-12-29 18:57:57下载
- 积分:1
-
KmeansinJava
Java代码实现数据挖掘算法,kmeans(Java code to implement data mining algorithms, kmeans)
- 2016-04-10 10:14:03下载
- 积分:1
-
infoboxAndTemplateExtractor
info-box extractor for Wikipedia
- 2015-07-31 07:36:38下载
- 积分:1
-
smithaaa
很好的一个直接画史密斯圆图的源代码,且圆图的精度很高(a good painting directly Smith Chart of source code, which plans a round of high accuracy)
- 2007-05-09 18:24:23下载
- 积分:1
-
MoveableShape_bemjh
用java实现接口,实现不同图形的移动效果,主要是演示接口的用法(achieved using java interface, graphics achieve different effects of the movement, the main interface is a demonstration of the usage)
- 2006-12-30 10:18:52下载
- 积分:1
-
lombok-plugin-0.25-EAP
说明: idea 的lombcok插件 上行下效寻寻x(lombok plugins in idea)
- 2020-01-18 11:24:29下载
- 积分:1
-
apache-ant-1.7.0-bin
说明: ant1.7 , built utils
- 2020-06-22 18:00:01下载
- 积分:1
-
simplejavaprogramming
题目:打印出杨辉三角形(要求打印出10行题目:输入3个数a,b,c,按大小顺序输出。
题目:输入数组,最大的与第一个元素交换,最小的与最后一个元素交换,输出数组。
题目:有n个整数,使其前面各数顺序向后移m个位置,最后m个数变成最前面的m个数
题目:编写一个函数,输入n为偶数时,调用函数求1/2+1/4+...+1/n,当输入n为奇数时,调用函数1/1+1/3+...+1/n(Topic: Print out the Triangle (10 lines required to print out the Title: Enter the 3 numbers a, b, c, according to the order of output. Title: input array, the largest exchange with the first element, the smallest and the last element of the exchange, output array. Topic: There are n integers, to move back in front of the order m of the number of positions, finally becoming the foremost m m the number of number of topics: write a function, the input n is even, the call function evaluation 1/2+1/4+...+ 1/n, when the input n is odd, call the function 1/1+1/3+...+ 1/n)
- 2011-05-09 19:11:58下载
- 积分:1