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  • The VGA example generates a 320x240 diffusion-limited-aggregation (DLA) on Altera DE2 board. A DLA i

    The VGA example generates a 320x240 diffusion-limited-aggregation (DLA) on Altera DE2 board. A DLA is a clump formed by sticky particles adhering to an existing structure. In this design, we start with one pixel at the center of the screen and allow a random walker to bounce around the screen until it hits the pixel at the center. It then sticks and a new walker is started randomly at one of the 4 corners of the screen. The random number generators for x and y steps are XOR feedback shift registers (see also Hamblen, Appendix A). The VGA driver, PLL, and reset controller from the DE2 CDROM are necessary to compile this example. Note that you must push KEY0 to start the state machine.

    标签: diffusion-limited-aggregation DLA generates 320x240

    上传时间: 2014-01-16

    上传用户:225588

  • Grass 5.00 start for idiot, one of the best documentation for Grass beginner users. And this is the

    Grass 5.00 start for idiot, one of the best documentation for Grass beginner users. And this is the Chinese edition, for the ones prefer mother language. From the beginning for Linux participation to the BIOS update, and finally the details setup for GRASS Linux users. Moreover, with my suggestion is that Ubuntu 8.10 is the best package for GRASS 5.00, but it is totally up to you how to choose a platform for this talent open source application.

    标签: Grass documentation for the

    上传时间: 2016-12-06

    上传用户:jjj0202

  • 小型公司工资管理系统 工资的计算方法: A 经理:固定月薪为8000; B 技术员:工作时间*小时工资(100元每小时); C 销售员:销售额*4%提成; D 销售经理:底薪(5000)+所

    小型公司工资管理系统 工资的计算方法: A 经理:固定月薪为8000; B 技术员:工作时间*小时工资(100元每小时); C 销售员:销售额*4%提成; D 销售经理:底薪(5000)+所辖部门销售额总额*0.5%;

    标签: 8000 5000 销售 100

    上传时间: 2013-12-18

    上传用户:qilin

  • 溫度華氏轉變攝氏 #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void)

    溫度華氏轉變攝氏 #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void) { int a=73,b=85,c=66 { if (a>=90) printf("a=A等級!!\n") else if (a>=80) printf("73分=B等級!!\n") else if (a>=70) printf("73分=C等級!!\n") else if (a>=60) printf("73分=D等級!!\n") else if (a<60) printf("73分=E等級!!\n") } { if (b>=90) printf("b=A等級!!\n") else if (b>=80) printf("85分=B等級!!\n") else if (b>=70) printf("85分=C等級!!\n") else if (b>=60) printf("85分=D等級!!\n") else if (b<60) printf("85分=E等級!!\n") } { if (c>=90) printf("c=A等級!!\n") else if (c>=80) printf("66分=B等級!!\n") else if (c>=70) printf("66分=C等級!!\n") else if (c>=60) printf("66分=D等級!!\n") else if (c<60) printf("66分=E等級!!\n") } system("pause") return 0 }

    标签: include stdlib stdio gt

    上传时间: 2014-11-10

    上传用户:wpwpwlxwlx

  • 溫度華氏轉變攝氏 #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void)

    溫度華氏轉變攝氏 #include <stdio.h> #include <stdlib.h> enum x {A,B,C,D,E} int main(void) { int a=73,b=85,c=66 { if (a>=90) printf("a=A等級!!\n") else if (a>=80) printf("73分=B等級!!\n") else if (a>=70) printf("73分=C等級!!\n") else if (a>=60) printf("73分=D等級!!\n") else if (a<60) printf("73分=E等級!!\n") } { if (b>=90) printf("b=A等級!!\n") else if (b>=80) printf("85分=B等級!!\n") else if (b>=70) printf("85分=C等級!!\n") else if (b>=60) printf("85分=D等級!!\n") else if (b<60) printf("85分=E等級!!\n") } { if (c>=90) printf("c=A等級!!\n") else if (c>=80) printf("66分=B等級!!\n") else if (c>=70) printf("66分=C等級!!\n") else if (c>=60) printf("66分=D等級!!\n") else if (c<60) printf("66分=E等級!!\n") } system("pause") return 0 }

    标签: include stdlib stdio gt

    上传时间: 2013-12-12

    上传用户:亚亚娟娟123

  • Problem B:Longest Ordered Subsequence A numeric sequence of ai is ordered if a1 < a2 < ... &l

    Problem B:Longest Ordered Subsequence A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).

    标签: Subsequence sequence Problem Longest

    上传时间: 2016-12-08

    上传用户:busterman

  • 两台处理机A 和B处理n个作业。设第i个作业交给机器 A 处理时需要时间ai

    两台处理机A 和B处理n个作业。设第i个作业交给机器 A 处理时需要时间ai,若由机器B 来处理,则需要时间bi。由于各作 业的特点和机器的性能关系,很可能对于某些i,有ai >=bi,而对于 某些j,j!=i,有aj<bj。既不能将一个作业分开由两台机器处理,也没 有一台机器能同时处理2 个作业。设计一个动态规划算法,使得这两 台机器处理完成这n 个作业的时间最短(从任何一台机器开工到最后 一台机器停工的总时间)。研究一个实例:(a1,a2,a3,a4,a5,a6)= (2,5,7,10,5,2);(b1,b2,b3,b4,b5,b6)=(3,8,4,11,3,4)

    标签: 处理机 机器

    上传时间: 2014-01-14

    上传用户:独孤求源

  • DS1302读写程序。功能:向串口调试工具输入b

    DS1302读写程序。功能:向串口调试工具输入b,窗口立刻显示从DS1302读出的时间值。用户只需更改管脚定义,即可在自己的板子上运行。该程序在STC12C5A56S2单片机上已通过硬件调试,晶振24M。

    标签: 1302 DS 读写程序 串口调试工具

    上传时间: 2016-12-14

    上传用户:thuyenvinh

  • 1.B树的实现 2.ElfHash的实现 3.三种排序方式(插入

    1.B树的实现 2.ElfHash的实现 3.三种排序方式(插入,归并,快速)

    标签: ElfHash 排序 方式

    上传时间: 2013-12-29

    上传用户:exxxds

  • 一本c++学习的必备之书《Essential C++》By Stanley B. Lippman

    一本c++学习的必备之书《Essential C++》By Stanley B. Lippman,

    标签: B. Essential Stanley Lippman

    上传时间: 2014-01-20

    上传用户:水中浮云