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Description: public class CircleMenuCanvas extends Canvas implements Runnable{
double pi = Math.PI
public final int MENURIGHT = 1
public final int MENULEFT = 0
Image menuImage[] = new Image[6]
int []jiaodu = {330,30,90,150,210,270}
String menuName[] = {"新游戏","继续游戏","游戏设置","高分榜","游戏帮助","退出游戏"}
int x = getWidth()/2
int y = getHeight()/2
int count = 0
int local
int index = 0
Font f
boolean running = false
/**
* 构造方法
*
*/-public class CircleMenuCanvas extends Canvas implements Runnable (double pi = Math.PI public final int MENURIGHT = 1 public final int MENULEFT = 0 Image menuImage [] = new Image [6] int [] jiaodu = (330,30,90150210270) String menuName [] = (
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Size: 18432 |
Author: 么么 |
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Description: Linux 下 的superpi源码, 使用FFT算法实现可以做测试使用.-A good superpi source code running at linux/unix platform, also you can port to any hardware.Realized with FFT algorithm.
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Size: 143360 |
Author: Edwin |
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Description: hnbytfkmi tyyuik8u.ty ipo -[po"
{y-hik ply /pi[l
[
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Size: 4108288 |
Author: bab |
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Description: 本程序用于检测音频文件中是否具有DTMF信号，若有则将其检出。
程序首先使用Goertzel算法求出以FRAMESIZE（默认200）为大小的一帧数据在8个DTMF频点上的能量。
对Goertzel算法的改进，对于系数的计算不是采用2*cos[2*pi*k/N]，而是采用2*cos[2*pi*fn/fs],这样能够降低误差。
确定了8个频点的能量后运用一系列判决门限来确定有没有DTMF信号，以及信号是什么。
-This procedure used to detect whether an audio file with a DTMF signal, if so, what will be its detection. Procedures to derive the first use of Goertzel algorithm to FRAMESIZE [default 200] for the size of data in an eight point DTMF frequency energy. Of the Goertzel algorithm for the calculation of coefficients instead of 2* cos [2* pi* k/N], instead of using the 2* cos [2* pi* fn/fs], this can reduce the error. Identified eight frequency energy use after a series of sentences to determine the threshold has no DTMF signal, as well as the signal is.
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Size: 620544 |
Author: 李昊然 |
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Description: 在0 / 1背包问题中，需对容量为c 的背包进行装载。从n 个物品中选取装入背包的物品，每件物品i 的重量为wi ，价值为pi 。对于可行的背包装载，背包中物品的总重量不能超过背包的容量，最佳装载是指所装入的物品价值最高，即n ?i=1pi xi 取得最大值。约束条件为n ?i =1wi xi≤c 和xi?[ 0 , 1 ] [ 1≤i≤n]。-At 0/1 knapsack problem, there is a need for a capacity of c for the loaded backpack. N items from a backpack loaded, select the items, each item i the weight of wi, the value of pi. Feasible for loading backpack, backpack items total weight of the backpack should not exceed capacity, the best means by loading articles into the highest value, that is, n? I = 1pi xi to obtain the maximum value. Constraint condition for n? I = 1wi xi ≤ c and xi? [0, 1] [1 ≤ i ≤ n].
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Size: 1024 |
Author: samuel |
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Description: 计算PI的值的示例代码。仅供参考.无版权责任。-Demo to calc PI value.
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Size: 1024 |
Author: yiwei |
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Description: YSA PI is realized this mdl
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Size: 6144 |
Author: Alper |
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Description: 1． 在No.1图形窗口中绘制 y=sin(x)在[0,2*pi]内的曲线。要求曲线的颜色为绿色，线型为
点划线，用*标示坐标点，在x轴的附近用 黑体 标注 ‘x轴’字样，在图形的上方加上标题 ‘正弦函数’，严格控制x,y轴分度相等，并开启网格。
2． 在No.2图形窗口中创建四个子窗口，在第一、二子窗口中用不同的方法同时绘制 y=x^2,y=-x^2,y=x^2*sin(x) 在[0,2*pi]内的曲线，并要给出标注 在第三个子窗口中绘制 三维曲线
3． 把No.3图形窗口分成五个子窗口，分别用plot3 mesh meshc meshz surf 来绘制 z=x*exp(-x^2-y^2) 在 -5=<x,y<=5 内的空间曲面图形，说明他们的区别，其中要求在用surf绘制的窗口内加入位置为[1，0.5，2]的光源，加入颜色标尺，采用spring色系
-1． 在No.1图形窗口中绘制 y=sin(x)在[0,2*pi]内的曲线。要求曲线的颜色为绿色，线型为
点划线，用*标示坐标点，在x轴的附近用 黑体 标注 ‘x轴’字样，在图形的上方加上标题 ‘正弦函数’，严格控制x,y轴分度相等，并开启网格。
2． 在No.2图形窗口中创建四个子窗口，在第一、二子窗口中用不同的方法同时绘制 y=x^2,y=-x^2,y=x^2*sin(x) 在[0,2*pi]内的曲线，并要给出标注 在第三个子窗口中绘制 三维曲线
3． 把No.3图形窗口分成五个子窗口，分别用plot3 mesh meshc meshz surf 来绘制 z=x*exp(-x^2-y^2) 在-5=<x,y<=5 内的空间曲面图形，说明他们的区别，其中要求在用surf绘制的窗口内加入位置为[1，0.5，2]的光源，加入颜色标尺，采用spring色系
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Size: 1024 |
Author: 李子木 |
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Description: COMMANDE D un convertiseur AC/DC 脿 deux niveaux par hysteresis
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Size: 11264 |
Author: paradox |
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Description: pi的蒙特卡罗估计 matlab源程序有5个pi的估计，每一个都是基于500次的重复随机实验-Monte Carlo estimate pi source matlab has 5 pi estimates are based on each of 500 random repeat the experiment
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Size: 1024 |
Author: shilei |
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Description: 以cos（2*pi*k*t/N）信号空间，k=0,1,……N-1, 取N= 4，8，16，32，64等基信号作为传输信号，通过计算机仿真正交信号的误码率。-To cos (2* pi* k* t/N) signal space, k = 0,1, ... ... N-1, take N = 4,8,16,32,64, such as the base signal as a transmission signal, through computer simulation orthogonal signal BER.
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Size: 54272 |
Author: caomin |
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Description: vb2005编写的计算器。
直接输入字符串，如:
4+pi^2+sin(3*cos(2)*e^2)+2.3E13
常数：pi,e
科学计数表示：1E4(1*10^4)
函数支持常用函数:sin,cos,tan,atan,asin-vb2005 Calculator
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Size: 25600 |
Author: GeYuhang |
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Description: 一个利用OpenMP并行计算pi值的程序,双核CPU可以提高计算速度一倍,减少时间一半儿.-OpenMP parallel computing using a pi value of procedures, dual-core CPU can improve computing speed doubled, to reduce the half time.
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Size: 66560 |
Author: ifelse |
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Description: 操作系统编程，利用pipe管道实现进程间通信，linux环境下gcc编译-Operating system programming, the use of pipe pipeline implementation process communication, linux environment gcc compiler
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Size: 1024 |
Author: |
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Description: 需对容量为c 的背包进行装载。从n 个物品中选取装入背包的物品，每件物品i 的重量为wi ，价值为pi 。对于可行的背包装载，背包中物品的总重量不能超过背包的容量，最佳装载是指所装入的物品价值最高。-sdad
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Size: 1024 |
Author: 张慧 |
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Description: 基于DSP的永磁交流伺服控制系统开发
本论文在分析了PMSM的结构、运动原理及数学模型的基础上，系统地阐述了PMSM交流伺服系统矢量控
制的基本原理、坐标变换方法及空间矢量脉冲宽度调制(PWM)波的生成算法，论述了PI控制算法和
速度位置计算方法。在MATLAB仿真软件环境下建立了交流伺服系统的仿真模型，并对模型进行了仿真-DEVELOPMENT OF THE PERMANENT AC SERVO CONTROL
SYSTEM BASED ON DSP
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Size: 3135488 |
Author: yjh |
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Description: 1． 0-1背包问题
在0 / 1背包问题中，需对容量为c 的背包进行装载。从n 个物品中选取装入背包的物品，每件物品i 的重量为wi ，价值为pi 。对于可行的背包装载，背包中物品的总重量不能超过背包的容量，最佳装载是指所装入的物品价值最高
-1. 0-1 knapsack problem at 0/1 knapsack problem, there is a need for a capacity of c for the loaded backpack. N items from a backpack loaded, select the items, each item i the weight of wi, the value of pi. Feasible for loading backpack, backpack items total weight of the backpack should not exceed capacity, the best load is the value of items packed by the highest
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Size: 144384 |
Author: 松柏长青 |
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Description: 描写一段基于CS8955的RDS解码方案，支持ＰＩ，ＰＴＹ，ＰＳ、ＲＴ等功能，所有信息通过串口打印。-Description section of the CS8955-based RDS decoder program to support the PI, PTY, PS, RT and other functions, all of the information through the serial port to print.
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Size: 250880 |
Author: cheuna |
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Description: 里面有设计简单的马达控制应用，也有用了PI算法的，也有使用空间矢量算法的。-There are simple motor control applications, it also uses the PI algorithm, but also has the use of space vector algorithm.
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Size: 3072 |
Author: 林国龙 |
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Description: 在0 / 1背包问题中，需对容量为c 的背包进行装载。从n 个物品中选取装入背包的物品，每件物品i 的重量为wi ，价值为pi 。对于可行的背包装载，背包中物品的总重量不能超过背包的容量，最佳装载是指所装入的物品价值最高。-At 0/1 knapsack problem, there is a need for a capacity of c to load the backpack. N items from a selected items into the backpack, each item i the weight wi, the value of pi. Feasible for loading backpack, backpack in the total weight of items should not exceed the capacity of backpack, the best means by loading items into the highest value.
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Size: 172032 |
Author: 松柏长青 |
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