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Description: 图是用牛顿法求方程z3-1=0的根所得到的“项链”。以上程序采用浮点运算，速度是比较慢的。如果每迭代一次就计算一次距离，则速度会更慢。为了使不同吸引区域有不同的颜色，并且能够表现收敛速度，必须做更多的距离计算，但得到的图形也更美丽。特别是将分界处的“项链”放大来看，会得到精致的结构。-Fig Newton method is used Equation z3-1 = 0 has been the root of the "necklace." The procedure is used floating-point operations, speed is relatively slow. If every iteration on a calculated distance, it would grow more slowly. In order to attract different regions have different colors, and be able to demonstrate the convergence rate and must do more distance, but the graphics are more beautiful. In particular, the boundaries of the "necklace" to enlarge it, will be refined structure.
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Size: 2093 |
Author: 刁轩 |
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Description: J2ME编写的飞行设计类手机游戏需要在诺机亚的模拟器下才能够正常执行.-J2ME prepared by the design of flight mobile games category in the Asian Connaught machine simulator can be under normal execution.
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Size: 390821 |
Author: 胡峻 |
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Description: motorola Z3 mobile phone
在EE论坛上下载到的摩托罗拉公司的Z3手机原理图，总共三个pdf文件，对于手机开发的同志们值得参考的。-motorola Z3 mobile phone in EE forum download to the public Motorola Secretary Z3 phone diagram, a total of three pdf document on the development of the mobile phone comrades worth considering.
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Size: 587672 |
Author: 黄冠 |
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Description: 51单片机指令教程，很漂亮的PPT教程，初学者的好帮手哦！！值得下载
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Size: 127245 |
Author: wilson |
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Description: 用VC++做的 正则投影变换
从x，y，z3个方向看观察的不同效果
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Size: 1748 |
Author: 杨洁 |
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Description: 图是用牛顿法求方程z3-1=0的根所得到的“项链”。以上程序采用浮点运算，速度是比较慢的。如果每迭代一次就计算一次距离，则速度会更慢。为了使不同吸引区域有不同的颜色，并且能够表现收敛速度，必须做更多的距离计算，但得到的图形也更美丽。特别是将分界处的“项链”放大来看，会得到精致的结构。-Fig Newton method is used Equation z3-1 = 0 has been the root of the "necklace." The procedure is used floating-point operations, speed is relatively slow. If every iteration on a calculated distance, it would grow more slowly. In order to attract different regions have different colors, and be able to demonstrate the convergence rate and must do more distance, but the graphics are more beautiful. In particular, the boundaries of the "necklace" to enlarge it, will be refined structure.
Platform: |
Size: 2048 |
Author: 刁轩 |
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Description: J2ME编写的飞行设计类手机游戏需要在诺机亚的模拟器下才能够正常执行.-J2ME prepared by the design of flight mobile games category in the Asian Connaught machine simulator can be under normal execution.
Platform: |
Size: 390144 |
Author: 胡峻 |
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Description: motorola Z3 mobile phone
在EE论坛上下载到的摩托罗拉公司的Z3手机原理图，总共三个pdf文件，对于手机开发的同志们值得参考的。-motorola Z3 mobile phone in EE forum download to the public Motorola Secretary Z3 phone diagram, a total of three pdf document on the development of the mobile phone comrades worth considering.
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Size: 587776 |
Author: 黄冠 |
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Description: 51单片机指令教程，很漂亮的PPT教程，初学者的好帮手哦！！值得下载-51 Single-chip instruction manuals very beautiful PPT tutorials, beginners a good helper Oh! ! Worth downloading
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Size: 126976 |
Author: wilson |
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Description: 用VC++做的 正则投影变换
从x，y，z3个方向看观察的不同效果-With VC++ Make the canonical projection transformation from the x, y, z3 direction to see the different effects observed
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Size: 1024 |
Author: 杨洁 |
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Description: 中芯通讯Z32-U FLASH 32位控制器详细文档-SMIC communications Z32-U FLASH 32 digital controller detailed documentation
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Size: 622592 |
Author: 刘会忠 |
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Description: 3
摘 要:研究了具有深度运动模糊效果的图像的复原算法.采用对运动模糊图像的傅里叶频谱进行
Radon 变换来估计运动模糊方向 ,在此方向上计算运动模糊图像的自相关来估计运动模糊长度 ,再
以运动模糊方向和运动模糊长度为参量结合超分辨力图像复原处理算法对比较严重的运动模糊图像进行复原.结果表明 ,该综合性算法能够较为精确地估算出运动模糊图像的模糊参量并取得较好的复原效果.-3 Abstract: The effect of motion blur with the depth of the image restoration algorithm. The use of motion-blurred images of the Fourier spectrum of the Radon transform to estimate the direction of motion blur in the direction of motion-blurred images to calculate the auto-correlation to estimate the motion blur length, then the direction of motion blur and motion blur parameters combined length of super-resolution image restoration algorithm to deal with more serious motion-blurred image recovery. The results show that the algorithm can be integrated more accurately estimate the motion-blurred images of the fuzzy parameters and obtain a better recovery results.
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Size: 965632 |
Author: zy |
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Description: 两层前馈网络，采用Hebb算法，实现机组协调控制-Two-tier feed-forward networks, Hebb algorithm used to achieve coordinated control
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Size: 1024 |
Author: 张宇 |
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Description: 数码管依次显示0到5（通过改变也可现实其他值），间隔500ms-Digital tube display followed by 0-5 (by changing the other values can also be a reality), 500ms interval
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Size: 13312 |
Author: 周利均 |
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Description: Z3惯导系统主要元部件,讲解了惯性系统元部件的知识-Z3 inertial navigation system is mainly yuan components, explained the components of the knowledge of inertial system yuan
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Size: 343040 |
Author: 秦金彪 |
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Description: 设有数组X和Y。X数组中有X1,…,X10；Y数组中有Y1,…,Y10。试编制程序计算Z1=X1+Y1,Z2=X2+Y2,Z3=X3-Y3,Z4=X4-Y4,Z5=X5-Y5,Z6=X6+Y6,Z7=X7-Y7,Z8=X8-Y8,Z9=X9+Y9,Z10=X10+Y10,结果存入Z数组。-Equipped with an array of X and Y. X array are X1, ..., X10 Y array are Y1, ..., Y10. Test preparation program to calculate Z1 = X1+ Y1, Z2 = X2+ Y2, Z3 = X3-Y3, Z4 = X4-Y4, Z5 = X5-Y5, Z6 = X6+ Y6, Z7 = X7-Y7, Z8 = X8- Y8, Z9 = X9+ Y9, Z10 = X10+ Y10, the results into an array Z.
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Size: 1024 |
Author: 陈玉 |
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Description: 主要 文本格式化
数据结构 课题源码
绝对好用
可以格式化文本
还有其他很多源码
我一个学期的结晶
如 约瑟夫环
进制转换
z3回文判断等等
共有18个课题-The main subject of text formatting source data structure can be absolutely easy to use formatted text there are many other source I have one semester of crystals, such as Joseph Central hex conversion z3 palindrome judgments, etc.
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Size: 17408 |
Author: 大龙虾 |
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Description: 设在内存区域中有数组X和Y。X数组有X1，X2，…，X10；Y数组有Y1，Y2，…，Y10，编程序实现下面的计算：
Z1=X1－Y1，Z2=X2＋Y2，Z3=X3－Y3，Z4=X4＋Y4，Z5=X5＋Y5，Z6=X6－Y6，Z7=X7－Y7，Z8=X8＋Y8，Z9=X9－Y9，Z10=X10＋Y10。计算结果Z存放在数组Z中。
为了便于检查结果，设数组X的内容为db 22,11,22,11,11,22,22,11,22,11,0,0,0,0,0,0存放于ds：0的内存中，为了便于计算，把数组X中的数补足16个；数组Y存放于ds：16开始的16个内存中，内容为11,88,11,88,88,11,11,88,11,88,0,0,0,0,0,0，而数组Z存放于ds：32的内存空间中，且数组值初始化为0。
数组X、Y、Z都是连续存放的！-Memory regions are located in an array of X and Y. X arrays are X1, X2, ..., X10 Y array with Y1, Y2, ..., Y10, programmed to achieve the following calculation: Z1 = X1-Y1, Z2 = X2+ Y2, Z3 = X3-Y3, Z4 = X4+ Y4, Z5 = X5+ Y5, Z6 = X6-Y6, Z7 = X7-Y7, Z8 = X8+ Y8, Z9 = X9-Y9, Z10 = X10+ Y10. The results stored in an array Z in Z. In order to facilitate the inspection results, set the contents of the array X as db 22,11,22,11,11,22,22,11,22,11,0,0,0,0,0,0 stored in ds: 0 Memory , in order to facilitate the calculation, the number of array X in the complement of 16 array Y stored in the ds: 16 of 16 starting in memory, says 11,88,11,88,88,11,11,88,11, 88,0,0,0,0,0,0, and stored in the array Z ds: 32 of the memory space, and the array value is initialized to 0. Array X, Y, Z are continuous storage!
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Size: 5120 |
Author: 杜宾 |
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Description: ARM嵌入式系统软件实时时钟的设计,pdf格式-ARM embedded system software design of real-time clock
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Size: 69632 |
Author: 张东 |
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Description: Crackme keygen based on z3
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Size: 7270400 |
Author: joelkurtz
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