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Description: BlackFin 561的Filter的源码，详细介绍应用ADI533芯片构筑嵌入式多媒体系统的源码，更多详细内容可以在www.tayitone.com 上获取，或EMAIL给咱家-BlackFin 561 of the Filter source, detailed application ADI533 build chips embedded multimedia system source code, more details can www.tayitone.com access, or email to the window
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Size: 287797 |
Author: 李长宏 |
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Description: BlackFin 561的Filter的源码，详细介绍应用ADI533芯片构筑嵌入式多媒体系统的源码，更多详细内容可以在www.tayitone.com 上获取，或EMAIL给咱家-BlackFin 561 of the Filter source, detailed application ADI533 build chips embedded multimedia system source code, more details can www.tayitone.com access, or email to the window
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Size: 287744 |
Author: |
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Description: 这个是balckfin533/561的MPEG和Mjpeg的源代码，可以在BF533和BF561上跑，本人测试通过的-This is the MPEG and MJPEG balckfin533/561 source code, you can BF533 and BF561 on the run, I passed the test
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Size: 71680 |
Author: 郑英杰 |
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Description: 这个是balckfin533/561的视频输入和输出的源代码，可以在BF533和BF561上跑，本人测试通过的-Balckfin533/561 This is a video input and output of the source code, you can BF533 and BF561 on the run, I passed the test
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Size: 459776 |
Author: 郑英杰 |
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Description: 这个是balckfin533/561的视频输入和输出（针对LCD）的源代码，可以在BF533和BF561上跑，本人测试通过的-Balckfin533/561 This is a video input and output (for LCD) source code, you can BF533 and BF561 on the run, I passed the test
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Size: 15360 |
Author: 郑英杰 |
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Description: 生成多项式为（561，753）的（2，1，8）卷积码的编码、译码、ＡＷＧＮ信道下的误码率性能的测试程序-Q matrix to solve the problem, according to input parameters can be calculated as a Q band random matrices, satisfy the following conditions, each row of the matrix, each row, each diagonal are a number of 1
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Size: 8192 |
Author: zhangyang |
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Description: Balckfin533/561 This is a video input and output (for LCD) source code, you can BF533 and BF561 on the run, I passed the test
Platform: |
Size: 1626112 |
Author: iwan |
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Description: 若一个数（首位不为零）从左向右读与从右向左读都一样，我们就将其称之为回文数。
例如：给定一个10进制数56，将56加65（即把56从右向左读），得到121是一个回文数。
又如：对于10进制数87：
STEP1：87+78 = 165
STEP2：165+561 = 726
STEP3：726+627 = 1353
STEP4：1353+3531 = 4884
在这里的一步是指进行了一次N进制的加法，上例最少用了4步得到回文数4884。
写一个程序，给定一个N（2<=N<=10或N=16）进制数M，求最少经过几步可以得到回文数。
-If a number (the first non-zero) reading from left to right and from right to left the same time, we will call it palindrome. For example: Given a 10-decimal number 56, will be 56 plus 65 (that is read from right to left 56), by 121 is a palindrome. Another example: For 10 decimal number 87: STEP1: 87+78 = 165 STEP2: 165+561 = 726 STEP3: 726+627 = 1353 STEP4: 1353+3531 = 4884 step here is to a N band in addition, the patients received at least step 4 with a palindrome 4884. Write a program that, given a N (2 < = N < = 10 or N = 16) hexadecimal number of M, after seeking at least steps can be palindromes.
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Size: 2048 |
Author: 谢某 |
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Description: MPC561的can调试例程相关的资料，希望能对从事这方面开发的朋友有些须帮助-CAN measurement and calibration procedures and lower computer detailed documentation. can C language-CAN Calibration
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Size: 160768 |
Author: 毛毛 |
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Description:
若一个数（首位不为零）从左向右读与从右向左读都一样，我们就将其称之为回文数。
例如：给定一个10进制数56，将56加65（即把56从右向左读），得到121是一个回文数。
又如：对于10进制数87：
STEP1：87+78 = 165
STEP2：165+561 = 726
STEP3：726+627 = 1353
STEP4：1353+3531 = 4884
在这里的一步是指进行了一次N进制的加法，上例最少用了4步得到回文数4884。
写一个程序，给定一个N（2<=N<=10或N=16）进制数M，求最少经过几步可以得到回文数。
(十六进制数使用大写ABCDEF)
输入
N M
M的位数小于10-If a number (the first non-zero) reading from left to right and from right to left the same time, we will call it a palindrome. For example: Given a decimal number 10, 56, will be 56 plus 65 (that is, to 56 read from right to left), received 121 is a palindrome. Another example: For the 10 number 87: STEP1: 87+78 = 165 STEP2: 165+561 = 726 STEP3: 726+627 = 1353 STEP4: 1353+3531 = 4884 step here is the band conducted a N in addition, the patients received at least step 4 with the palindrome 4884. Write a program that, given a N (2 < = N < = 10 or N = 16) hexadecimal number M, seeking at least after a few steps can be palindrome. (Hexadecimal number using uppercase ABCDEF) Enter the number of bits is less than 10 NM M output
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Size: 1024 |
Author: ss |
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Description: 功能：release捕捉异常
编译选项中加入 /EHa即可有在release下有效
代码：VC6++
作者:依星
联系方式:
QQ:34596561
MSN:airen3339@hotmail.com
E-mail:airen3339@163.com
制作:2012-12-Function: release catch exceptions
Compiler option added/EHA can be effective in the next release
Code: vc6++
Author: Iridium
Contact:
QQ: 34,596,561
MSN: airen3339@hotmail.com
E-mail: airen3339@163.com
Production :2012-12
Platform: |
Size: 1692672 |
Author: 李明宇 |
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Description: 求1-20000之间的平方回文数
回文数也叫对称数，是指这类数的各位数字具有对称性，例如12121，通常，可以通过使用任何一个数字与其倒序数相加，再依次执行下去获得，例如：96+69 165，165+561 726，726+627 1353，1353+3531 4884。但是也有某些数字目前为止还不能验证是否可以通过上述操作获得回文数，例如数字196。平方回文数是指这类数字即是回文数，又是某个自然数的平方数，例如121，就是11的平方数，同时也是回文数。可以编写程序，通过遍历获得平方回文数。-Find the number of square returns between 1-20000
The number of palindromes is also known as symmetry, that is, the number of each digit of such a number of symmetry, such as 12121, usually, you can use any number and its inverse number is added, followed by the implementation of the next to get, for example: 96+69 165 , 165+ 561 726, 726+ 627 1353, 1353+ 3531 4884. However, there are certain numbers that have not been verified so far as to obtain the number of palindromes, such as the number 196, by the above operation. Square palindrome is the number of such palindrome that is the number, but also the square of a natural number, such as 121, is the square of 11, but also palindrome number. You can write a program to get the square palindrome by traversing.
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Size: 2048 |
Author: 水云间 |
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