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Description: I ve written some many years ago dynamic Huffman algorithm to compress and decompress data. It is mainly targeted to data with some symbols occuring more often than the rest (e.g. having some data file consisted of 3 different symbols and their total number of occurence in that file s1(1000), s2(200), s3(30) so the total length of file is 1000+200+30=1230 bytes, it will be encoded assigning one bit to s1 and 2 bits to s2, s3 so the encoded length will be 1*1000+2*(200+30)=1460 bits=182 bytes). In the best case the file consisted of just one symbol will be encoded with compression ratio as 1:8. Huffman coding is used in image compression, however in JPEG2000 arithmetic codec is imployed.
Platform: |
Size: 7401 |
Author: 毛磊 |
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Description: 组数游戏
n个正整数n<=20,联成一排，组成最大多位数。程序输入：n，程序输出：n个数连成的多位数。
提示
eg 输入 123 2 33 1006 12
先找出最大数字为四位，再将所有数字变成四位数：
1230 2000 3300 1006 1200
然后排序
3300 2000 1230 1200 1006
将后面加上的0去掉得
332123121006
自己编的,一次作业而已,全用的for while循环，很基础。
Platform: |
Size: 1102 |
Author: 张超 |
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Description: 组数游戏的第二种方法，同学编的：
组数游戏
n个正整数n<=20,联成一排，组成最大多位数。程序输入：n，程序输出：n个数连成的多位数。
提示
eg 输入 123 2 33 1006 12
先找出最大数字为四位，再将所有数字变成四位数：
1230 2000 3300 1006 1200
然后排序
3300 2000 1230 1200 1006
将后面加上的0去掉得
332123121006
Platform: |
Size: 1580 |
Author: 张超 |
Hits:
Description: I ve written some many years ago dynamic Huffman algorithm to compress and decompress data. It is mainly targeted to data with some symbols occuring more often than the rest (e.g. having some data file consisted of 3 different symbols and their total number of occurence in that file s1(1000), s2(200), s3(30) so the total length of file is 1000+200+30=1230 bytes, it will be encoded assigning one bit to s1 and 2 bits to s2, s3 so the encoded length will be 1*1000+2*(200+30)=1460 bits=182 bytes). In the best case the file consisted of just one symbol will be encoded with compression ratio as 1:8. Huffman coding is used in image compression, however in JPEG2000 arithmetic codec is imployed.
Platform: |
Size: 7168 |
Author: 毛磊 |
Hits:
Description: 组数游戏
n个正整数n<=20,联成一排，组成最大多位数。程序输入：n，程序输出：n个数连成的多位数。
提示
eg 输入 123 2 33 1006 12
先找出最大数字为四位，再将所有数字变成四位数：
1230 2000 3300 1006 1200
然后排序
3300 2000 1230 1200 1006
将后面加上的0去掉得
332123121006
自己编的,一次作业而已,全用的for while循环，很基础。-Group a few games n positive integers n
Platform: |
Size: 1024 |
Author: 张超 |
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Description: 组数游戏的第二种方法，同学编的：
组数游戏
n个正整数n<=20,联成一排，组成最大多位数。程序输入：n，程序输出：n个数连成的多位数。
提示
eg 输入 123 2 33 1006 12
先找出最大数字为四位，再将所有数字变成四位数：
1230 2000 3300 1006 1200
然后排序
3300 2000 1230 1200 1006
将后面加上的0去掉得
332123121006-Group a few games of the second approach, the students made: the group a few games n positive integers n
Platform: |
Size: 1024 |
Author: 张超 |
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Description: ads1230驱动程序,注释非常详细,编写风格清晰,欢迎下载-driver program
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Size: 1024 |
Author: yunfeng |
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Description: Turbo C2.0中文版 提供一个中文环境-Turbo C2.0 to provide a Chinese version of the Chinese environment
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Size: 2657280 |
Author: 牛猛猛 |
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Description: 介绍了ADS1230芯片的特性及其在称重系统中的硬件电路设计.-Application of ADS1230 in the balance system
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Size: 22528 |
Author: 古月雨山石 |
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Description: 网络商城系统
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Size: 15023104 |
Author: lilovecher |
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Description: 函数功能：
--- --- --- --- --- --- --- --- --- --- ----
用4个按键控制四个数码管的显示
四个按键分别k1，k2，k3，k4。
实现功能：
1. K1 单击，数码管上显示的数加1，加到9999后，再单击，显示0000
2. K2 单击 数码管上显示的数减1，减到0000后，再单击，显示9999
3. K1双击 数码管上显示的数加5，加到超过9999后，再双击，显示超出的数，如当前显示为9998，双击后，显示4
4. K2双击 数码管上显示的数减5，加到超过0后，再双击，显示超出的数与9999的和，如当前显示为1，双击后，显示9995
5. K1连击 数码管上的以步长为10的值增加，加到9999或超过，均显示超出的数，同上k1双击处理。
6. 如当前值为1200，则连击时，显示值为 1210 1220 1230……..
7. K2连击 数码管上的以步长为10的值增加，减到0或超过，均显示超出的数，同上k2双击处理。
8. 如当前值为1200，则连击时，显示值为 1190 1180 1170……..
9. K3 功能键 ，类似于shift
当按住k3,再按下k1,k2时，数码管显示以步长100的速度增减，显示处理同上
10. 无击 10秒后，数码显示为0000
11. K4 三击 数码管显示-Function:
----------------------------------------------------------------
With four key control of four digital display
Four buttons respectively, erp k1, k2 and k4.
Functions:
1 K1 click on display, digital pipe number 1, add to 9999, then click, display 0000
2 K2 click on the number of digital pipe showed reduced to 1, and then click 0000, display 9999
3 K1 double-click on the display of digital tube number 5, to more than 999, after that the number and double-click beyond that, if present, for 9998, 4. After double-click
4 K2 double-click on the number of digital tube that 5, to reduce after more than 0, then double-click the number and beyond, display, such as the current and 999 for 1, after that, 9995 double-click on
5 combo digital pipe K1 by incrementing 10, plus the value increases to 9999 or over, all show beyond number, alexandrine K1 double-click processing.
6 if the value is 1200, when the display value, for the combo 1210 1220 1230... ..
7. K2 combo digita
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Size: 3072 |
Author: leiyin |
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Description: Jinshan drug gangsters application device
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Size: 52224 |
Author: 贺强 |
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Description: 比如输入的是123，2，33，1006，12这样几个数字。
先找出最大的数字的位数是4位，再将所有 的数字变成4位数：
1230 2000 3300 1006 1200
然后进行排序：
-itoa () function takes three parameters: The first argument is the number to be converted, the second parameter is to be written to convert the results of the target string, the third parameter is used to transfer digital base (hex). In the above example, the conversion base 10, means 10 to convert hex. 10: decimal 2: binary ...
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Size: 2886656 |
Author: 杨峰 |
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Description: ADS1230驱动文件。此驱动用于MSP430单片机。-ADS1230 driver file. This driver for MSP430 microcontroller.
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Size: 1024 |
Author: 张兴伟 |
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Description: 区间相交问题
给定 x 轴上 n 个闭区间。去掉尽可能少的闭区间，使剩下的闭区间都不相交。
算法设计： 对于给定的 n 个闭区间，计算去掉的最少闭区间数。-Given X axis n on a closed interval. Remove as much as possible closed interval less, so that the rest of the closed interval are disjoint.
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Size: 406528 |
Author: wjf |
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Description: 平衡车程序,平衡车控制器，中央处理器平衡车
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Size: 2669568 |
Author: 陆浩 |
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Description: poj 1230
In modern day magic shows, passing through walls is very popular in which a magician performer passes through several walls in a predesigned stage show. The wall-passer (Pass-Muraille) has a limited wall-passing energy to pass through at most k walls in each wall-passing show. The walls are placed on a grid-like area. An example is shown in Figure 1, where the land is viewed from above. All the walls have unit widths, but different lengths. You may assume that no grid cell belongs to two or more walls. A spectator chooses a column of the grid. Our wall-passer starts from the upper side of the grid and walks along the entire column, passing through every wall in his way to get to the lower side of the grid. If he faces more than k walls when he tries to walk along a column, he would fail presenting a good show. For example, in the wall configuration shown in Figure 1, a wall-passer with k = 3 can pass from the upper side to the lower side choosing any column except column 6. -In modern day magic shows, passing through walls is very popular in which a magician performer passes through several walls in a predesigned stage show. The wall-passer (Pass-Muraille) has a limited wall-passing energy to pass through at most k walls in each wall-passing show. The walls are placed on a grid-like area. An example is shown in Figure 1, where the land is viewed from above. All the walls have unit widths, but different lengths. You may assume that no grid cell belongs to two or more walls. A spectator chooses a column of the grid. Our wall-passer starts from the upper side of the grid and walks along the entire column, passing through every wall in his way to get to the lower side of the grid. If he faces more than k walls when he tries to walk along a column, he would fail presenting a good show. For example, in the wall configuration shown in Figure 1, a wall-passer with k = 3 can pass from the upper side to the lower side choosing any column except column 6.
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Size: 488448 |
Author: cam |
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Description: 编写程序，令数码管的显示顺序为：0123,1230,2301,3012-Write a program to make digital tube display order: 0123,1230,2301,3012
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Size: 4096 |
Author: liu |
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Description: Java Web+Servlet+Ajax+MySQL的学生信息管理系统-Java Web+Servlet+Ajax+MySQL
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Size: 1305600 |
Author: kaijun |
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