-
Using java to develop horse line issue and realize the different ways in each ha...
使用java开发的马行线问题,实现方式各不不同
在半张中国象棋的棋盘上,一只马从左下角跳到右上角,只允许往右跳,不允许往左跳,问能有多少种跳步方案。(绝对原创)-Using java to develop horse line issue and realize the different ways in each half of Chinese chess chessboard, a horse jumping from the upper right corner of the lower-left corner, allowing only the right jump, jump to the left are not allowed to ask how many step-by-step program of dancing. (Absolute original)
- 2022-06-03 08:04:09下载
- 积分:1
-
坦克游戏的源码,并可以地图编辑
坦克游戏的源码,并可以地图编辑-tanks games source, and can map editors
- 2022-02-04 01:45:36下载
- 积分:1
-
digital image processing application
digital image processing application
- 2023-01-30 22:45:04下载
- 积分:1
-
一种小小的乘法器,基于VC++,可以说是一个比较简单的练习MFC与C++的小程序。...
一种小小的乘法器,基于VC++,可以说是一个比较简单的练习MFC与C++的小程序。-A small multiplier, based on VC++, Can be said to be a relatively simple exercise MFC and C++ Small procedures.
- 2023-04-11 10:10:04下载
- 积分:1
-
介绍如何用单片机配置FPGA,仅供参考,如有疑问请联系我。
介绍如何用单片机配置FPGA,仅供参考,如有疑问请联系我。-how to use microcontroller FPGA configuration, for reference purposes only, if any questions, please contact me.
- 2022-03-16 18:15:24下载
- 积分:1
-
预测控制是适用于含有大滞后环节、积分环节的复杂工业系统控制。GPC为广义预测控制...
预测控制是适用于含有大滞后环节、积分环节的复杂工业系统控制。GPC为广义预测控制-predictive control is applicable to lag links with big, integral part of the complex industrial control systems. GPC for GPC
- 2022-08-12 19:13:56下载
- 积分:1
-
RSS的开源项目.很方便的解析和生成标准的RSS的XML文件.
RSS的开源项目.很方便的解析和生成标准的RSS的XML文件.-RSS revenue item. Very convenient analytical and generate standard RSS XML file.
- 2022-04-08 16:51:41下载
- 积分:1
-
function viewer function Viewer
函数查看器 函数查看器-function viewer function Viewer
- 2022-07-17 09:02:46下载
- 积分:1
-
Please read your package and describe it at least 40 bytes.
System will automat...
Please read your package and describe it at least 40 bytes.
System will automatically delete the directory of debug and release, so please do not put files on these two directory
- 2022-08-14 07:19:32下载
- 积分:1
-
problem with M
排列问题
M个1,N个0的排列(高效率版)
排列数为:c(m+n,n)
对n个0,m个1,我的想法是这样的:
每个排列可以分三段:
全0列,全1列, 子问题列
设各段长:r,s,t .子问题列就是 (n,m) = (n-r,m-s),其中0
- 2022-11-27 09:05:03下载
- 积分:1