搜索资源列表
railkk
- 这是个火车转轨的问题, 用堆栈实现的一个解决方法-This is a transition of the train, using the stack to achieve a solution
xsxxcx
- 基于java技术开发的网站管理系统，是一个经过完善设计并适用于各种服务器环境的易用、高效、快速、安全、优秀的网站解决方案-based on java technology development of the web site management system Following is a complete design and applied to a wide variety of server environment-friendly, efficient, quick, safe, e
TJU1127
- 同济大学ACM题库TJU1127的题解代码.-TJU1127 Tongji University ACM questions in the code that solution.
TJU1103
- 同济大学ACM题库TJU1103的题解代码.深度优先搜索的一个好例子.-TJU1103 Tongji University ACM questions in the code that solution. Depth-first search of a good example.
qzy
- 利用全主元高斯消去法对线性方程组的求解过程进行数值模拟与计算。-use all PCA Gaussian elimination of linear equations, the solution process simulation and numerical calculation.
maze8
- 可求出迷宫所有解和最优解，步骤：先创建迷宫：0表示通路，1表示障碍；再按m键计算；q键退出。-maze can be obtained for all solutions and the optimal solution, steps : first create the maze : 0 indicates that access, said one obstacle; Key then calculated m; q Key withdrawal.
SIPnat
- 非常实用的SIP穿越NAT解决方案，初学者也能看懂的文章-very practical SIP NAT traversal solution, beginners can also read the article
AI-widefirst
- 八数码宽度优先搜索算法,可以找到最优解,但会消耗更多的资源.-eight digital width first search algorithm to find the optimal solution, but it will consume more resources.
NonLinearEquation
- 通过此程序能够得到非线性方程组的数值解，程序中有详细的说明-through this procedure can be obtained by numerical solution procedure is described in detail in
smias
- Short Message Internet Access Solution（SMIAS）是结合中国移动通信总公司的“梦网计划”应运而生的一套基于短信的移动互联网应用解决方案。 本系统提供了面向服务提供商（ASP或ICP）的开发接口，服务提供商可以使用这些接口API实现与福建移动通信公司SMIAS系统的接入。 -Short Message Internet Access Solution (S 5d) with China Mobile Communication Corporation
Geddes
- L为桩长，r为计算点与桩轴线的水平距离，z为计算点距离地面深度。都取正值 ％f1,f2,f3依次是点荷载、均布荷载以及三角形荷载下的Geddes应力解函数 ％Ip,Ir,It依次是对应三个荷载下的应力系数，负表示拉，正表示压-pile length L, r points and to calculate the pile axis horizontal distance, z to calculate the distance depth of the ground. Take t
banana098429383342134123
- 香蕉问题的一个解决方法。个人兴趣。1234567890-banana issue a solution. Personal interests. 1234567890
8QUEENS-BEST
- 8皇后问题最好的算法,请登陆本站才能下载哦!这是最优的解法哦!-8 Queens best algorithm, please log onto the site can download! This is the best solution!
ActualJPEG
- 一个实际的JPEG压缩解压缩源码，每个M文件都带有详细的说明。-a practical solution for JPEG compression source, each M documents with detailed explanations.
defuzz
- 源代码有关MATLAB的解模糊代码。好好学习把 -MATLAB source code for the solution of the fuzzy code. Learning to make good
EightQueues
- 八皇后VC图形演 算法思想：回溯法，先在第1行放上一个皇后，然后在第2行合适的位置放上一个皇后，依次类推，如果8行都放满了，说明找到了一个解，如果第好第i行的皇后后，第i+1行找不到合适的位置，这时就回到第i行，把第i行的皇后放到下一个位置，继续尝试下一行。如此反复，知道找到所有的解。注意，这种算法找的解可能有等价的，某些解可由别的解经过旋转棋盘得到。-eight Queen's VC graphics algorithm thinking : A Retrospective, th
1800Decorations
- 该源码是一个问题的解决方法。问题是给你个长为L的串，串中可以出现n种字符，还给出m个子串，求有多少个长为n的只由这些字串组成的串。输入例子：4 5 6 ABB BCA BCD CAB CDD DDA 结果为2.而5 4 5 E D C B A的结果为625-source of the problem is a solution. The problem is that you have the head of the L series, the series can occur n ch
ffmpeg-20050624
- FFmpeg is a complete solution to record, convert and stream audio and video. It includes libavcodec, the leading audio/video codec library. FFmpeg is developed under Linux, but it can compiled under most operating systems, including Windows.-Sorenson
Dezip
- java的解压缩程序,希望大家批评指正,多发表意见-java solution compression program, we hope to correcting the criticism, express their views
Dynamic-encrypt-instruction
- 指令动态加密法。这儿讲述的是用单条指令加密法，再用 int 1 单步中断解下一条指令的第一字节，由于用另外程序解密时无法预知指令长, 所以不能用编程的方法解密，只能用手工一条一条地解。具体实现见注释，这种加密法的麻烦只处就是加密时也要一句一句来。-directive dynamic encryption method. Here is on the order of the single-encryption method, and then a single step int slips of