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求对称三对角阵的全部特征值与特征向量
- 利用C语言编译实现对对称三对角阵求解其全部特征值与特征向量-C language compiler to achieve symmetric diagonal matrix solving all their eigenvalues and eigenvectors
cholesky分解法
- cholesky算法，解三对角矩阵的常用方法-cholesky algorithm, and three diagonal matrix method used
C语言实现追赶法
- C语言实现追赶法 可方便三对角线矩阵的运算-C language to catch up with the law will facilitate the three diagonal matrix arithmetic
n后问题
- 在 n 行 n 列的国际象棋棋盘上，最多可布n个皇后。 若两个皇后位于同一行、同一列、同一对角线上，则称为它们为互相攻击。 n皇后问题是指找到这 n 个皇后的互不攻击的布局。 n 行 n 列的棋盘上，主次对角线各有2n-1条。-n n trip out of the international chess board, the maximum n Queen's cloth. If two at the same Queen's visit, the same s
matrilass
- if(max<=1e-5) //三角矩阵的对角元等于0则无解 { cout<<\"no inverse array\\n\" exit(0) } if(line!=i) { swap(p,i,line) swap(q,i,line) } for (k=0 k<N k++) -if (maxlt; = 1e-5) / / triangular matrix diagonal yuan equals 0 (coutlt
zhuigan
- 追赶法，可以求解三对角 矩阵，非常可靠，经过本人多次验证-chasing method can solve three-diagonal matrix, very reliable, after I repeatedly verified
奇数阶幻方
- 幻方即1至n*n自然数排成的每行、每列、每条对角线数字之和均相等的方阵。这是一个流行的解法。 含源文件及说明-that is a magic square to n * n natural numbers line up every trip, every out, and each diagonal figures and have the same matrix. This is a popular method. Sources with papers and notes
duijiaoxian
- 用VC编写程序绘制对角线，运行程序包中后缀名为.dsw文件即可-VC mapping procedure to prepare the diagonal, running package suffix named. Dsw document can be
dayinmofang
- 打印“魔方阵”。所谓“魔方阵”是指这样的方阵，它的每一行、每一列和对角线之和均相等。例如，三阶魔方阵为: 8 1 6 3 5 7 4 9 2 要求打印出由1到n*n的自然数构成的魔方阵。-Print "magic matrices." The so-called "magic matrix" refers to the matrix, which each line, each diagonal out and the sum of all the sam
ImageProcessing5
- 图像高斯模糊处理，利用log边缘算子法对图像进行边缘提取，对象进行对角镜处理等-Gaussian blur image processing, the use of log marginal operator on image edge detection, object processing, such as diagonal mirror
nmagic
- N阶磨方阵算法代码。用于生成N阶行、列和对角线之和相等的方阵。-N-order square algorithm code mill. Used to generate N-order rows, columns and diagonal and equal to the square.
jiugeqi
- C++下实现的简单九格棋人机对弈，黑白任一方只要连成一行或列及对角线即可获胜-C++ 9 under a simple man-machine chess game grid, black and white as long as either linked or columns and diagonal line to win
eightqueens
- 八皇后问题 任何两个皇后都满足：1、不在同行， 2、不在同列， 3、不在同一对角线-Eight Queen' s any question the Queen' s are both met: one, not peer, 2, not in the same column, 3, is not the same diagonal
Exercise5_26
- 三字棋游戏，两个游戏者在3*3的网络中轮流做标记，一个人用x，一个人用o。如果一个游戏着在网络的水平、垂直或者对角线方向上作了3个连续的标记，游戏就以这个游戏者得胜而告终。当网格所有的单元格都填满了标记而没有一个游戏者获胜的话，就出现平局。编写一个玩三字棋游戏：1、程序提示第一个游戏者输入x标记，接着提示第二个游戏者输入o标记。每输入一个标记后，程序刷新棋盘并显示游戏状态（获胜、平局或者未结束）2、为了输入标记，通过输入对话框提示用户输入标记的行和列坐标。-Words chess games,
queen
- N皇后问题，N皇后放置过程的现实。 由八皇后问题扩展开来，在N*N的棋盘上摆放N个皇后，使其不能互相攻击，即任意两个皇后都不能处于同一行、同一列或同一斜线上。求可能的解决方案，并显示每一种解决方案的皇后放置过程。-N Queen' s problem, N Queen placed the reality of the process. Expansion by opening eight to question Queen' s, at the N* N chessboard
LCS
- LCS问题就是求两个字符串最长公共子串的问题。解法就是用一个矩阵来记录两个字符串中所有位置的两个字符之间的匹配情况，若是匹配则为1，否则为0。然后求出对角线最长的1序列，其对应的位置就是最长匹配子串的位置。 最长递增子序列LCS的实现C源码！-LCS problem for two strings is the longest common sub-string problem. Solution is to use a matrix to record the location of two
source
- 实现一个三子棋博弈系统，支持图形用户界面和智能博弈，电脑和用户双方对战，轮流放X或O在3*3的棋盘上，能够在水平、垂直或对角线上连成三子者获胜。-Implementation of a chess game three sub-systems to support graphical user interface and smart game, against both computer and user, alternately place X or O in the 3* 3 on the
Three-diagonal-matrix
- 三对角阵的LU分解和三对角方程组的求解.可以用于油藏数值模拟-Diagonal matrix of the LU decomposition of tridiagonal equations and can be used in numerical reservoir simulation
diagonal
- C语言计算对角线之和，适合初学者使用，代码简单易懂。-C diagonal calculation combined, suitable for beginners to use, code easy to understand.
HMM-with-skips-and-single-Diagonal-Gaussian-Sourc
- HMM with skips and single Diagonal Gaussian SourceForge 1.03