Bayesian Filter.贝叶斯（Bayesian）滤波器的C++类库。包括卡尔曼滤波（kalman filter）、粒子滤波(particle filter)等。
Bayesian Filter. Bayesian (Bayesian) filters C Class. Including Kalman filter (Kalman filter). particle filter (particle filter). (2006-10-03, C/C++, 124KB, 下载1186次)
提供水蒸气热力性质计算源代码（C++基础）
provide steam thermodynamic properties calculation of the source code (C basis) (2006-03-16, Visual C++, 303KB, 下载192次)
用VB编写的多项式拟合程序
Public Function funPolynomial(Num As Long, x() As Single, y() As Single, Degree As Integer, AA() As Single) As Long
多项式曲线拟合 y=a0+a1*x+a2*x^2+an*x^n
Num为输入数据点个数
x()为输入数据点横坐标组成的数组
y()为输入数据点纵坐标组成的数组
Degree为要拟合的多项式曲线次数
AA() 为待求系数，为输出项
failed to translate (2009-10-16, Visual Basic, 5KB, 下载117次)
用最小二乘法拟合曲线y=a0+a1*x+a2*x^2+a3*x^3+...+an*x^n 的vc源码
using the method of least squares fitting curve y = a0 a1 a2* x* x* a3 ^ 2 x ^ 3 ... an* x ^ n vc FO (2007-06-01, Visual C++, 2KB, 下载96次)
//=== === === === === === === === === === === =====
//函数说明
//函数名称：Correlation
//函数功能：计算最小二乘法拟合的多项式的相关系数
//使用方法：int M------ 拟合多项式的阶数（已知条件）
// double *b--- 拟合曲线的系数,排列顺序为由高阶到低阶（已知条件）
// double *x--- 结点x轴数据（已知条件）
// double *y--- 结点y轴数据（已知条件）
// double *Yg-- 结点估计值，个数为m（过程变量）
// int m------ 结点个数（已知条件）
//注意事项：多项式阶数最高为10,多项式的形式为 y = a0 + a1x +a2x2
//=== === === === === ==== === === === === === === =// letter Description// function name : Correlation// Function functions : Calculation least squares polynomial fitting of the correlation coefficient// Use : int M------ polynomial fitting stage (known condition)//* b--- double fitting song The coefficient, Higher-order the order on grounds of low order (known condition)//* double-node x x axis data (known to be pieces)// double* y--- node y-axis data (known condition)//* double FSL-- Nodes estimates, Number m (process variables)// int m------ node number (known condition)// Note : polynomial order of a maximum of 10. polynomials in the form of y = a0 a1x a2x2 //==================== ======================================== (2006-10-27, C/C++, 1KB, 下载75次)
（1）利用多项式拟合的两个模块程序求解下题：
给出 x、y的观测值列表如下：
x 0 1 2 3 4 5
y 2.08 7.68 13.8 27.1 40.8 61.2
试利用二次多项式y=a0+a1x+a2x2进行曲线拟合。
（1）多项式拟合方法：假设我们收集到两个相关变量x、y的n对观测值列表：
x x0 x1 x2 x3 x4 x5
y y0 y1 y2 y3 y4 y5
我们希望用m+1个基函数w0(x),w1(x),…,wm(x)的一个线形组合
y=a0w0(x)+a1w1(x)+…+amwm(x)
来近似的表达x、y间的函数关系，我们把几对测量值分别代入上式中，就可以得到一个线形方程组：
a0w0(x0)+a1w1(x0)+…+amwm(x0)=y0
a0w0(x1)+a1w1(x1)+…+amwm(x1)=y1
… …
a0w0(xn)+a1w1(xn)+…+amwm(xn)=yn
只需要求出该线形方程组的最小二乘解，就能得到所构造的的多项式的系数，从而解决问题。
err (2007-11-16, Visual C++, 1KB, 下载70次)
钢筋混凝土简支梁基于材料非线性和几何非线性的ANSYS建模，得出位移，速度，加速度的响应，加载方式是地震波。
Reinforced concrete beams based on nonlinear material and geometric nonlinear ANSYS modeling, draw displacement, velocity, acceleration response, load the way seismic waves. (2011-11-12, Fortran, 2KB, 下载42次)
使用FFT实现的两个多项式相乘的算法。
输入文件：
第一行为(n-1)
第二行为第一个多项式系数序列 第三行为第二个多项式系数序列
系数序列的格式为：an,an-1,an-2 ,…, a1,a0
输出文件：result5.txt
格式为结果多项式的系数序列。
序列格式为：an,an-1,an-2 ,…, a1,a0（此n与输入中的n不同）
Implementation of both the use of FFT polynomial multiplication algorithms. Input file: the first act (n-1) the second act of the first polynomial coefficient sequence of the third act of the second polynomial coefficient sequence coefficient sequence format: an, an-1, an-2, ..., a1, a0 output file: result5.txt format for the results of polynomial coefficient sequence. Sequence format: an, an-1, an-2, ..., a1, a0 (the n with the input of the n different) (2009-03-15, Visual C++, 2KB, 下载34次)
此程序为二进制9-7小波提升格式的变换，算法中只有整数加法和移位
this procedure to upgrade binary 9-7 wavelet transform format, the algorithm only integers and translocation (2006-05-30, Visual C++, 2KB, 下载33次)
求满足 M > N > 0 的两个正整数之和的最大公约数.
Order to meet the M> N> 0, two positive integers and the greatest common divisor. (2009-11-09, Pascal, 1KB, 下载30次)
验证哥德巴赫才:任何一个充分大的偶数N(≥4),可以用两个素数之和表示.
如 4=2+2 6=3+3 8=3+5 98=17+79
Verify Goldbach before: Any one sufficiently large even number N (≥ 4), can be expressed the sum of two primes. Such as 4 = 2+2 6 = 3+3 8 = 3+5 98 = 17+79 (2009-11-09, Pascal, 1KB, 下载24次)
DLT645 检验和、+/-33计算小软件，附带源码。
DLT645 inspection and+/-33 computing software, with source code. (2012-07-07, C++ Builder, 489KB, 下载22次)
用C语言编写的有限差分法，求解一维非稳态温度场的小程序
Using C language finite difference method, solving the problem of two-dimensional steady-state temperature field (2012-02-24, Visual C++, 1KB, 下载21次)
采用NLJ随机搜索的方法辨识一个以状态方法表示的非线性系统。选其初值 a1(0) =50 , a2(0) =100 , a3(0) =100 , a4(0) =50 , a5(0) =10 , 选范围为 r(1)(i)=0.5 a(0)(i) , 取数据长度 L ＝40, t =0.005 , 性能指标 J＝ 。迭代计算结果得 a 的估计值 1=17.6043243, 1=17.5977, 2=72.9573, 3=51.3014, 4=22.9889, 5=5.99965, J = 0.000000916 。
Using NLJ random search method to identify a method that the state of the nonlinear system. Election of its initial value a1 (0) = 50, a2 (0) = 100, a3 (0) = 100, a4 (0) = 50, a5 (0) = 10, choose the range of r (1) (i) = 0.5 a (0) (i), check data length L = 40, t = 0.005, performance index J =. Iterative calculation results have a estimated value of 1 = 17.6043243, 1 = 17.5977, 2 = 72.9573, 3 = 51.3014, 4 = 22.9889, 5 = 5.99965, J = 0.000000916. (2008-09-12, Visual C++, 222KB, 下载20次)
灰色残差GM（1,1）模型，使用前，建立a1.txt文件。第一列年份，第二列统计数据
Grey residual GM (1,1) model, prior to use, to establish a1.txt file. Year of the first column, the second column statistics (2013-06-19, matlab, 3KB, 下载18次)
最佳矩阵连乘 给定n个矩阵｛A1,A2,…An｝，其中Ai与A i+1是可乘的，i=1，2…，n-1。考察这n个矩阵的连乘积A1A2…An。矩阵A和B可乘的条件是矩阵A的列数等于矩阵B的行数。若A是一个p×q矩阵，B是一个q×r矩阵，则其乘积C=AB是一个p×r矩阵,需要pqr次数乘。
由于矩阵乘法满足结合律，故计算矩阵的连乘积可以有许多不同的计算次序。例如，设3个矩阵{A1,A2,A3}的维数分别为10×100，100×5，和5×50。若按加括号方式((A1A2)A3)计算，3个矩阵连乘积需要的数乘次数为10×100×5+10×5×50=7500。若按加括号方式(A1(A2A3))计算，3个矩阵连乘积总共需要10×5×50+10×100×50=75000次数乘。由此可见，在计算矩阵连乘积时，加括号方式，即计算次序对计算量有很大影响。
矩阵连乘积的最优计算次序问题，即对于给定的相继n个矩阵｛A1,A2,…An｝(其中矩阵Ai的维数为pi-1×p，i=1,2，…，n)，确定计算矩阵连乘积A1,A2,…An的计算次序，使得依此次序计算矩阵连乘积需要的数乘次数最少。
best matrix continually multiply given n matrix (A1, A2, ... An), Ai and A is a mere i, i = 1, 2 ..., n-1. N explore the link matrix product ... An A1A2. Matrices A and B can either condition is out of the matrix A few matrix B is the number of rows. If A is a p q matrix B is a q-r-matrix, its product C = AB is a p r matrix, the number required by pqr. Because matrix multiplication meet the law of combination, it's even calculated matrix product can be calculated in many different priorities. For example, the matrix-based 3 (A1, A2, A3) dimension of 10 100, 100 5 5 and 50. If bracketed by the way ((A1A2) A3), even three product matrix multiplication in the number of 10 100 10 5 50 = 7,500. If bracketed by the way (A1 (A2A3)), three matrix product even need a total of 10 5 50 (2005-11-28, C++ Builder, 6KB, 下载18次)
试写一算法,实现顺序表的就地逆置,即利用原表的存储空间将线性表(a1,a2,...,an)逆置为(an,an-1,...,a1).
try to write an algorithm to achieve the order form in situ reverse home, namely the use of the original table of linear storage space (a1, a2 ,..., an) inverse home (an, an-1 ,..., a1). (2005-06-24, C/C++, 1KB, 下载17次)
计算机安全学中椭圆曲线公钥系统的c源码
输入与输出：
请输入椭圆曲线方程y^2=x^3+cx+d(mod p)中c,d,p的值：8,10,23
椭圆曲线方程为y^2=x^3+8x+10(mod23)
请输入所取明文x的x1,x2：19,13
请输入选择的椭圆曲线上的点a0的x,y：7,8
请输入私钥a=17
选取t=3
加密的结果是(y0,y1,y2) = ((22,22),20,18)
实施解密:
(c1,c2)=(18,12)
x1=19
x2=13
Computer Security Studies in the elliptic curve public key system, c-source input and output: Please enter the elliptic curve equation y ^ 2 = x ^ 3+ cx+ d (mod p) in the c, d, p values: 8,10,23 elliptic curve equation y ^ 2 = x ^ 3+8 x+10 (mod23) Please enter the plaintext x by taking the x1, x2: 19,13 Please enter a choice of elliptic curve point a0 of the x, y: 7,8 Please enter a private key a = 17 select t = 3 encrypted result is (y0, y1, y2) = ((22,22), 20,18) implementation of the decryption: (c1, c2) = (18,12) x1 = 19 x2 = 13 (2009-12-24, Visual C++, 12KB, 下载15次)
Matrix Chain Multiplication is perhaps the quintessential example of dynamic programming. The problem can be stated as follows: given a chain <A1, A2,..., An> of n matrices, where for i = 1, 2,...,n, matrix Ai has dimension pi-1 x pi, fully parenthesize the product A1A2...An in a way that minimizes the number of scalar multiplications.
Matrix Chain Multiplication is perhaps the quintessential example of dynamic programming. The problem can be stated as follows: given a chain <A1, A2,..., An> of n matrices, where for i = 1, 2,...,n, matrix Ai has dimension pi-1 x pi, fully parenthesize the product A1A2...An in a way that minimizes the number of scalar multiplications. (2011-04-18, Java, 1KB, 下载13次)
（1）Msls分三步对系统和噪声模型进行辨识，采用脉冲序列作为辅助系统模型，用 计算输出数据 ；用原输出数据 计算 ，用递推最小二乘方法分别对系统参数和模型参数进行估计。
（2）M.dat,wnoise1.dat分别为M和白噪声序列。Wnoise1.dat的长度为700，wnoise2.dat的长度为1000。Msls6.c为N=600的程序，Msls8.c为N=800的程序。
（3）程序运行后，生成的两个h文件为产生的脉冲响应函数。Msls6.dat为msls6.c的参数估计结果，msls8.dat为msls8.c的参数辨识结果。分别如下所示：
a1=0.906331 a2=0.160170 a3=0.025525 b1=0.704475 b2=-1.497551 c1=1.009114 c2=0.446890
a1=0.906347 a2=0.159066 a3=0.024650 b1=0.700720 b2=-1.493327 c1=1.008787 c2=0.425714
（4）由数据结果可以看出，采用msls辨识方法估计精度要比els法的估计精度差一些。尤其是噪声参数c2的估计误差不在1％以内。这是由于msls法计算上较为简便，计算上的简化就带来了估计精度上的误差。由N=600和N=800相比较，可以看出当N增大时，误差有所减小。理论上当N趋于无穷时， 。
err (2008-09-12, Visual C++, 270KB, 下载13次)