给定n个矩阵构成的一个链<A1,A2,...,An>，矩阵Ai的维数为pi-1 x pi，其中i=1,2,...,n 。
{ 3x5,5x2,2x7,7x4,4x8 }
该矩阵链的乘积为： A1A2...An。
通过在乘积A1A2...An的不同位置加上括弧，以先计算其中几个的乘积，再合并计算，以减少总共的计算代价。
Given n matrix consisting of a chain < A1, A2,..., An >, the dimensions of the matrix Ai for PI- 1 x PI, I = 1, 2,..., n.
{3 x 5, 5 x2, 2 x7, 7 x4, 4 by 8}
The product of matrix chain for: A1A2...The An.
Through the product A1A2...An the different position of the bracket, by calculating the product of several of them first, then merge computation, in order to reduce the calculation of total price. (2014-01-15, Visual C++, 819KB, 下载3次)
MIT教授编写的大名鼎鼎的翼型性能计算软件
Airfoil performance computing software (2013-08-20, Visual C++, 3811KB, 下载6次)
C语言实现动态规划 矩阵连乘问题！
问题：给定n个矩阵{A1,A2,…，An}，其中Ai与Ai+1是可乘的，i=1,2,…,n-1。计算这n个矩阵的连乘积。由于矩阵乘法满足结合律，故计算矩阵的连乘积可以由许多不同的计算次序。计算次序不同计算连乘积的计算量也不同。运用动态规划法思想给出你认为合理的计算矩阵连乘积最少次数的方案。
C language dynamic programming matrix continually multiply problems!
The problem: given n matrix {A1, A2, ..., An}, where Ai and Ai+1 is multiplicative, i = 1,2, ..., n-1. Calculating these n matrix product. Law, the calculation of the matrix with the product of order by many different computing Since matrix multiplication meet combined. Calculating order of different computing connected multiplied by the amount of calculation is also different. Given that you think that the reasonable calculation matrix with the product of at least the number of programs using dynamic programming thought. (2013-03-20, Visual C++, 1KB, 下载3次)
3、 矩阵（三）
编写C++程序完成以下功能：
（1） 用类来实现矩阵，定义一个矩阵的类，属性包括：
矩阵大小，用 lines, rows（行、列来表示）；
存贮矩阵的数组指针，根据矩阵大小动态申请（new）。
（2） 矩阵类的方法包括：
构造函数，参数是矩阵大小，需要动态申请存贮矩阵的数组；
析构函数，需要释放矩阵的数组指针；
拷贝构造函数，需要申请和复制数组；
输入，可以从cin中输入矩阵元素；
输出，将矩阵格式化输出到cout；
矩阵相加的函数，实现两个矩阵相加的功能,结果保存在另一个矩阵类，但必须矩阵大小相同；
矩阵相减的函数，实现两个矩阵相减的功能,结果保存在另一个矩阵类，但必须矩阵大小相同。
（3） 定义三个矩阵：A1、A2、A3；
（4） 初始化A1、A2；
（5） 计算并输出A3 = A1加A2，A3=A1减A2；
（6） 用new动态创建三个矩阵类的对象：pA1、pA1、pA3；
（7） 初始化pA1、pA2；
（8） 计算并输出pA3=pA1加pA2，pA3=pA1减pA2；
（9） 释放pA1、pA1、pA3。
3, matrix, (c)
Write a C++ program to perform the following functions:
(1) matrix class, define a matrix of class attributes include:
matrix size, lines, rows (rows and columns to represent)
storage matrix array pointer, dynamic application (new) according to the size of the matrix.
(2) matrix class method comprising:
constructor parameter is the size of the matrix, dynamic application memory matrix array
destructor release matrix array pointer
copy constructor, and the need to apply for and copy an array
The input from cin input matrix elements
output, the the matrix formatted output to cout
Function matrix sum, the sum of two matrices, and the results are stored in another matrix class, but must be the same as the size of the matrix
matrix function subtraction, subtract two matrices, and save the results in another matrix class, but it must be the same as the size of the matrix.
(3) d (2012-10-07, Visual C++, 1KB, 下载4次)
其中一个是是变步长复化梯形公式例程，该程序是用变步长复化梯形公式计算定积分的C语言程序。另一个是1.1.2 变步长复化抛物线公式例程，该程序是用变步长复化抛物线公式计算定积分的C语言程序。
One is the variable step size compound trapezoid formula routines, the program is the variable step size compound trapezoid formula to calculate definite integral C language program. The other is a1.1.2 variable step size composite parabola formula routines, the program is the variable step size composite parabola formula to calculate the definite integral of the C language program. (2012-06-13, Visual C++, 6KB, 下载8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)
给定n个矩阵A1，A2，…，An，其中，Ai与Aj+1是可乘的，i=1，2，…，n-l。
你的任务是要确定矩阵连乘的运算次序，使计算这n个矩阵的连乘积A1A2…An时总的元素乘法次数达到最少。
Given n matrices A1, A2, ..., An, which, Ai and Aj+1 is multiplicative in, i = 1,2, ..., nl. Your task is to determine the matrix multiplication operation with the order to calculate the product of n matrices even when A1A2 ... An element of the total number of multiplications to a minimum. (2011-11-04, Visual C++, 135KB, 下载3次)
给定n个矩阵{A1,A2,…,An}，其中Ai与Ai+1是可乘的，i=1,2,…,n-1。要算出这n个矩阵的连乘积A1A2…An。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。这种计算次序可以用加括号的方式来确定。若一个矩阵连乘积的计算次序完全确定，也就是说该连乘积已完全加括号，则可以依此次序反复调用2个矩阵相乘的标准算法计算出矩阵连乘积。本文的功能是采用动态规划算法，给出矩阵的一种最优的加括号方式，是计算量最小。
Given n matrices {A1, A2, ..., An}, where Ai and Ai+1 is a mere of, i = 1,2, ..., n-1. To calculate the n matrix with the product A1A2 ... An. Since matrix multiplication to meet with law, so the calculation of the matrix with the product can have many different calculation order. This calculation can be bracketed in order to determine ways. Even if the product of a matrix calculation to determine the complete sequence, that the product is fully bracketed with, so you can call the order of 2 repeated the standard matrix multiplication algorithm to calculate the matrix with the product. This feature is the use of dynamic programming algorithm, a matrix is given in parentheses best way is to calculate the minimum amount. (2011-06-30, Visual C++, 13KB, 下载6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)
矩阵运算：求逆、转置、相乘等，然后通过矩阵运算求多项式纠正的值，多项式如：X=a0+a1*x+a2*y+a3*x*x+a4*x*y+a5*y*y
Y=b0+b1*x+b2*y+b3*x*x+b4*x*y+b5*y*y
Matrix operations: inversion, transpose, multiply, etc., and then through the matrix for computing the value of correct polynomial, polynomials, such as: X = a0+ a1* x+ a2* y+ a3* x* x+ a4* x* y+ a5* y* y Y = b0+ b1* x+ b2* y+ b3* x* x+ b4* x* y+ b5* y* y (2009-06-10, Visual C++, 2KB, 下载12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)
Problem B:Longest Ordered Subsequence
A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 < i2 < ... < iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, e. g., (1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, e. g., (1, 3, 5, 8).
Problem B: Longest Ordered SubsequenceA numeric sequence of ai is ordered if a1 <a2 <... <aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK), where 1 <= i1 <i2 <... <iK <= N. For example, sequence (1, 7, 3, 5, 9, 4, 8) has ordered subsequences, eg, ( 1, 7), (3, 4, 8) and many others. All longest ordered subsequences are of length 4, eg, (1, 3, 5, 8). (2008-11-16, Visual C++, 1KB, 下载4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)
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)
设有n种不同面值a1, a2,…, an的邮票，规定每封信最多贴m张邮票。对于给定的m，n，求出最大的邮资连续区间。
With a face value of n different types a1, a2, ..., an of the stamps, each letter provides a maximum of m stamps affixed. For a given m, n, obtained the largest continuous interval of postage. (2008-06-02, Visual C++, 165KB, 下载7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)
对于给定的n 个整数a0, a1,., an-1 组成的序列，计算相应的亲兄弟元素位置序列。
For a given n integers a0, a1,., An-1, composed of sequences, calculation of the corresponding elements of the location of sequence brothers. (2007-11-13, Visual C++, 1KB, 下载4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)
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)
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)