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mathvalue_1
- 从人口普查统计，已知某国新生儿母亲的年龄累计分布为N=N(t)，其中t为母亲年龄，N为新生儿的母亲年龄低于或等于t的新生儿数目。下表给出了间距为5的数据。 t 15 20 25 30 35 40 45 50 N 0 7442 26703 41635 49785 50209 50226 50230 取边界条件为N’(15)=N’(50)=0,利用三次样条插值求出15至50岁之间，每一年龄（即x=15,16,……，50）所对应的新生儿累计数目N，并绘出插值函数的图形。
TVAL3
- %TVDENOISE Total variation grayscale and color image denoising % u = TVDENOISE(f,lambda) denoises the input image f. The smaller % the parameter lambda, the stronger the denoising. % % The output u approximately minimizes the Rudin-Osher-Fatemi (ROF)
17-2
- 编译：mpicc intersect.c –o intersect 运行：可以使用命令 mpirun –np SIZE intersect来运行该串匹配程序，其中SIZE是所使用的处理器个数，本实例中使用了SIZE=3个处理器。 mpirun –np 3 intersect 运行结果： 输入： please input first polygon please input the count of vetex:3 please input the vertex
LinearBPM
- 这是一个简单的FFT为无电位1 +1 +1 Ð 薛定谔方程的光束传播方法。如果该软件灵活，允许引入的术语（如果是极少数需要包括色散效应高阶导数）。例如，如果一个人渴望解决的一个方程的形式： （四/ dz的+ Ð ^ 2/dx ^ 2 - 0.25 * Ð ^ 3/dx ^ 3）== 0 Ÿ 它可以解决使用以下代码： Ž = linspace（0,1,512） x = linspace（-5,5,1024） psi0 =进出口
spline_VB
- 三次样条函数 Type coordinateX Y x As Double y As Double End Type Public Const dotSize = 25 Public Const H = 4000 Public Const W = 4-Cubic Spline
LMS-RLSAdaptiveFilter
- 数字信号处理，LMS和RLS实例：给定正弦信号s(n)，现在我们获得得是受影响的数据x(n)=s(n)+v(n) ， v(n)为方差1.25的告示白噪声信号，请设计一个滤波器，使其输出与s(n)的均方误差最小，并给出用LMS和RLS算法的自适应求解方法的MATLAB仿真。-Digital signal processing, LMS and RLS instance: Given a sinusoidal signal s (n), now we get the data have affect
choice-of-radio-freq
- C,C++ Questions 1. Base class has some virtual method and derived class has a method with the same name. If we initialize the base class pointer with derived object,. calling of that virtual method will result in which method being called? a.
Virtual-router
- C,C++ Questions 1. Base class has some virtual method and derived class has a method with the same name. If we initialize the base class pointer with derived object,. calling of that virtual method will result in which method being called? a.
Index
- C,C++ Questions 1. Base class has some virtual method and derived class has a method with the same name. If we initialize the base class pointer with derived object,. calling of that virtual method will result in which method being called? a.
DBMM
- 单表密码 定义： 1、 明表： A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 2、 明文 they will arrive tomorrow 3、 密钥 K=Monday 4、 密码实现 1） 将明文与密钥转为数字串： K=(12,14,13,3,0,24)
bemlabecom
- Boundary element method function [index,g,h,a,unb,x,u,un,uin]=bemlabecom(n,c,index,in,cin) [index,g,h,a,unb,x,u,un,uin]=bemlabecom(16,[0 0 0.25 0 0.5 0 0.75 0 1 0 1 0.25 1 0.5 1 0.75 1 1 0.75 1 0.5 1 0.25 1 0 1 0 0.75 0 0.5 0 0.25 0 0],[1 0 1 0
bp_application_2
- 用BP神经网络实现输入x=0.3*sin(i*pi/50)+0.4*sin(i*pi/25)到输出y(n)=1/(1+3*y(n-1)+x(n)*x(n))的近似模拟，其中y（0）=0.15-BP neural network input x = 0.3* sin (i* pi/50)+0.4* sin (i* pi/25) to the output y (n) = 1/(1+3* y-(n-1)+approximate simulation of x (n)* x (n)), where y
newton
- f(x)=1/(1+25*x^2),-1<x<1,在n=5,10,15,20时 牛顿插值误差计算，样条误差计算-F (x) = 1/(1+ 25* x ^ 2), 1<x<1, when n = 5,10,15,20 Newton interpolation error calculation, spline error calculation
CPSO
- 用CPSO算法求解带约束的函数优化问题。目标函数为f=(x(3)+2)*x(2)*x(1)^2，求其最小值；约束为g(1)=1-x(2)^3*x(3)/(71785*x(1)^4) g(2)=(4*x(2)^2-x(1)*x(2))/(12566*(x(2)*x(1)^3-x(1)^4))+1/(5108*x(1)^2)-1 g(3)=1-140.45*x(1)/(x(2)^2*x(3)) g(4)=(x(1)+x(2))/1.5-1 决策变量取值范围：x1-[0.05,
matlab
- 三条样条插值函数程序x=[1 4 9 16 25 36 49 64 81] y=[1 2 3 4 5 6 7 8 9] -Three spline interpolation procedure x = [1 4 9 16 25 36 49 64 81] y = [1 2 3 4 5 6 7 8 9]
matlab
- 一个二维静电场，电位函数为φ（x, y），将正方形分成25个正方形网格。有16个内部网格点。假定16个网格点的初值都定为零，试用超松弛法确定16个内部网格点的电位值。-A two dimensional electrostatic field, the potential function for phi (x, y), the square into 25 square grid. There are 16 internal grid point. Assume that 16 mesh po
TRADITIONAL-Z-SOURCE-CONVERTER
- The Z-source converter (ZSC) is a new topology in power conversion, which has unique features that can overcome the limitations of VSI and CSI [5]. Figure 2.1 shows the ZSC implemented as a 3-phase DC/AC converter (inverter). Although DC/AC conver
sgu262
- sgu262：Symbol Recognition(状压DP) 题目大意： 给出k个n∗ m的01矩阵Si，求出一个1尽量少的n∗ m的01矩阵P，满足k个矩阵与该矩阵的交互不相同，也就是说通过该矩阵能表示出给出的k个矩阵。-262. Symbol Recognition time limit per test: 0.25 sec. memory limit per test: 65536 KB input: standard output:
EKF
- 扩展卡尔曼滤波，非线性EKF的仿真与用法。描述了两个非线性方程x=0.5 x + (25 * x / (1 + x^2)) + 8 * cos(1.2*(t-1)) y = (x^2 / 20)的EKF的用法。简单实用，有备注。对于卡尔曼滤波初学者很有用。-Extended Kalman filter, nonlinear EKF simulation and usage. Describes two nonlinear equations x = 0.5 x+ (25* x/(1+ x ^ 2
UKF
- 很标准的无味卡尔曼滤波，UKF的仿真与用法。描述了两个非线性方程x=0.5 x + (25 * x / (1 + x^2)) + 8 * cos(1.2*(t-1)) y = (x^2 / 20)的UKF的用法。简单实用，有备注。对于卡尔曼滤波初学者很有用。-Very standard unscented Kalman filter, UKF simulation and usage. Describes two nonlinear equations x = 0.5 x+ (25* x/(1+