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Description: Cubic.java是JAVA经典游戏范例，适合初学者学习使用-Cubic.java game is JAVA classic example of learning to use for beginners
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Size: 29994 |
Author: 肖苏 |
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Description: Cubic.java是JAVA经典游戏范例，适合初学者学习使用-Cubic.java game is JAVA classic example of learning to use for beginners
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Size: 29696 |
Author: 肖苏 |
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Description: 初学JAVA，求给2D绘制曲线的源代码，要三次样条的-Beginner JAVA, for drawing 2D curves to the source code, it is necessary to Cubic Spline
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Size: 1024 |
Author: ouyangfeng |
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Description: java网页上传程序试例 -java test page Uploader Example
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Size: 5581824 |
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Description: 数值算法,三次样条算法源码,动态演示过程,JAVA源码-Numerical algorithm, cubic spline algorithm source, dynamic presentation process, JAVA source
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Size: 1024 |
Author: 张辉 |
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Description: java语言编程实现：利用命令行参数解一元三次方程。
包含相关代码和运行结果
-java programming language: the use of command-line parameter cubic equation solution of one dollar. Contains the relevant code and operating results
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Size: 8192 |
Author: zhou |
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Description: 实现离散数据的三次样条插值的java语言实现-Realize discrete data of the cubic spline interpolation of the java language
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Size: 1024 |
Author: yangyanlong |
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Description: 三次样条插值，需要三次样条插值的可以下载这个包，也许对你有帮助-cubic spline curve
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Size: 2048 |
Author: shanshan |
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Description: Cubic Spline Interpolation
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Size: 15360 |
Author: EMIC |
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Description: Cubic spline algorithm approximating function y=f(x) - ((-2)*x*(2*x+3))/(x*x+4*x+5) in interval [-2 2]-Cubic spline algorithm approximating function y=f(x)- ((-2)*x*(2*x+3))/(x*x+4*x+5) in interval [-2 2]
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Size: 4096 |
Author: Vaidas Zid |
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Description: 用两种方法实现贝赛尔曲线的画法。Java库函数和贝赛尔曲线原理公式画法，用这两种方法画一个小动物。-Two methods to achieve Bezier curve drawing method. Java library functions and principles of Bezier curve formula for painting, using these two methods to draw a small animals.
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Size: 5120 |
Author: chen |
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Description: java计算器，能够算加减乘除，及开方立方的计算及其他功能-java calculator, to count addition, subtraction, and square root calculation and other functions of cubic
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Size: 44032 |
Author: 陈泽 |
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Description: 用JAVA编写的一个界面程序，实现了二分法、牛顿法、高斯法、SOR迭代法、三角分解法、三次样条插值曲线、曲线拟合的最小二乘法、数值积分Romberg算法、常微分方程的初值解法 改进Euler法、矩阵的特征值和特征向量 反幂法-An interface with a JAVA program written to achieve a dichotomy, Newton method, Gauss law, SOR iteration method, triangular decomposition, cubic spline interpolation curve, least squares curve fitting, numerical integration Romberg method, ordinary differential Solutions to improve the initial value equation Euler method, eigenvalues and eigenvectors of inverse power law.
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Size: 5521408 |
Author: Lynn |
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Description: 求花朵数的JAVA代码实现
一个N位的十进制正整数，如果它的每个位上的数字的N次方的和等于这个数本身，则称其为花朵数。
例如：
当N=3时，153就满足条件，因为 1^3 + 5^3 + 3^3 = 153，这样的数字也被称为水仙花数（其中，“^”表示乘方，5^3表示5的3次方，也就是立方）。
当N=4时，1634满足条件，因为 1^4 + 6^4 + 3^4 + 4^4 = 1634。
当N=5时，92727满足条件。
实际上，对N的每个取值，可能有多个数字满足条件。
程序的任务是：求N=21时，所有满足条件的花朵数。注意：这个整数有21位，它的各个位数字的21次方之和正好等于这个数本身。
如果满足条件的数字不只有一个，请从小到大输出所有符合条件的数字，每个数字占一行。因为这个数字很大，请注意解法时间上的可行性。要求程序在3分钟内运行完毕
-An N-bit decimal integer, if each of its N-bit numbers on the power and is equal to the number itself, called it the number of flowers.
For example:
When N = 3, the 153 to meet the conditions, because 1 ^ 3+ 5 ^ 3+ 3 ^ 3 = 153, that figure is also known as Narcissus number (where "^" means power, 5 ^ 3 said 5 3 power, which is cubic).
When N = 4, the 1634 to meet the conditions, because 1 ^ 6 ^ 4+ 4+ 4+ 4 ^ 3 ^ 4 = 1634.
When N = 5, the 92,727 to meet the conditions.
In fact, for each N value, there may be multiple numbers to meet the conditions.
Program s mission is: find N = 21, all meet the conditions of flower number. Note: This integer has 21 bits, each digit of its 21 th power, and exactly equal to the number itself.
If the figures do not meet the conditions of only one, make small to large output of all eligible numbers, each number per line. Because of this high number, please note that the feasibility of solution time. Claims process has finished running in 3 m
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Size: 2048 |
Author: 谷琼 |
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Description: 根据（0.0,0.0）（0.5,1.6）（1.0,2.0）（6.0,2.0）（7.0,1.5）（9.0,0.0）六个点使用三次样条差值求多项式并作图-According to (0.0,0.0) (0.5,1.6) (1.0,2.0) (6.0,2.0) (7.0,1.5) (9.0,0.0) using cubic spline six point margin and a polynomial mapping
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Size: 2048 |
Author: 回忆 |
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Description: 对一元一次，一元二次一元三次方程的求解,此题是java程序设计之网络编程的实验的源代码-On a dollar again, a quadratic equation Cubic, this problem is the network programming java programming source code for experiments
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Size: 1024 |
Author: 吴燕杰 |
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Description: Bspline曲线生成程序Catmull-Rom Spline, Lagrange, Natural Cubic Spline, and NURBS方法获得B样条曲线-Implementation of various mathematical curves that define themselves over a set of control points. The API is written in Java. The curves supported are: Bezier, B-Spline, Cardinal Spline,
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Size: 481280 |
Author: zhuwh |
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Description: 1. 最小二乘法拟合数据成直线。
2. 三次样条插值。
3. 两个3D演示程序（用程序直接在平面上画的三维图像）。-自动翻译的：1 a linear least-squares fitting the data. 2 cubic spline interpolation. 3 two 3D demo program (application directly on the painting in three-dimensional image plane).
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Size: 322560 |
Author: zbj |
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Description: java语言，找出0到1000之间的所有水仙花数，水仙花数就是各位数字立方和等于该数本身的数-java language, find out all the between 0 and 1000 the number of daffodils, narcissus and the number is, you figure cubic number equal to the number itself
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Size: 2048 |
Author: lishaoyang |
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Description: 内附三次样条插值、拉格朗日插值、积分、傅里叶变换、定态薛定谔方程求解程序,各程序有详细的说明-Included cubic spline interpolation, Lagrange interpolation, integration, Fourier transform, Schrodinger equation solver, a detailed description of each program
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Size: 6144 |
Author: 申小二 |
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