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[Mathimatics-Numerical algorithms] file_173 DL : 0: 平面内两点间的距离公式-plane the distance between two points formula
Date : 2008-10-13 Size : 95.75kb User : gg
[Mathimatics-Numerical algorithms] file_173 DL : 0: 平面内两点间的距离公式-plane the distance between two points formula
Date : 2025-12-29 Size : 95kb User : gg
[Mathimatics-Numerical algorithms] LS DL : 0: 利用Fortran程式計算平面座標上各點的最適線，其座標上各點到此線段的距離為最小，此法稱為最小方差法，InputX及InputY分別為平面座標上各點的X座標及Y座標，輸入完成後，執行LS.EXE，接著在輸入所要計算的線段的最高次方，以及座標點的個數即可，接著程式便會產生一個OUTPUT，內容從上而下分別為最高次方項至常數項，舉例來說，若最高次方項為2。則計算出的結果第一項便為y=ax+b的a，第二行便為b`.-Fortran program using the calculation plane coordinates at various points along the optimal line, its coordinates at various points along this line segment the smallest distance, this method is called minimum variance method, InputX and InputY were points on the plane coordinates X and Y coordinates coordinates, input is completed, the implementation of LS.EXE, and then calculated in the input segment to be the highest-th power, as well as the number of coordinates of points you can, then the program will generate an OUTPUT, the content from the top, respectively the highest-th power entry to the constant term, for example, if the maximum entry of 2-th power. Then the calculated results of the first will be for the y = ax+ b of a, the second line will be for the b `.
Date : 2025-12-29 Size : 132kb User : 言靈
[Mathimatics-Numerical algorithms] PPingRelation DL : 0: 摘要针对参数样条曲面变形或近似展开时变形或近似展开前后点的对应关系的确定问题，提出了一种算法:根据参数样条曲面与其参数平面上点的一一对应关系，用参数平面将变形或近似展开前后的曲面联系起来，通过由坐标反解参数和由参数正解坐标，建立起其上点的对应关系。 -Abstract The deformation parameters of spline surfaces or near or near the commencement of deformation around the point to start the corresponding relationships to identify problems, propose an algorithm: According to the parameters of spline surface and its parameters-one correspondence between points on the plane, with the parameter plane will be deformation started around or near the surface linked by the anti-solution parameters and coordinates are solutions of the coordinates by the parameters and establish the correspondence between the points.
Date : 2025-12-29 Size : 125kb User : 周保兴
[Mathimatics-Numerical algorithms] 5_9_4 DL : 0: 设计并实现平面内点类POINT。要求计算两点之间距离-Design and implementation of point-plane type POINT. Requires calculating the distance between two points
Date : 2025-12-29 Size : 865kb User : 张婧
[Mathimatics-Numerical algorithms] Princess-Dido-story DL : 0: 狄多公主圈地的故事 . 为了建设城市，她需要从当地一片森林中的拥有者那里买一些木材，交易条件是这样的，在有限面积的森林内，散布着一些树木，而不同的树可得到的木材体积也不相同。狄多需要划出一个圆形区域，区域内的树木皆归她所有，当然，树木分布位置并不规则。请大家帮助狄多，在树木总数与分布情况，每棵树的可得到的木材体积已知的情况下，算一下最大可圈出的木材体积是多少。可以将树木看做平面内的N个点，以点的权值M0，M1，...Mn表示每棵树可出产的木材体积。圆的半径为R(注：在圆边上的点不算圈入圆内)，求以圆圈地，最大可以圈到多大的权值？-Princess Dido story enclosure can be seen as the plane tree of N points to the right point value M0, M1, ... Mn that can be produced per tree wood volume. Circle of radius R (Note: the point is not in the round edge of the circle into the circle), seeking to circle, the maximum you can ring to much weight?
Date : 2025-12-29 Size : 1kb User : NoKoo
[Mathimatics-Numerical algorithms] a25 DL : 0: 给出平面上的N 个二维点，求出距离最小的2 个点对。本题中距离定义为他们的直线距离。例如(0,0) (3,4)的距离为5.-Plane gives the N-dimensional point, the minimum distance obtained two points pairs. This question is defined as the distance of their straight-line distance. For example, (0,0) (3,4) distance is 5.
Date : 2025-12-29 Size : 57kb User : 廖sir
[Mathimatics-Numerical algorithms] lagrange DL : 0: 拉格朗日插值法可以找到一个多项式，其恰好在各个观测的点取到观测到的值。这样的多项式称为拉格朗日（插值）多项式。数学上来说，拉格朗日插值法可以给出一个恰好穿过二维平面上若干个已知点的多项式函数。(Lagrange interpolation can find a polynomial, which happens to take the observed value at each observation point. Such polynomials are called Lagrange polynomials. Mathematically, Lagrange interpolation can give a polynomial function that happens to cross just a few known points in a two-dimensional plane.)
Date : 2025-12-29 Size : 2kb User : shiwel