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Description: 3n+1的算法
acm程序设计大赛的练习题目-3n algorithm acm a program design contest entitled to practice
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Size: 2925 |
Author: 幻帝 |
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Description: 求对给定数据集估计PI 的值
具体过程是：
给定数据集中任意取两个数组成一对数的取法有
( )
2
C2 n 1 n n
−
= 种，求出这
( )
2
n − 1 n
对数中
两个数互素的对数Count 。再利用公式
2
( 1)
6
2 −
=
n n
Count
π
推出
Count
3n(n − 1)
π = 。-right for the given data set is estimated that the value of PI specific process : determined to concentrate on data from two arbitrary number of components of a number of copying it () C2 2 1 n n n
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Size: 762880 |
Author: rcponder |
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Description: 3n+1的算法
acm程序设计大赛的练习题目-3n algorithm acm a program design contest entitled to practice
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Size: 3072 |
Author: 幻帝 |
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Description: Selective Harmonic Elimination Pulses for three-phase Inverter
The model shows the simulation for Six Harmonic Removal Pulses (i.e., N=7)
Selected Harmonics made zero are:
5th, 7th, 11th, 13th, 17th and 19th
(i.e., N=7 means six (i.e., N-1) dominant harmonics are made zero)
In this simulation the dominant harmonics will be:
(3N+2) and (3N+4) i.e., 23rd and 25th Harmonics
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Size: 72704 |
Author: QQ |
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Description: 刚性位置变化计算，输入物体在空间坐标系中的坐标，以及旋转向量和平移向量，计算其在摄像机坐标系中的坐标。- [Y,dYdom,dYdT] = rigid_motion(X,om,T)
Computes the rigid motion transformation Y = R*X+T, where R = rodrigues(om).
INPUT: X: 3D structure in the world coordinate frame (3xN matrix for N points)
(om,T): Rigid motion parameters between world coordinate frame and camera reference frame
om: rotation vector (3x1 vector) T: translation vector (3x1 vector)
OUTPUT: Y: 3D coordinates of the structure points in the camera reference frame (3xN matrix for N points)
dYdom: Derivative of Y with respect to om ((3N)x3 matrix)
dYdT: Derivative of Y with respect to T ((3N)x3 matrix)
Definitions:
Let P be a point in 3D of coordinates X in the world reference frame (stored in the matrix X)
The coordinate vector of P in the camera reference frame is: Y = R*X+ T
where R is the rotation matrix corresponding to the rotation vector om: R = rodrigues(om)
Important function called within that program:
rodrigues.m: Computes the
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Size: 1024 |
Author: shuang |
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Description: 弹力测试程序,测试范围为0-0.3牛顿,用于汽车电子测试行业，精度较高-spring tester，tester from 0-0.3N for pump tester mts
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Size: 175104 |
Author: 王明 |
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Description: This pascal source solve ``The 3n + 1 problem this mathematical problem is avalible in UVa Online Judge it is a recursive function.-This pascal source solve ``The 3n + 1 problem this mathematical problem is avalible in UVa Online Judge it is a recursive function.
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Size: 3072 |
Author: ucbicpc |
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Description: 3N层架构框架搭建
4N层数据访问层抽象工厂设计
2标准三层架构实现-4N 3N tier framework to build data access layer abstraction layer 2 standard three-tier plant design to achieve
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Size: 49152 |
Author: 安信 |
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Description: Euler工程，找最长的数列！
定义一个正整数数列，其迭代公式为：
n = n/2 (当n为偶数)
n = 3n+1 (当n为奇数)
比如从n=13开始，计算这个数列得：
13 ->40->20->10->5->16->8->4->1
这个数列一共有10项。
要求在小于1百万的所有起始数中，哪个数能产生最长的数列。
-Euler project, find the longest series! Define a positive integers, the iterative formula: n = n/2 (when n is even) n = 3n+1 (when n is odd), n = 13 for example, from the beginning, this series was calculated: 13-> 40-> 20-> 10-> 5-> 16-> 8-> 4-> 1 this series a total of 10 items. Requires less than 1 million threshold in all, which few can produce the longest series.
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Size: 3072 |
Author: superman |
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Description: 数学中有三叶玫瑰线（方程为ρ=Asin(3β)、四叶玫瑰线（方程为ρ=Asin(2β)等曲线，这些曲线的极坐标方程很简单，基本形式均为：ρ=Asin(nβ)，即任意一点的极半径ρ是角度β的函数；其直角坐标方程为：x=Asin(nβ)cos(β)，y =Asin(nβ)sin(β)。
在程序中控制角度β使其从0变化到2π，描出极半径ρ所对应的点，这样就可以绘出漂亮的玫瑰线；当然，n不同所描出的曲线的形状也就不同。
出于好奇，笔者又编写了一些方程，如：ρ=A[sin(nβ)＋3 sin(3nβ)]、ρ=Asin(nβ)exp(-kβ)等等，结果发现每种方程都能绘出形状各异的漂亮曲线。今介绍其中的六种，暂命名为：玫瑰线、玫瑰环、万寿菊、大丽花、蓬蒿菊、令箭荷花。-1. Algorithm Overview
If using an event loop to control the angle β from 0 to 2π, then draw the flower is static. If the timer event so that β increases every time a certain value, and draw the corresponding line segment, then draw the flower is dynamic. If you add a timer event in the time factor, that equation becomes: x = Asin (nβ) cos (β+ t), y = Asin (nβ) sin (β+ t) changing the value of t, and Each t draw β from 0 to 2π corresponding to the graphics, then draw the flowers will spin. If the line drawn between the endpoints, then draw the flower is hollow if drawn from the center line to the end, the draw of the flower is solid.
2. Interface design
Start VB6.0, add the following controls: used to select the pattern of the combo box Combo1 (the nickname as a list in front of the six items), the effect of the combo box for selecting Combo2 (list of contents as follows: static, dynamic and rotary) , n used to change the scroll bar HScroll1, n used to display the label Label1,
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Size: 3072 |
Author: 安多 |
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Description: 某公司收到若干报价，然后报价由低到高进行排序。设最低报价为F,最高报价为G。n=0.2*(G-F)
设A，B，C，D，E五个等距区间并取：
A=[F,F+n） B=[F+n,F+2n) C=[F+2n,F+3n) D=[F+3n,F+4n) E=[F+4n,G)
所有的报价都按各自的大小，分别列入上面五个区间。各自区间内的最小报价为该区间的代表报价。如果某区间内没有报价 则以小于该区间的且与该区间相邻的区间内的最高报价代表该区间报价，如果该区间与与之相邻的较小区间内都没有报价则该区间不参加最后计算。取各区间的代表值， 求出其平均值 做为评标基准报价
-A company received a number of quotations, and quotations from low to high order. Minimum price for the F, the highest offer for the G. n = 0.2* (GF) Let A, B, C, D, E, and five equidistant intervals obtained: A = [F, F+ n) B = [F+ n, F+2 n) C = [F+ 2n, F+3 n) D = [F+3 n, F+4 n) E = [F+4 n, G) All quotes are by their size, five were included in the above range. The minimum offer their own range representative of the range quoted. If a range does not offer less than the range of places and range of adjacent intervals with the highest offer within the offer on behalf of the interval, if the interval and the adjacent smaller range do not offer the range not to participate the final calculation. Take the representative value of each interval, calculate the average price as a benchmark evaluation
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Size: 393216 |
Author: 郭军 |
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Description: 某公司收到若干报价，然后报价由低到高进行排序。设最低报价为F,最高报价为G。n=0.2*(G-F)设A，B，C，D，E五个等距区间并取：A=[F,F+n） B=[F+n,F+2n) C=[F+2n,F+3n) D=[F+3n,F+4n) E=[F+4n,G)所有的报价都按各自的大小，分别列入上面五个区间。各自区间内的最小报价为该区间的代表报价。如果某区间内没有报价 则以小于该区间的且与该区间相邻的区间内的最高报价代表该区间报价，如果该区间与与之相邻的较小区间内都没有报价则该区间不参加最后计算。取各区间的代表值， 求出其平均值 做为评标基准报价
-A company received a number of quotations, and quotations from low to high order. Minimum price for the F, the highest offer for the G. n = 0.2* (GF) Let A, B, C, D, E, and five equidistant intervals obtained: A = [F, F+ n) B = [F+ n, F+2 n) C = [F+ 2n, F+3 n) D = [F+3 n, F+4 n) E = [F+4 n, G) All quotes are by their size, five were included in the above range. The minimum offer their own range representative of the range quoted. If a range does not offer less than the range of places and range of adjacent intervals with the highest offer within the offer on behalf of the interval, if the interval and the adjacent smaller range do not offer the range not to participate the final calculation. Take the representative value of each interval, calculate the average price as a benchmark evaluation
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Size: 394240 |
Author: 郭军 |
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Description: Dsdv-throughput 0f 3 nodes-Dsdv-throughput 0f 3 nodes
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Size: 101376 |
Author: Rev |
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Description: PKU 1207 The 3n+1 problem
1250 Tanning salon
1258 Agri-net-PKU 1207 The 3n+1 problem
1250 Tanning salon
1258 Agri-net
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Size: 919552 |
Author: Philip |
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Description: 本站不提供任何文件的下载服务)
仿QQ皮
文件大小3N-QQ imitation skin
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Size: 930816 |
Author: webe7 |
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Description: 这是3n+1问题。在ACM比赛上常常出现的问题。-This is 3n+1 problem. ACM game in question often arise.
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Size: 200704 |
Author: 蔡完锡 |
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Description: 这个程序用来解决经典的3N加1问题，可以将结果输出出来-This program is used to solve the problem of classic 3 n+ 1, the result can be output
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Size: 1024 |
Author: cheng |
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Description: 3n+1猜想对於每一个正整数，如果它是奇数，则对它乘 3 再加 1，如果它是偶数，则对它除以 2，如此循环，最终都能够得到 1-3n+1 guess for each positive integer, if it is odd, then multiply it 3 plus 1, even if it is, then it is divided by 2, and so on, eventually able to get a
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Size: 457728 |
Author: clh |
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Description: PAT基础题1001.害死人不偿命的(3n+1)猜想 简洁易懂 可以给大家提供一些解题思路 扩展思维-PAT basic question 1001. Killed attractiveness of (3n+1) guess concise and can give you some idea of solving the expansion of thinking
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Size: 1024 |
Author: happy |
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Description: Messing around with entropy and trajectory paths of the Hailstone Sequence.
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Size: 22826 |
Author: nVasquez |
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