specter x2

Mean value algorithm of MA, EMA, SMA, DMA, TMA, WMA6MA, EMA, SMA, DMA, TMA, WMA6 average algorithms are often used in various indicator formulas, but most beginners may not understand their specific differences, organized as follows.MA (x,n) simple arithmetic averageThe N-day moving average of x, not the weight, the average calculation. The algorithm is:(X1+X2+X3+.....+XN)/nFor example, MA (c,20) represents the average closing price of 20th. C means c

This article mainly introduces the implementation of the Golden section in Python, involving techniques related to the calculation of Python's mathematics, the need for friends can refer to the ? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 This is the ; > "' A,b = bracket (f,xstart,h) finds the brackets (a,b) of a minimum point of the user-supplied scalar function f (x). The search starts downhill from Xstart and a step length h. x,fmin = Search (f,a,b,tol=1.0e-6)

clip method. The usage is simple:Ctx.save () Ctx.beginpath () Ctx.arc (X2,Y2,A,0,2*MATH.PI); Ctx.clip () Ctx.clearrect (0,0,canvas.width,canvas.height); Ctx.restore ();The above code realizes the erase of the circle area, that is, to implement a circular path, then use this path as the clipping region, then clear the pixels. One thing to note is that you need to save the drawing environment first, to reset the drawing environment after you have clear

HTML5Canvas gradient is a color pattern used to fill or stroke a graph. Gradient color is a transition from different colors, rather than a single color. The gradient is divided into linear gradient and radial gradient according to the type. Let's take a look at several examples of canvas fading ...,. HTML5 CANVAS: Draw gradient HTML5 Canvas gradient is a color pattern used to fill or stroke a graph. Gradient color is a transition from different colors, rather than a single color. Let's take a

/*The problem of 10^8 violence can be answered but slow ..... There is a small bug that the person card int but did not pay attention to short with short can not hash directly to engage49428K 313MS */#include#include#include#defineBase 12500000using namespacestd;intA1,a2,a3,a4,a5,x1,x2,x3,x4,x5,ans; Shortf[12500000*2+ -];intMain () {scanf ("%d%d%d%d%d",a1,a2,a3,a4,a5); for(x1=- -; x1 -; x1++)if(x1) for(x2=

E-tiling_easy versionThe main idea: there is a grid size of 2 x N, now need to use 2 specifications of the Dominoes paved, the domino specifications are 2 x 1 and 2 x 2, please calculate how many kinds of laying method.A simple DP question: dp[i+2]=dp[i+1]+2*dp[i]; Initial conditions: dp[1]=1;dp[2]=3;At first did not consider clearly, the recursive relationship written dp[i+2]=3*dp[i];Explain Dp[i+2]=dp[i+1]+2*dp[i]DP[I+2] In contrast to dp[i] just a 2x2 square, the current i+1 lattice are fille

Poj_1410 Notes: ① The coordinates of the rectangular vertices must be re-sorted and used again. It is only a reference for discuss to say that the top is not strictly above the top, so we need to re-sort and re-use them in the lower order. ② Because the inside of the rectangle is also part of the rectangle, the line segment is considered to be at the intersection of the rectangle in the rectangle. ③ When determining whether the four sides of a line segment and a rectangle have intersecti

random order, so that X is used as a random variable, and the value is equal to the number of times a new office is hired. Then E [x] = Σ xpr {x = x}, but this computation will be very troublesome. We define n variables corresponding to whether or not each candidate is hired, and XI is the indicator random variable corresponding to the event where I candidate is hired. X = X1 + X2 +... + Xn.E [XI] = Pr {Xi} = 1/I, because the applicant is random, so

Main features of this algorithm The problem of two-dimensional cropping is converted into a secondary one-dimensional cropping problem, and the problem of cropping is transformed into solving a set of inequalities; The improvement of the Cohen-Sutherland encoding algorithm is only applicable to the shortcomings of the line segments on the same side (or left, or right, or top or bottom) of the window. The algorithm is divided into one and two dimensions. The former is the basis of th

Implementation of the Golden Segmentation Method in python This article mainly introduces the implementation of the Golden division method in python, involving the related skills of Python mathematical computation. For more information, see ? 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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 '''A, B = bracket (f, xStart, h) Finds the brackets (a, B) of a minimum point of User-supplied scalar

HDU 4553 line segment tree dual-Keyword interval merge Use two keywords to record the dash time and Goddess time. Update the time if the silk is about Update the Time of the diaosi and goddess if the goddess appointment Learning, then both are cleared # Include "stdio. h "# include" string. h "struct Data {int l, r, x1, x2, l1, l2, r1, r2;} data [400010]; int Max (int a, int B) {if ( = X) return data [k]. l; if (data [k * 2]. x1> = x) return qu

Reprinted please indicate the source, thank you http://blog.csdn.net/acm_cxlove/article/details/7854526By --- cxlove Question: it is similar to gss3, but it is a little troublesome to query. Pay attention to the details. The start and end of the interval are within the specified interval. Http://www.spoj.pl/problems/GSS5/ The update part is similar to the query part. It mainly divides each query into clear points. I am divided into three situations First, there is no overlap between the two inte

is the clip method. The usage is simple:　Ctx.save () Ctx.beginpath () Ctx.arc (x2,y2,a,0,2*Math.PI); Ctx.clip () ctx.clearrect (0,0, Canvas.width,canvas.height); Ctx.restore ();　　The above code realizes the erase of the circle area, that is, to implement a circular path, then use this path as the clipping region, then clear the pixels. One thing to note is that you need to save the drawing environment first, to reset the drawing environment after you

again we see some incredible links in math. We are now re-examining and defining the game's outcome.(1) If it is time for someone to take a match and the match is gone, the person loses and is set to P-pattern(2) If it is time for someone to take a match, he can finish the match, then the person wins, set to N-Pattern(3) If it is time for someone to take a match, he can turn the current pattern into a p pattern, then the pattern is N-pattern(4) If it's time for someone to take a match, he can't

of n*1 for the maximum value of the xend:x range of values function Description : The item that does not exist is filled in []. function function : Linear programming to find the optimal value. Example 1: Find the maximum value of f=3*x1+6*x2+2*x3 The conditions to be met are 3*x1+4*x2+x3≤2 X1+3*x2+2*x3≤1 and X1, x2,

Directly on the code, using hist to draw frequency histogram and frequency histogram ... %rand fs=1000; n=10000; T=0:1/FS: (N-1)/fs; X1=rand (1,length (t)); Subplot (221) plot (t,x1), Xlabel (' time (s) '), Ylabel (' A random signal with a 10,000-point sample ') title (' A random signal with a 10,000-point sample: 0-1 uniformly distributed white noise '); % RANDN usage x3=mean3+sqrt (VAR3). *RANDN (1,n)%fs=1000; n=100000; %T=0:1/FS: (N-1)/fs x2=rand

# Include # Include DoubleF (doublex){Returnpow (x, 4) + x * 2 + 4;} Doublegolden (doublea, doubleb, doublee){Inti = 0;Doublex1, x2, f1, f2; X2 = a + 0.618 * (B-);F2 = F (x2 );X1 = a + 0.382 * (B-);F1 = F (x1 );While (fabs (B-a)> e){Printf ("\ n % d \ t % lf ", I ++, a, B, x1, x2, f1, f2, fabs (B-));If (f1 {B =

1. Minimum value of a constrained unary FunctionThe standard form of a single variable function for minimum value is Min f (x) sub. To X1 lt; X lt; x2.Use the fmin function in MATLAB 5.x to calculate its minimum value.Function fminbndFormat x = fminbnd (fun, x1, x2) % returns the value of X when the function fun gets the minimum value in the range of the independent variable X, fun is the expression strin

; For (I = 0 ; I Scanf ( " % Lf " , X [I], Y [I]); For (I = N; I 2 * N; I ++)X [I] = x [I-n], Y [I] = Y [I-n]- 1 ;} Double Det ( Double X1, Double Y1, Double X2, Double Y2){Return X1 * Y2-X2 * Y1;} Double Getx ( Double X1, Double X2, Double T1, Double T2){ Return (FABS (T2) * X1 + FABS (T1) * x2)/(FABS (T1) + FABS

Let's take a look at the derivation of the formula and some key points. [two-dimensional interpolation is used here, and so on] 1. interpolation problem: a pile of discrete data is often obtained during the experiment. Now I want to use a mathematical formula to simulate this pile of discrete data. What should we do? mathematicians raised the interpolation question. The interpolation problem is such that a pile of data points (x0, y0), (x1, Y1), (X2,