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　　Feedforwardautoencoder. M: function [activation] = feedForwardAutoencoder(theta, hiddenSize, visibleSize, data)% theta: trained weights from the autoencoder% visibleSize: the number of input units (probably 64) % hiddenSize: the number of hidden units (probably 25) % data: Our matrix containing the training data as columns. So, data(:,i) is the i-th training example. % We first convert theta to the (W1, W2, b1, b2) matrix/vector format, so that this % follows the notation convention of t

= shockwaveflashobjects_ishockwaveflashsetmovie (* movie, null, PATH); $ M1 h; E8 S0 H8 L .? 8 A5 m3 _If (errorcode 7 P % @ 0 z, H/I6 ~ 0 ^, p-@ goto error;J) I ^ $ D.] 7 B9 ^ % F Step 3: start playing the file % G0 W9 u! ~ 'W) n' K, RErrorcode = shockwaveflashobjects_ishockwaveflashplay (* movie, null );S 'i0 y R * C # X if (errorcode Goto error; H7 {$ X3 S. T * V0 ~ % L $ uStep 4: Close the opened object handle;: R0 '.', I) g )~! EErrorcode = ca_discardobjhandle (movie); $ B "^ (] % F % v $

, vector A × vector B = vector CSuppose vector A turns to vector B at an angle less than 180 degrees.Point the four fingers in the right hand to the direction of vector A, and the four fingers in the right hand bend to represent the direction of the above rotation, then the straight thumb points to their cross-Multiplied vector CIf the direction of vector C is the same, it is on the same side; otherwise, it is on both sides. Note: The crossover calculation formula! Http://hi.baidu.com/foresh

Set a = (A1, A2, A3), B = (B1, B2, B3 ), A (x1, Y1, Z1), B (X2, Y2, Z2 ), Then a + B = (A1 + B1, A2 + B2, A3 + B3 ); A-B = (a1-b1, a2-b2, a3-b3 ); λ A = (λ A1, λ A2, λ A3) (λ ε r ); A · B = a1b1 + A2B2 + a3b3; A between B a1 = λ B1, a2 = λ B2, A3 = λ B3 (λ ε r ); A ⊥ B a1b1 + A2B2 + a3b3 = 0; = (X2-x1, y2-y1, z2-z1 ). The angle formula of the vector is used to obtain the angle of the isosurface lin

'. there's a single blank line after each level. input is terminated by three zeroes for L, R and C. OutputEach maze generates one line of output. If it is possible to reach the exit, print a line of the form Escaped in x minute (s ). Where x is replaced by the shortest time it takes to escape. If it is not possible to escape, print the line Trapped! Sample Input 3 4 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0 Sample Output Escaped in 11 minute(s).

image is black and white, though any of the colors can be used. The color used for the object (s) in the image is the foreground color while the rest of the image is the background color. [1] in the document-scanning industry, which is often referred to as "bi-tonal".Binary images is also called bi-level or two-level. This means the pixel is stored as a and a single bit-i.e., a 0 or 1.A binary image can be stored in memory as a bitmap, a packed array of bits. A 640x480 image requires 37.5 KiB o

The longest common subsequence:Given a sequence x={x1,x2,x3...xm}, another sequence Z={Z1,Z2,Z3...ZK} is called a subsequence of x when the following conditions exist, i.e. there is a strictly incrementing x subscript sequence Solving the longest common subsequence problem with dynamic programming method:1. Characterization of the longest common sub-sequenceMake X={X1,X2,X3...XM},Y={Y1,Y2,Y3...YN},Z={Z1,Z2,

; R:Array[0..5000000] ofNode AA,A,FB,FB2:Array[0.. to,0.. to] ofLongint; FF,Z,Z2:Array[0.. to,0.. to,1..4] ofLongint; F:Array[0.. to,0.. to,1..4,1..4] ofLongint; procedure bfs(x,y:longint); //Wide searchvarI,x1,y1:longint;beginwei:=1; tou:=0; R[wei].x:=x;//r.x is used to record the horizontal axis that can be movedR[wei].y:=y;//OrdinateR[wei].cs:=fb[x,y];//A few steps can be moved to whileTou Do beginInc (TOU); fori:=1 to 4 Do //Can I move to a

-like array.Open an array of dp[i][j] to the number of the first, the remaining J energy of the longest ascending subsequence.The largest DP value in a number smaller than a[i] is found in the first J tree array, and the tree array can be updated at the same timeThen in the j+1 tree array to find the largest DP value in a number smaller than b[i], you can also update the J Tree array, because it consumes 1 of the energyRead the bin God code, learned a bitQuestion D,Test instructions is very simp

. reference: http://bbs.scmlife.com/thread-22570-1-1.html When using git merge, it could be one of the following three modes9 m _7 c% N) J (B. M3 m5 P6 U4 G0 n S "L3 F # I1.Fast forwardWhen the last commit of the 2 branch to be merged is a linear relationship' c v:y-n+ p n+ B + v# ROr, a branch has no commit information since it was last updated.Git moves the pointer directly, does not have a real merge operation, and does not have a corresponding merge commit message# S7 O ' L "R,

to dividing all y into y1 parts, C (y-1,y1-1)For each of the multi-plug y, we fortress a Z to prevent y adjacent, then after the end of our z is z2=z-z1-y2, and then insert these z y1 sequence of two ends, the scheme is C (2*Y1,Z2)At this point, the assignment ends#include #include#include#include#includeusing namespaceStd;typedefLong Longll;intn=1000000;intmo=1000000007;intjc[1000011],fc[1000011];intans,t

-S) realview-eb ARM RealView Emulation Baseboard (ARM926EJ-S) realview-eb-mpcore ARM RealView Emulation Baseboard (ARM11MPCore) realview-pb-a8 ARM RealView Platform Baseboard for Cortex-A8realview-pbx-a9 ARM RealView Platform Baseboard for C Ortex-A9lm3s811evb Stellaris connector Gumstix Connex (PXA255) verdex Gumstix Verdex (PXA270) akita Akita PDA (PXA270) spitz Spitz PDA (PXA270) borzoi Borzoi PDA (PXA270) terrier Terrier PDA (PXA270) tosa Tosa PDA (metric) beagle Beagle board (OMAP3530) beag