squares on checkerboard

There are partial least squares in the Sklearn Library of Python3.You can see the following library description: http://scikit-learn.org/stable/modules/generated/sklearn.cross_decomposition.PLSRegression.htmlProgram Source: Source: 78467689 modified to run on Python3My Data a.csvMy Data ron.csv#-*-coding:utf-8-*-#It seems that this program is suitable for the Python2 version. Source: 78467689#This can now be used, mainly in the 60th row of data conver

body{Background:pink; } #c1 {background:white; } HTML5 Canvas Moving small squares

Control the movement of small squares with keyboard Elements [Ctrl + A select all Note: If you need to introduce external Js, You need to refresh it to execute]

#include 5 6 #defineN 57 8 intMain ()9 {Ten intA[n][n] = {0}; One intCount =1; A introw =0, cul = N/2; - while(Count N) - { theA[row][cul] =count; - inti =Row; - intj =Cul; - if(i = =0) + { -i = N-1; + } A Else at { -i = i-1; - } -j = (J +1) %N; - if(a[i][j]!=0|| (row==0cul==n-1)) - { ini = row +1; -j =Cul; to } +row =i; -Cul =J; thecount++; * } $ Panax Notoginseng for(inti =0; i

Title DescriptionGiven a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, ...) which sum to n .For example, given n = A, return 3 because 12 = 4 + 4 + 4; Given n = 13, return 2 because = 4 + 9.Credits:Special thanks to @jianchao. Li.fighter for adding the problem and creating all test cases.Ideas:In the subject or BFS, the recursive object is the complete square number set, and the number set to be searched LIST[0...I] (=n-table[0...i])1. Find out the to

Topic:Given a positive integer n, find the least number of perfect square numbers ( For example, 1, 4, 9, return return 2 because 13 = 4 + 9. Analysis: DP problem, set DP[N] as the input positive integer n, the function returns the result. Find the following rules:Dp[0] = 0dp[1] = 1dp[2] = 2dp[3] = 3dp[4] = 1 = dp[0] + 1 = dp[4-2*2] + 1dp[5 ] = 2dp[6] = 3dp[7] = 4dp[8] = 2 = dp[4] + 1 = dp[8-2*2] + 1dp[9] = 1 = dp[0] + 1 = DP [9-3*3] + 1The code is as follows: Public intNumsqu

= =2)returnDist2 (p[0], p[1]); theP.push_back (p[0]);//lest the mold be taken94 intAns =0; the for(intU =0, V =1; U ) the { the //a straight line to the side p[u]-p[u+1]98 for(;;) About { - //Stop rotation when area (P[u], p[u+1 [], p[v+1]) 101 //( P[u+1]-p[u], P[v+1]-p[u])-Cross (P[u+1]-p[u], P[v]-p[u]) 102 //According to Cross (A, b)-Cross (A,C) = Cross (a,b-c)103 //Simplification cross (P[u+1]-p[u], p[v+1]-

.--------------------------------------------------------------------------------------------------------------- ---------------------------------Code1 /*2 id:icedrea13 Prob:palsquare4 lang:c++5 */6 7#include 8#include 9 using namespacestd;Ten One intB; A Charto[ -]={'0','1','2','3','4','5','6','7','8','9', - 'A','B','C','D','E','F','G','H','I','J' }; - the stringChangeintx) - { - stringnum; - while(x) {num=to[x%b]+num; x/=B;} + returnnum; - } + A BOOLIspal (intx

area.Cross (convex[p+1]-CONVEX[P],CONVEX[Q]-CONVEX[P]))//find the farthest point of the corresponding P (due to the single-peak function, so the result is like a double pointer)Q= (q+1)%N; Ans=max (Ans,max (DIS (convex[p],convex[q)), Dis (convex[p+1],convex[q+1]))); } returnans;} Point Poi[max],convex[max];DoubleSolve (intN//Find the square of the farthest point pair{ Doublefpp=0; intM=convexhull (Poi,n,convex);//first find the convex bagFpp=rotatestuck (CONVEX,M);//rotation jam to the he

Test instructions: Give a number, the rule that the eight numbers around after the removal of the numbers are not desirable, the largest number and ideas can be obtained: State compression DP, each row of the method is the state, obviously, due to the rule of the number of restrictions, the desirable state is not 1#include using namespacestd;intdp[ -][ the],sta[ the],len,n;intg[ the][ the];voidinit () { for(intI=0;i1 the); i++)if(! (i (i1)) ! (i (i>>1))) Sta[len++]=i;//preprocess all the desirab

Evaluation Portalinput3 6 ( 5) 4 145 2 215 6 3 157 2 0 0output67Practice:It's like a note, but it's a little different.The most basic approach, open four-dimensional violence DP, can update the updateWe can also save space time by updating the state of each layer and using the Manhattan distance to represent another dimension in one dimension.Code "1"#include #includeusing namespacestd;intmp[ One][ One],f[ One][ One][ One][ One];intMain () {intN,x,y,c; scanf ("%d",N); while(SCANF ("%d%d%d",

I accidentally a small "bug", but let the animation become super Brilliant = =So, do not be afraid of the bug, who knows the bug will not open a fancy flower?DOCTYPE HTML>HTMLLang= "en"> Head> MetaCharSet= "Utf-8"> title>title> style>Body{text-align:Center; }Canvas{Border:1px dashed Green; } style> Head> Body> CanvasID= "CVS"width= "$"Height= "All">Canvas> Script> varCVS=document.getElementById ("CVS"); varContext=Cvs.g

; attypedefstructPoint { - intx, y; - Point () {} -Point (intXxintyy): X (xx), Y (yy) {} - }point; - intN, ans; in Point P[MAXN]; - to BOOLCMP (Point A, point B) { + if(a.x = = b.x)returnA.y b.y; - returnA.x b.x; the } * $ BOOLBsintXinty) {Panax Notoginseng intLL =0, rr =n, mm; - while(LL RR) { theMM = (ll + RR) >>1; + if(p[mm].x = = x p[mm].y = = y)return 1; A Else if(CMP (P[MM], point (x, y))) ll = mm +1; the Elserr = mm-1; + - } $ retur

3 Sample output 17 24 1 8 1523 5 7 14 164 6 13 20 2210 12 19 21 311 18 25 2 9 Analysis: According to the meaning of the topic, if a[i][j]=t, then the next to fill in the coordinates of the t+1 first consideration is x=i-1,y=j+1, but if i==0j==2*n-1 or a[x][y] has been occupied, should be placed in a[i][ J], just below.So the code is:1#include 2#include 3 intMainintargcChar*argv[])4 {5 intn,i,j,t;6 int**A;7 LongNn,len;8 intx, y;

Function [a, B, c, d] = get_ls_plane (data) % A * x + B * Y + C * z + D = 0 planedata = data; % in the SVD transformation of the covariance matrix, the singular vector corresponding to the minimum singular value is in the direction of the plane xyz0 = mean (planedata, 1); centeredplane = bsxfun (@ minus, planedata, xyz0 ); [~,~, V] = SVD (centeredplane); A = V (1, 3); B = V (2, 3); C = V (3, 3); D =-dot ([a B C], xyz0); End[MATLAB] ls (Least Squares)

I'll tell you the conversion. I forgot that I had the digital logic textbook in mind./*id:modengd1prog:palsquarelang:c++*/#include 　　Usaco palindromic Squares

#import "ViewController.h"@interface Viewcontroller ()@end@implementation Viewcontroller-(void) Viewdidload {[Super Viewdidload];int arr[4][4]={1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1};Nsmutablearray *varray=[[nsmutablearray alloc]init];//Create a dynamic arrayfor (int i=0; i{int row =i/4; The number you want is in that line.int column=i%4;//want the number in that columnUIView *v=[[uiview Alloc]initwithframe:cgrectmake (column*54, row*54, 50, 50)]; Create a block and replace it with a pointer V, create

Recently in the CF related papers, "collaborative Filtering for implicit Feedback Datasets" thought very well, very easy to understand, but from the objective functionHow to derive the formula for the renewal of Xu and Yi is not well described, so write it downDerivation:The first derivative of Xu:Where Y is the item matrix, n*f, each line is a item_vec,c^u is a diagonal matrix of n*n dimension,Each element on the diagonal is c_ui,p (u) is a n*1 column vector, and its element i is p_ui.Then the

); - End;WuyiExit (c[t]maxlongint); the End; - proceduredeal; Wu begin -i:=t;l:=Maxlongint; About whileI Do $ begin -l:=min (e[i]^.w,l); -i:=e[i]^.anti^.g; - End; ADec (flow,l); i:=T; + whileI Do the begin - ifE[i]^.w ThenDec (e[i]^.w,l); $ ifE[i]^.anti^.w ThenInc (E[I]^.ANTI^.W,L); thei:=e[i]^.anti^.g; the

ConceptThe least squares polynomial curve fitting, according to the given m points, does not require this curve to pass through these points precisely, but rather the approximate curve y=φ (x) of the curve y=f (x).principle[principle part by individual according to the information on the Internet to summarize, hope to everyone can be practical]Given the data point Pi (xi,yi), i=1,2,..., m. The approximate curve y=φ (x) is obtained. It also minimizes t