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1. The load is gradually reduced while the load is maintained and maintained. The measured data, N, and Matlab are used to process the data to obtain the working characteristics and mechanical characteristics. U= 220; If = 0.058; If2 = 0.1; IA = [1.2 1.1 1 0.95 0.9 0.85 0.81 0.75 0.7 0.65 0.6 0.55 0.53 0.29]; N = [1600 1634 1656 1666 1672 1680 1688 1700 1712 1722 1739 1748 1755 1824]; If = [0.83 0.75 0.65 0.6 0.56 0.52 0.48 0.43 0.38 0.33 0.29 0.24 0.22 0]; T2 = [1.08 1 0.88 0.83 0.78 0.74 0.7 0

Copy codeThe Code is as follows: public class MirrorView extends View { Paint m_paint; Int m_nShadowH; Drawable m_dw; Bitmap m_bitmap; // Xxxx Matrix mMatrix; Int shadowHeight; Public MirrorView (Context context, Bitmap bitmap ){ Super (context ); M_bitmap = bitmap; _ Init (); // TODO Auto-generated constructor stub } Private void _ Init (){ // M_dw = new BitmapDrawable (BitmapFactory. decodeResource (getResources (), R. drawable. icon )); M_dw = new BitmapDrawable (m_bitmap ); M_dw.setBounds (0

When the plane is set to (NX,NY,NZ,D), the column main sequence reflection matrix with this planar mirror is as follows:The derivation is as follows:One, the expression of the platform:, over point P, the plane of the normal vector n, can be expressed as:Np+d=0where D is the forward distance from the plane to the origin point. If the plane faces the origin, D is positive and D is negative if the plane is back to the origin point.So the plane can be re

(i,s,t) for (int i=s;i#defineren for (int i=first[x];i!=-1;i=next[i])using namespaceStd;inlineintRead () {intx=0, f=1;CharC=GetChar (); for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1; for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;}Const intmaxn=1010;CharA[MAXN][MAXN];intN,D[MAXN][MAXN],VIS[MAXN][MAXN],IDX[MAXN],TP;structPoint {intx, y; BOOL operatorConstpoint ths) {returnd[x][y]d[ths.x][ths.y];}; Queue3];intGetfront () {intc=-1; if(q[0].size ()) c=0; if(q[1].size () (c0|| q[1].

Topic Portal1 /*2 bfs+ Simulation: dp[i][j][p] means to go to i,j, the direction of the number of steps of P;3 BFS in 4 different situations, and finally in the end 4 direction to find the minimum value:)4 */5#include 6#include 7#include 8#include 9#include string>Ten#include One using namespacestd; A - Const intMAXN = 1e2 +Ten; - Const intINF =0x3f3f3f3f; the intdir[4][4][2] = { -{{0, -1}, {0,1}, {-1,0}, {1,0}}, -{{1,0}, {0, -1}, {0,1}, {-1,0}}, -{{-1,0}, {1,0}, {0, -1}, {0,1}}, +{{0,1}, {-1,

. If you can eliminate the output "YES", you cannot output "no". Sample Input3 41 2 3 40 0 0 04 3 2 141 1 3 41 1 2 41 1 3 32 1 2 43 40 1 4 30 2 4 10 0 0 021 1 2 41 3 2 30 0Sample OutputYesnonononoyesLinks: http://acm.hdu.edu.cn/showproblem.php?pid=1175The puzzle: is the transition two times can go from the beginning to the end. So use dire to record the direction, TER records the number of turns, DFS can.WA several times, the reason is restores the starting point when restores to 0, restores the

Data:34 4#####JF ##.. ##.. #3 3FFFFjfFFF3 3.... J....Well, the third data card for half a day ....Two BFS pre-processing the earliest fire time of each point./************************************************author:p k29created time:2015/8/19 19:26:13File name:4.c pp*************************************************/#include#include#include#include#include#include#include#includeSet>#include#includestring>#include#include#include#include#include#include#defineull unsigned long Long#definell Long L

Code One:#include #include#includeusing namespacestd;Const intmax_n=205;Const intinf=0x30303030;CharMap[max_n][max_n];intn,m;intSx,sy;intGx,gy;intdx[4]={1,0, -1,0};intdy[4]={0,1,0, -1};typedef pairint,int>P;intDist[max_n][max_n];intBFsintYintx) { for(intI=0; i) for(intj=0; j) Dist[i][j]=INF; DIST[SY][SX]=0; Queueque; Que.push (P (SY,SX)); while(!Que.empty ()) {P pos=Que.front (); Que.pop (); inty=Pos.first; intx=Pos.second; if(y==gyx==GX) { Break; } for(in

coordinates of the root node of the newly added set in UnionSet.The Code is as follows: # Include # Include # Include # Include Using namespace std; Const int MAX_N = 40010;Int nN, nM;Int nSets [MAX_N];Int nX [MAX_N];Int nY [MAX_N];Char szInput [MAX_N] [100]; Void MakeSets (int nNum){For (int I = 0; I {NSets [I] = I;NX [I] = nY [I] = 0;}} Int FindSets (int nI){If (nSets [nI]! = NI){Int nPre = nSets [nI];NS

(); BOOLJudgeintRintc);intBFS ();voidinput () {scanf ("%d%d", row, col); for(inti =0; i ) { for(intj =0; J ) {cin>>Maze[i][j]; } } //All locations Initialized for(inti =0; i ) { for(intj =0; J ) {D[i][j]=INF; } } }BOOLJudgeintRintc) { return(R >=0 r 0 C Col) (maze[r][c]! ='1');//can go}intBFS () {//set the starting point to the queue and the distance of this location to 0Que.push (P (0,0)); Queuestring>path; Path.push (""); d[0][0] =0; Min=""; while(!Que.empty

Copy Code code as follows: public class MirrorView extends View { Paint M_paint; int m_nshadowh; Drawable M_dw; Bitmap M_bitmap; Xxxx Matrix Mmatrix; int shadowheight; Public MirrorView (context, Bitmap Bitmap) { Super (context); M_bitmap = bitmap; _init (); TODO auto-generated Constructor stub } private void _init () { M_dw = new Bitmapdrawable (Bitmapfactory.decoderesource (Getresources (), R.drawable.icon)); M_dw = new Bitmapdrawable (M_BITMAP); M_dw.setbounds (0,0,m_dw.

Python + matplotlib implement the example code of the Box Bar Chart, pythonmatplotlib Demo result: Complete code: import matplotlib.pyplot as pltimport numpy as npfrom matplotlib.image import BboxImagefrom matplotlib._png import read_pngimport matplotlib.colorsfrom matplotlib.cbook import get_sample_dataclass RibbonBox(object): original_image = read_png(get_sample_data("Minduka_Present_Blue_Pack.png", asfileobj=False)) cut_location = 70 b_and_h = original_image[:, :, 2]

Summary of str_replace function usage in PHP. During this time, I saw the str_replace explanation in the book PHP and MySQLWeb development. I wrote a small note: Three of str_replace can be imported using arrays, but the explanation is relatively simple, so during this time, I read the book "PHP and MySQL Web development" and saw the str_replace explanation. I wrote a small note: Three of str_replace can be passed in using arrays, but the explanation is relatively simple, so I decided to test th

In actual program development, string replacement is a very frequent task, and the str_replace function is also very useful. During this time, I read the book "PHP and MySQL Web development" and saw the str_replace explanation. I wrote a small note: Three of str_replace can be passed in using arrays, but the explanation is relatively simple, so I decided to test the execution result of the function when each parameter is passed into the array. Function prototype: mixed str_replace (mixed needle

[nd],nd); V[d][x][y]=false; if(npos==-2)returnpos[d][x][y]= (x1) *m+y; Else returnpos[d][x][y]=NPOs;}voidSPFA (intLintR) { for(intI=0; i0; for(intI=1; i; for(intI=1; i1]; for(intI=tot;i;--i) q[cnt[dp[l][r][tmp[i].first][tmp[i].second]]--]=Tmp[i]; for(intI=1; ii) Q2.push (Q[i]); intX,y,nx,ny; while(!q2.empty () | | |Q1.empty ()) { if(Q1.empty ()) {x=Q2.front (). First; Y=Q2.front (). Second; Q2.pop (); } Else if(Q2.empty

, 42, 43, 44, 45, 46, 47,48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59,60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71;}2. Contents of simple_xy-wr.c File/* The NetCDF package.Copyright 2006 University Corporation for atmospheric research/unidata.See COPYRIGHT "File for conditions".This was a very simple example which writes a 2D array ofSample data. To handle this in NetCDF we create the sharedDimensions, "x" and "Y", and a NetCDF variable, called "Data".This example demonstrates the NetCDF C A

);d [sx][sy]=0; - Wuyi while(Que.size ()) the { -Node now=Que.front (), NEX; Wu /*For (i=1;i - printf ("%d", D[1][i]); About printf ("\ n"); $ printf ("%d\n", now.y);*/ - Que.pop (); - - for(i=0;i4; i++) A { + intnx=now.x+dx[i],ny=now.y+Dy[i]; the if(11'#') - { $ intValue=D[NOW.X][NOW.Y]; the if(mp[nx][

(intj=0; j) Dp[i][j]=inf; }}intMain () {intT; intSx,sy; intFlag; Charts[3]; scanf ("%d",t); while(t--) {flag= -1; scanf ("%d%d",n,m); Init (); for(intI=0; i){ for(intj=0; j) {scanf ("%s", TS); G[I][J]= ts[0]; if(g[i][j]=='2') {SX=i; Sy=J; } } } /*printf ("debug\n"); for (int i=0;i*/step Tp,fro; Tp.x=SX; TP.Y=Sy; TP.T=0; Tp.limit=0; Q.push (TP); G[sx][sy]='1'; Dp[sx][sy]=0; intNx,ny; while(!Q.e

; - Q.pop (); - for(intI=1; i4; i++){ - intnx=x+mx[i],ny=y+My[i]; - if(nx>0 nx0 nyW) in if(!vis[nx][ny] (mp[nx][ny]==0|| mp[nx][ny]==4)){ -dis[nx][ny][mode]=dis[x][y][mode]+1; tovis[nx][n