blueant q1

Turn from: http://www.cnblogs.com/tgycoder/p/5106463.html four, quaternion class static method There are 9 static methods in Quaternion: Angle method, Dot method, Euler method, Fromtorotation method, inverse method, Lerp method, Lookrotation method, Rotatetowards method and Slerp method. The use of static methods is to call their static methods directly with class names, such as Quaternion.angle (Q1,Q2), and to do analysis under these static methods.

#pragmaComment (linker, "/stack:1024000000,1024000000")#include#include#include#include#include#include#includeSet>#include#include#include#include#include#include#include#includetypedefLong Longll;using namespacestd;//freopen ("d.in", "R", stdin);//freopen ("D.out", "w", stdout);#defineSspeed ios_base::sync_with_stdio (0); Cin.tie (0)#defineMAXN 1000006#defineMoD 1000000007#defineEPS 1e-9#definee exp (1.0)#definePI ACOs (-1)Const DoubleEP = 1e-Ten ;intNum;//const int INF=0X7FFFFFFF;Constll inf=

, just +1 . } returni;}structnode* Node_new (intID) {structnode* q = (structnode*)malloc(sizeof(structNode)); memset (q,0,sizeof(structNode)); Q->_id =ID; returnq;}voidtestfunction () {structnode* q1 = node_new (1); structnode* q2 = node_new (2); structnode* q3 = node_new (3); structnode* Q4 = node_new (4); structnode* Q5 = Node_new (5); Puts ("# # #有节点的链表:"); printf ("count =%d\n", Chain_count (NULL)); Chain_print (NULL); Puts ("\n## #1个节点的链

, partition sales_200910 tablespace partition, partition sales_200911 tablespace partition, partition sales_200912 tablespace partition, partition tablespace partition, partition sales_201002 tablespace partition ); --- bitmap create bitmap index index_sales_data_sales_type on sales_data (sales_type) local; -- bitmap connects to create bitmap index index_sales_data_city sales_data (city. city_id) from sales_data, city where sales_data.city_id = city. city_id local; run SQLSql code select city, s

Q is in P1, p2 is inside the rectangle of the diagonal vertex 4. Determine whether two line segments are Intersection We can determine whether two line segments overlap in two steps: (1). Rapid Rejection Test Set the rectangle with the linestring P1P2 as the diagonal line to R and the rectangle with the linestring q1q2 as the diagonal line to T. If the R and t do not intersect, the two linestring obviously do not; (2). Cross-Site Test If the two line segments overlap, the two lines must cross e

; front + = 1; // pull shit return myqueue-> data [myqueue-> front-1]; // return the file you pulled} void printfQ (Q * myqueue) // print all elements of the queue {printf (); if (myqueue-> front = myqueue-> rear) {printf (your stomach is empty);} else {for (int I = myqueue-> front; I // Main. c # Include # Include # Include queue. hvoid main () {Q Q1; // create A struct variable init ( Q1);

quaternary element used to store rotation informationArcball. Drag ( mousept, thisquat );// Convert the Quaternary element into a rotating MatrixMatrix3fsetrotationfromquat4f ( thisrot, thisquat );Matrix3fmulmatrix3f ( thisrot, lastrot); // cumulative rotation result// Get our final rotation resultMatrix4fsetrotationfrommatrix3f ( transform, thisrot );}Else // undraggedIsdragging = false;}M_opengldisplay.displayscene (m_p3dmodel );//M_completed = true;}Cview: ontimer (nidevent );} The

refer to the following figure: 　　 Determine if the point is on a line segment: The point q is the p1p2 of the line, and the basis of the judge points in the line is:And Q is in the rectangle with the P1,P2 as the diagonal vertex. The former ensures that the Q-point is on a straight line p1p2, the latter is to ensure that the Q-point is not in the line p1p2 extension or reverse extension, for this step can be judged by the following procedure: On-segment (PI,PJ,PK) if min (xi,xj) In particul

Whether the segment P1P2 has an intersection with the segment q1q2. P1P2 to intersect with Q1Q2, point P1 and Point P2 have to be on both sides of the online segment Q1Q2 (the same point Q1 and Point Q2 are on both sides of the online segment p1p2). The cross product of a straight line is used to solve. That(P1-Q1) x (q2-q1) * (Q2-

Requirement 1:Use the for loop to create ten files in the/root/scripts/practice/q1/oldboy directory. The names are as follows: Oldboy-1, Oldboy-2, ..... Oldboy-10. Script implementation: #! /Bin/bash # question1shopt-s expand_aliasesalias LL = "ls-L" CD/root/scripts/practice/q1 # first, determine whether the directory exists. If it does not exist, create the directory if [! -E oldboy] Then/bin/mkdir oldboy

changes to the z-axis, the y-axis to the x-axis, and the z-axis to the y-axis (regardless of direction). 2.2.34-dollar calculation of the design of Euler's anglefloat q0 = 1, q1 = 0, q2 = 0, q3 = 0; Defining four elementsvoid Imu_update (void){float norm;Float GX = Mpu6500_gyro.x*gyro_gr,gy = MPU6500_GYRO.Y*GYRO_GR,GZ = unit conversion between mpu6500_gyro.z*gyro_gr;//anglesfloat ax = Acc_avg.x,ay = Acc_avg.y,az = acc_avg.z;float q0q0 = q0 * Q0;float

To: Http://hi.baidu.com/zhang_bi/blog/item/57edb701a9da6b00728b65db.html There are two types of value assignment statements in OpenGL:Blocking value assignment statement ("= ")AndNon-blocking value assignment statement (". Correct use of the two assignment statements is very important for the design and simulation of the Tilde. The following example shows the difference between blocking and non-blocking assignment. Let's take a look at several paragraphsCodeAnd the corresponding circuit

, the vectors (P1-Q1) and (P2-Q1) are located on both sides of the vector (Q2-Q1), I .e. (P1-Q1) × (Q2-Q1) * (P2-Q1) × (Q2-Q1) Under the same principle, the specific implementatio

-Site Test If the two line segments overlap, the two lines must cross each other. If P1P2 cross-sets q1q2, the vectors (P1-Q1) and (P2-Q1) are located on both sides of the vector (Q2-Q1), I .e. (P1-Q1) × (Q2-Q1) * (P2-Q1) × (Q2-

Three or four yuan to Euler angle conversionAs a result of arctan and arcsin, this does not cover all orientations (the range of values for the corners has been met), so atan2 is needed instead of arctan.Iv. use in other coordinate systemsIn other coordinate systems, the above formula should be adjusted according to the definition of the axis. As in Direct3D, the x-axis of the Cartesian coordinate system changes to the z-axis, the y-axis to the x-axis, and the z-axis to the y-axis (regardless of

BZOJ1047[HAOI2007] the ideal squareTest instructionsThere is a matrix of a*b integers, which asks for a n*n square area, which minimizes the difference between the maximum and minimum values of all the numbers in the region. a,b≤1000,n≤100ExercisesDo 4 times the monotone queue. Using the monotone queue to find the maximum minimum value of row J of line I to row j+n-1 of line I, and then use this to find the maximum minimum value of row J of line I to column i+n-1 j+n-1. The last enumeration of t

very high, and the resource cost is relatively small. It is not necessary to use parallel operations.(2) read a parallel execution planLS @ LEO> select object_type, count (*) from leo_t group by object_type; object type grouping statistics35 rows selected.Execution Plan Parallel Execution Plan----------------------------------------------------------Plan hash value: 852105030Bytes ------------------------------------------------------------------------------------------------------------------|

:1. Serial + SyncThe tasks in the serial queue are executed one after the other, and the synchronization tasks are executed in the order of execution, so the conclusion is that the synchronization tasks in the serial queue will be executed one after the other.dispatch_queue_t q1 = dispatch_queue_create ("Serial queue", dispatch_queue_serial);Dispatch_sync (Q1, ^{[Nsthread Sleepfortimeinterval:5];NSLog (@ "S

1. How to determine whether a basic amplification circuit is a common-base, co-injection or a common set? Remove the input, the output side, the rest is the common end. The following figure: The UI is connected to the base pole, and the UO is the common collector. 2. Basic Amplification Circuit: 2-1. Co-emitting minimal signal amplifier Function: Amplify the small amplitude voltage signal Working principle: Q1 is connected to the common-injection

The DFA (deterministic finite automata), a deterministic finite automaton, refers to a unique next state for each state that is determined by the input alphabet, and the NFA (Nondeterminisic finite Automata) is an indeterminate finite automaton, which means that for any state, when the state obtains the input alphabet, it is possible that the resulting state is not one, but multiple, that is, a set of states. In a sense, DFA is a special form of NFA. The nfa-ε is a more empty jump operation, tha