Another common method to describe the azimuth is ouarla. This technique is named by the famous mathematician Leonhard Euler (1707-1783). He proves that the angle displacement sequence is equivalent to a single angle displacement.
What is ouarla
The basic idea of the orah angle is to break down the angle displacement into a sequence composed of three rotations round three mutually vertical axes. This sounds complicated, but it is actually very intuiti
Methods
The representation of different orientations is applicable to different situations. Below are some suggestions for choosing a proper format:
(1) ouarla is the easiest to use. When you need to specify the orientation of an object in the world, the orah angle can greatly simplify human-computer interaction, including entering the orientation directly on the keyboard and specifying the orientation in the Code (for example, setting the camera
numbers, even duplicates, because there is no mutex0112345678922334567891011121314151617181920212223242526272829456789101112131415161718191.3Method Three: By inheriting threading. Thread Creation ThreadsClass Mz_thread (threading. Thread): Def __init__ (self, num): Threading. Thread.__init__ (self) self.num = num def run (self): while True:print (' I am ', self.num) For I in range: t = mz_thread (i) T.start ()ResultsI am 9I am 0I am 8I am 1I am 6I am 7I am 5I am 2I am
. http://www.cnblogs.com/wu8685/archive/2010/12/21/1912347.htmlWhen an algorithm contains recursion, the problem of computational complexity or the problem of solving the recursive equation is transformed.There are several common solutions:Iterative method:The recursive equation is expanded iteratively to make it a non-recursive sum, and then the estimation of the solution of the Cheng equation is achieved by the estimation of the sum formula.ExampleThe calculation time of an algorithm is: t (n)
X axis of the object coordinate system and the second rotation round the X axis of the object coordinate system the same (the same means that in the world coordinate system, the two rotation axes are in the opposite direction), resulting in a loss of degrees of freedom of rotation.
In fact, a system that uses three quantities to represent the orientation of a three-dimensional space will encounter this problem, unless it is expressed by four quantities, such as the Quaternary element.
In a
TABLE.MAXN (table)
Returns the largest positive numerical index of the given table, or zero if the table has no positive numerical indices. (To-do it job this function does a linear traversal of the whole table.)
Returns the largest positive value index in the table.Description1. This interface is not the number of elements in the statistics table.2, if the table positive numerical index is not contiguous, 12 4, 3 is missing the element of index, the calculated value is MAXN 4.Loca
time of an algorithm is T (n) = 3 T (N/4) + O (N), where T (1) = O (1 ), after two iterations, the right side can be expanded:T (n) = 3 T (N/4) + O (N)= O (n) + 3 (O (N/4) + 3 T (N/42 ))= O (n) + 3 (O (N/4) + 3 (O (N/42) + 3 T (N/43 )))We can see from the above formula that this is a recursive equation. We can write the equation after iteration I:T (n) = O (n) + 3 (O (N/4) + 3 (O (N/42) +... + 3 (n/4I + 3 T (N/4I
prove.Second, iterative methodThe calculation time of an algorithm is: t (n) = 3T (N/4) + O (n), where T (1) = O (1), iteration twice can expand the right side to:T (n) = 3T (N/4) + O (n)= O (n) + 3 (O (N/4) + 3T (N/42))= O (n) + 3 (O (N/4) + 3 (O (N/42) + 3T (N/43))As can be seen from the above, this is a recursive equation, we can write the equation after the iteration i:T (n) = O (n) + 3 (O (N/4) + 3 (O (N/42) + ... + 3 (n/4i + 3T (n/
calculation time of an algorithm is T (n) = 3 T (N/4) + O (N), where T (1) = O (1 ), after two iterations, the right side can be expanded:T (n) = 3 T (N/4) + O (N)= O (n) + 3 (O (N/4) + 3 T (N/42 ))= O (n) + 3 (O (N/4) + 3 (O (N/42) + 3 T (N/43 )))We can see from the above formula that this is a recursive equation. We can write the equation after iteration I:T (n) = O (n) + 3 (O (N/4) + 3 (O (N/42) +... + 3 (n/4I + 3 T (N/
prove.Second, iterative methodThe calculation time of an algorithm is: t (n) = 3T (N/4) + O (n), where T (1) = O (1), iteration twice can expand the right side to:T (n) = 3T (N/4) + O (n)= O (n) + 3 (O (N/4) + 3T (N/42))= O (n) + 3 (O (N/4) + 3 (O (N/42) + 3T (N/43))As can be seen from the above, this is a recursive equation, we can write the equation after the iteration i:T (n) = O (n) + 3 (O (N/4) + 3 (O (N/42) + ... + 3 (n/4i + 3T (n/
O (N2) is a solution of T (N), and then proved by the mathematical induction method. Second, iterative methodThe calculation time of an algorithm is: t (n) = 3T (N/4) + O (n), where T (1) = O (1), iteration Two can expand the right side to:T (n) = 3T (N/4) + O (n)= O (n) + 3 (O (N/4) + 3T (N/42))= O (n) + 3 (O (N/4) + 3 (O (N/42) + 3T (N/43))As can be seen from the above, this is a recursive equation, we can write the equation after the iteration i:T (n) = O (n) + 3 (O (N/4) + 3 (O (N/42) + ...
Advantages compared with orah:
1. incremental Rotation
2. Avoid universal lock
3. There are two expressions for the given orientation, which are negative to each other (whether there are several expressions in orah)
Forty-seven: In which function is the Mobile Camera Action? Why is it in this function?
LateUpdate is called only after all updates are completed. It is suitable for Executing command scripts.
: with NR, but relative to the current fileIGNORECASE: If true, ignores case matchingOFMT: The output format of the number (the default value is%.6g)Rstart: The first position of a string that matches a match functionRlength: The length of a string matched by the match functionSUBSEP: Array subscript delimiter (default value is 34)The Begin:awk command executes the operation before execution, where the variable can be defined (built-in variable)After the End:awk command executes, it can be used
-------"); for(inti = 1; I if(i = = 3)Continue;System.out.println ("i="+ i);}}/*** Break to end the entire loop body*/ Public voidTestBreak1 () {System.out.println ("--------test break1-------"); for(inti = 1; I if(i = = 3) Break;System.out.println ("i="+ i);}}/*** Test a tagged break statement* tags can only be written in the loop body before, by the way to learn about the definition and use of statement tags in Java*/ PublicvoidTestBreak2 () {System.out.println ("--------test break2-------");
1:hiMain threadI am Thread 2:hello!I am Thread 1:hiMain threadI am Thread 2:hello!I am Thread 1:hiMain threadI am Thread 2:hello!I am Thread 1:hi...Two by inheriting the thread classThis class can be understood by looking at the thread code in the JDK1 Implementation steps:Override the Run () methodIt is OK to execute the start () method when calling.Test Case:Package languagelearning.multithread;public class MultithreadTest1 extends Thread {private int number;public void Run () {while (true) {
same as the first two pages introduced so several, the biggest difference lies in the layout of the changes-from the original "lying" in the Hba/raid card PCB, instead of "Stand up", that is, perpendicular to the disk backplane. Do not underestimate the difference, saying that it affects at least one of the connectors of the fate is not too much.
SATA style connectors, SAS 4i (SFF-8484) connectors, and mini SAS
2, 0x2a0010d8 in?? ()(GDB)p/x $r 2//View main function address$ = 0x2a0017c9(GDB) B *0x2a0017c9//main function on the next breakpoint, at this time libc.so is loaded, you can also b open/fputs/exit ...Warning:breakpoint address adjusted from 0X2A0017C9 to 0X2A0017C8.Breakpoint 3 at 0x2a0017c8(GDB) CContinuing.Warning:breakpoint 3 Address previously adjusted from 0X2A0017C9 to 0X2A0017C8.Breakpoint 3, 0x2a0017c8 in?? ()(GDB) display/4i $pc1:x/
]. Key I J --; // find 1st from the right to the leftLessLocation of standard value JIf (I {Tab-> r [I]. Key = tab-> r [J]. Key; I ++;} // Place element J on the left side and reset IWhile (tab-> r [I]. Key I ++; // find 1st from left to rightGreaterStandard Value position IIf (I {Tab-> r [J]. Key = tab-> r [I]. Key; j --;} // Place the I-th element on the right and reset J} While (I! = J );
Tab-> r [I] = tab-> r [0]; // place the standard value to its final position. This division ends.Quickso
]. Key I J --; // find 1st from the right to the leftLessLocation of standard value JIf (I {Tab-> r [I]. Key = tab-> r [J]. Key; I ++;} // Place element J on the left side and reset IWhile (tab-> r [I]. Key I ++; // find 1st from left to rightGreaterStandard Value position IIf (I {Tab-> r [J]. Key = tab-> r [I]. Key; j --;} // Place the I-th element on the right and reset J} While (I! = J );
Tab-> r [I] = tab-> r [0]; // place the standard value to its final position. This division ends.Quickso
array:This time, the first element 2 in the array is the primary element.2 (master) 8 7 1 3 5 6 4
Please take a closer look:2 8 7 1 3 5 6 4I-> (Looking for big data) (looking for small data)I. JJ. Find the first element less than 2 and assign it to (overwrite reset) element 2Get:1 8 7 3 5 6 4I jII. II found the first element greater than 2, 8 and assigned to (overwrite reset) the element referred to by J (
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