Use Java to solve outgoing problems in sequence

 

/**
* Using Java to solve outgoing problems in sequence-Author: Wen min, CTO of Chenzhou
*------------------------------------------------------------------------------
* Problem Description
* There are n people sitting around, numbered 1. N,
* When reporting data from 1 to M, this person is out; when reporting data from 1 to m, the other person is out.
* In this loop and so on, calculate:
* Number of the person who finally leaves the game
* Display the numbers of each reporter in sequence
* Solutions
* Develop an algorithm. The algorithm is located in an array. Each time it is estimated from N1, a non-zero one is found,
* Assume that a number is reported, and m is reported. If an employee is found
* Store the number at this position in the array to another array, and set it to 0 to indicate that the array is out.
* The algorithm is constantly called until the last speaker is found.

*/

Class test {
Int [] Nums;
Int [] numsbk;
Int N1; // the previous one at the start of each search
Int N2; // location where the employee ID is stored in the numsback Array
Int OK; // number of players found
Int incret; // condition for outbound traffic (when the number of outgoing traffic points is reached, the number of outgoing traffic is exceeded)
 
Final int inicnt = 15; // array size, change it to change the number of people
 
// Initialize data
Public test (){
Nums = new int [inicnt];
Numsbk = new int [inicnt]; // The numsbsk array will store each outbound number in sequence

// The Nums array stores the number of each person. The number starts from 1 to inicnt.
For (INT I = 1; I <= inicnt; I ++ ){
Nums [I-1] = I;
}
N1 =-1; // for the first time, it will start from 0. Therefore, set N1 to-1 so that it can increase by 1 to 0.
N2 = 0; // The employee ID found for the first time will be saved to the first position of numsbk, so N2 is set to 0
OK = 0; // no one is logged out yet, so OK is set to 0
Incret = 4; // This person is out of the game every time he points to 4.
}
 
Public void findnext (){
Int CNT = 0;
While (true ){
N1 ++;
If (n1 = inicnt) // find the one after the last one and it should be 0 (go back)
N1 = 0;

If (Nums [N1]! = 0) {// = 0, so it is out! = 0
CNT ++; // find one! A value of 0 indicates the number of vertices.
/**
* If the number of incres has been found, this person should be out.
* Store the employee number at numsbsk [n2], and then
* Add N2 to 1 to store the outgoing phone number next time.
* Setting Nums [N1] to 0 indicates that this person is out, and then
* "OK" and "1" indicate that another player has been found.
*/
If (CNT = incres ){
Numsbsk [n2] = Nums [N1];
N2 ++;
Nums [N1] = 0;
OK ++;
Break;
}
}
}


If (OK <inicnt) // if no reporter is found, call the API repeatedly to find the next reporter.
Findnext ();
Else {
System. Out. println ("final performer:" + numsbak [inicnt-1]);
System. Out. println ("----------------------------------------");
System. Out. println ("outgoing in turn :");
// The following figure shows the number of outgoing personnel in sequence.
For (INT I = 0; I <inicnt; I ++ ){
System. Out. println (numsbk [I]);
}
}

}
 
Public static void main (string [] ARGs ){
Test test = new test ();
Test. findnext ();
}
}