The application of the ACM learning process 16--list Linked list Joseph problem

Joseph problem is one of the application of list lists, the problem description: N people in a circle, everyone has a unique number, numbering from 1 to N, numbering from 1 to start off, and then report K, number of people out of K, his next from 1 start off, until all the people are out, Ask for the sequence of this column. Combined with the previous blog, here is a list of lists, giving a workable answer code.

Note: N is the number of people, p is the number to be reported.

#include <iostream>
#include <list>

using namespace std;

int main ()
{
	int n,p;
	List<int> l;

	while (cin>>n>>p)
	{
		l.clear ();
		for (int i=1;i<=n;i++)
		{
			l.push_back (i);
		}

		List<int>::iterator It,pdel;

		It=l.begin ();
		while (L.size ()!=1)
		{
			//numbered, it points to the element to be deleted for
			(i=1;i<p;i++)
			{
				it++;
				if (It==l.end ())
					it=l.begin ();
			}
			Delete processing
			cout<<*it<< "->";
			Pdel=it;

			it++;
			It= (It==l.end ())? L.begin (): it;

			L.erase (Pdel);
		}
		cout<<*it<<endl;
	}
	return 0;
}

Java version of the simple Joseph Ring:

public class Test {public static void main (string[] args) {cyclink cyclink=new cyclink ();
		Cyclink.setlength (5);
		Cyclink.createlink ();
		Cyclink.printlink ();
		CYCLINK.SETK (2);
		Cyclink.setm (2);
	Cyclink.play ();
	class Child {int number;
	Child Nextchild;
	public child (int number) {this.number=number;
	}//Ring List class Cyclink {//Ring list size//point to first child application cannot move int length=0;
	int k=0;
	int m=0;
	Child Firstchild=null;
	Child Temp=null;
	Sets the list size public void setlength (int length) {this.length=length;
	public void setk (int k) {this.k=k;
	public void Setm (int m) {this.m=m; }//Initialize ring list public void Createlink () {for (int i=1;i<=length;i++) {///Create First child if (i==1) {Baby ch
			   Ild=new Child (i);
			   This.firstchild=child;
			This.temp=child;
				else {Child child=new child (i);
				This.temp.nextchild=child;
				This.temp=child;
if (i==this.length) {//Create the last child This.temp.nextchild=this.firstchild;				}}} public void Printlink () {child temp=this.firstchild;
    		do {System.out.print (temp.number+ "");
    	Temp=temp.nextchild;
    	}while (Temp!=firstchild);
    System.out.println ();
    	public void Play () {child temp=this.firstchild;
    	Find the number of nodes for (int i=1;i<k;i++) {temp=temp.nextchild; } while (this.length!=1) {//number m for (int j=1;j<m;j++) {Temp=temp.nextchild
        	;
        	//Find out the previous node of the Circle child Temp1=temp;
        	while (temp1.nextchild!=temp) {temp1=temp1.nextchild;
        	///delete the node of M to Temp1.nextchild=temp.nextchild;
        	Temp=temp.nextchild;
        	This.printlink ();
    	this.length--;
    }//Last node System.out.println (temp.number); }
}

Joseph problem There are other variants, such as starting from the first m off or when a person out of the team again to set a number of the value of a count, and so on, only to be on the basis of a C + + program to make changes can be.