/*
author:jiangxin
Blog:http://blog.csdn.net/jiangxinnju
Function:method of Josephus question
*/
#include <iostream>
using namespace std;
struct node
{
int seq;
node *next;
};
typedef struct node NODE;
void test_Josephus()
{
/*如果共同拥有n人。从第s个人開始数数，每数到m该人出列，后面的人又一次開始数。知道所有人出列*/
int n,s,m;
NODE *head,*last,*current,*prev;
cout << "Input the n,s,m(separate with space):";
cin >> n >> s >> m;
for(int i=1;i<=n;i++) //建立循环链表
{
current = new NODE;
current->seq = i;
current->next = head;
if(i == 1)
{
head = current;
last = current;
}
else
{
last->next = current;
last = last->next;
}
}
current = head; //遍历循环链表，输出序列
do
{
cout << current->seq << " ";
current = current->next;
}while(current!=head);
current = head; //将current置于第s个位置
for(int i=1;i<s;i++)
{
current = current->next;
}
cout << endl;
for(int i=1;i<=n;i++) //共循环n轮，得到一个总体的输出序列
{
for(int j=1;j<m;j++)
{
prev = current;
current = current->next;
}
cout << current->seq << " ";
prev->next = current->next;
delete current;
current = prev->next;
}
}