/*
    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()
{
    /*假設共有n人，從第s個人開始數數，每數到m該人出列，後面的人重新開始數，知道全部人出列*/
    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++) //建立迴圈連結串列
    {
        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; //遍歷迴圈連結串列，輸出序列
    do
    {
        cout << current->seq << " ";
        current = current->next;
    }while(current!=head);

    current = head; //將current置於第s個位置
    for(int i=1;i<s;i++)
    {
        current = current->next;
    }
    cout << endl;

    for(int i=1;i<=n;i++) //共迴圈n輪，得到一個整體的輸出序列
    {
        for(int j=1;j<m;j++)
        {
            prev = current;
            current = current->next;
        }
        cout << current->seq << " ";
        prev->next = current->next;
        delete current;
        current = prev->next;
    }
}